In this block my partner (Mike Bosse) and I will apply the Laplace transform to the solution of linear differential equations involving higher order derivatives. Our specific tasks are to discuss the general solutions of the differential equations. In each case we are going to each plot solution curve for a variety of parameters and given initial conditions. The Specific first and second-order linear differential equations for this block are listed below are listed below:

(a) The first two first-order equations on p. 466

(b) Any two second-order equations on p. 466.

(c) Two examples of linear systems from pp. 470-471.

For each we need to find the general solution using Laplace transforms (not by any other method) and plot solutions over a range of t values, for a given initial condition.

PART A (FIRST ORDER EQUATIONS)

To begin, we need to describe what needs to be done in order to solve these equations. As you can see on Gary’s page, these are differential equations of the form where . A similar approach to the homogeneous case shows us that the Laplace transform of a solution must satisfy . Using inverse Laplace transforms we get  where .

Problem 1

First we need to break down the equation into two sides and take the Laplace of them:

Looking at this and the table for Laplace’s theorem, we can take apart the right side and plug it in (from the table) to the left side:

Now we are going to continue with the process of the Laplace transform:

For s > 3

Before we can use the solver, we are going to simplify the problem:

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

________________________________________________________________________________________

Final Solution:

Now we are going to plot this equation by using EZPLOT function in MATLAB:

________________________________________________________________________________________

________________________________________________________________________________________

As a result from the following command we get:

For this plot, x is equal to t. From this plot you can see that y increases at a rapid rate once t is greater than 4.

Problem 2

First we need to break down the equation into two sides and take the Laplace of them:

Looking at this and the table for Laplace’s theorem, we can take apart the right side and plug it in (from the table) to the left side:

Now we are going to continue with the process of the Laplace transform:

For all s

Before we can use the solver, we are going to simplify the problem:

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

________________________________________________________________________________________

Final Solution:

Now we are going to plot this equation by using EZPLOT function in MATLAB:

________________________________________________________________________________________

________________________________________________________________________________________

As a result from the following command we get:

For this plot, x is equal to t. From this plot you can see that y decreases exponentially as t becomes more neagative.

PART A (SECOND ORDER EQUATIONS)

To begin, we need to describe second-order differential equations which is an algebraic expression( $latex\mathscr{L}[f''(t)]=s^2\mathscr{L}[f(t)]-sf(0)-f’(0)$ ), in which we solve for . We then apply the inverse Laplace transform as before to find the solution of the linear differential equation, but before this happens we need to perfom the second order differential equation. Here are the two problems that we chose to work on below.

Problem 1

First we need to break down the equation into two sides and take the Laplace of them:

Looking at this and the table for Laplace’s theorem, we can take apart the right side and plug it in (from the table) to the left side:

Now we are going to continue with the process of the Laplace transform:

Before we can use the solver, we are going to simplify the problem:

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

Final Solution:

Now we are going to plot this equation by using EZPLOT function in MATLAB:

________________________________________________________________________________________

________________________________________________________________________________________

As a result from the following command we get:

For this plot, x is equal to t. From this plot you can see that y is sinusoidal where the maximum seems to be 5/2 and minimum value as -5/2.

Problem 2

First we need to break down the equation into two sides and take the Laplace of them:

Looking at this and the table for Laplace’s theorem, we can take apart the right side and plug it in (from the table) to the left side:

Now we are going to continue with the process of the Laplace transform:

Before we can use the solver, we are going to simplify the problem:

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

________________________________________________________________________________________

Final Solution:

Now we are going to plot this equation by using EZPLOT function in MATLAB:

________________________________________________________________________________________

________________________________________________________________________________________

As a result from the following command we get:

For this plot, x is equal to t. From this plot you can see that y is a cosine wave with the maximum value equal to 7 and a minimum value equal to -7.

PART C (LINEAR SYSTEMS)

To begin, we need to describe what needs to be done for Linear Systems. For Linear systems two algebraic equations are solved for or . This gives the solution as a system of x(t) and y(t).

Problem 1

and

If we divide the two into two components x and y so we can decipher them and then reconnect. First lets start our same process with the y(t) function first:

We need to break down the equation into two sides and take the Laplace of them:

Looking at this and the table for Laplace’s theorem, we can take apart the right side and plug it in (from the table) to the left side:

Now we are going to continue with the process of the Laplace transform:

Lets represent this as Equation 1

Since we have the x(t) function simplified we can move onto the y(t) function:

First we need to break down the equation into two sides and take the Laplace of them:

Looking at this and the table for Laplace’s theorem, we can take apart the right side and plug it in (from the table) to the left side:

Lets represent this as equation 2

If we now plug equation 1 into equation 2 we can solve for the system of equations:

Now we are going to continue with the process of the Laplace transform:

Before we can use the solver, we are going to simplify the problem:

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

Final Solution:

In order to find x(t) we first must plug equation 3 into equation 1

Before we can use the solver, we are going to simplify the problem:

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

Final Solution:

Now we are going to plot this equation by using EZPLOT function in MATLAB:

________________________________________________________________________________________

________________________________________________________________________________________

As a result from the following command we get:

For this plot, x is equal to t. From this plot you can see that y(t) and x(t) decreases exponentially, when -3>t>1.

Problem 2

, and ,

If we divide the two into two components x and y so we can decipher them and then reconnect. First lets start our same process with the y(t) function first:

We need to break down the equation into two sides and take the Laplace of them:

Looking at this and the table for Laplace’s theorem, we can take apart the right side and plug it in (from the table) to the left side:

Now we are going to continue with the process of the Laplace transform:

This is equation1

Now if we take and plug it into we can solve for

Now we are going to continue with the process of the Laplace transform:

Before we can use the solver, we are going to simplify the problem:

$latex\mathscr{L}[y]=\frac{-1}{s^2-3s+2}$ equation 2

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

Final Solution:

In order to find x(t) we first must plug equation 2 into equation 1

Before we can use the solver, we are going to simplify the problem:

Using MATLAB we can now get our final Solution

_______________________________________________________________________________________

Final Solution:

Now we are going to plot this equation by using EZPLOT function in MATLAB:

________________________________________________________________________________________

________________________________________________________________________________________

As a result from the following command we get:

For this plot, x is equal to t. From this plot you can see that y(t) and x(t) decreases exponentially as t increases.

Conclusion

Overall this was quite a Lenghthy block. With the knowledge that I gathered through all my classes so far though I was able to accomplish the task at ahnd.

Systems of Linear Differential Equations

This is a 3-week block in which we will be looking at the behavior of systems of linear differential equations. A system of two linear differential equations is a pair of linked differential equations of the form

where are constants.

A linear function on the Euclidean plane is a function of the form . A linear function is exactly a function that preserves vector addition and scalar multiplication: and for all vectors and all real numbers .

As a result, once we know what a linear function T does to the vectors (1,0) we know what T does to every vector. The reason is that .

In principle, we could just draw, for each vector (x,y) a directed line segment. However, a moment’s thought tells us that all these directed line segments would overlap in a mess, and give us no sensible picture at all. One way around this is to draw a very short directed line segment. So that these small directed line segments do not overlap:

To accomplish this we will be using MATLAB.

Finding Invariant Lines – Invariant lines are lines where points map back onto themselves (line of invariant points). It can also be a line where the points map back onto a (usually) different place on the line. In this block we will also be finding these.

In this part of the block my partner (Mike Bosse) and I will be accomplishing the following:

1.    (a) Plot the vector field for the linear function .

(b) Find any invariant lines, and write down the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow.

(c) Try to describe the flow geometrically, in words.

2. Repeat this for the linear functions for several choices of integers chosen randomly in the range .

To begin this part of the block my partner and I, decided to make one simple function that would help simplify this part of the Block. To begin we solve for the invariant lines algebraically.

=

Solve for x and y

Rearrange

Finding the Determinant

Using the Quadratic Equation, solve for

Once is found, plug it into either of the following equations from before, to find the Invariant line equation.

Since we have now found a generalized formula to solve for the invariant lines, we decided to make a function in MATLAB that would plot the vectors and variant lines. All we would have to do is enter what a b c and d would be. Also to make it a little more easier, if you say that a b c and d are all zero the function will choose random values from -5 to 5 for you. Below is the following function we created in MATLAB.

______________________________________________________________________________________

% This Function is For Block 5. This function is for a system
% of two linear differential equations. dx/dt=ax+by and
% dy/dt=cx+dy. a=a b=b c=c d=d and g is whether or not you want
% to display invariant lines. The main point of this function is
% that it determines your invariant lines for you

function Block5_1(a,b,c,d,g)

% IF a,b,c,d = 0 then give them random values.
if (a==0 && b==0 && c==0 && d==0)
a=fix(10*rand())-5
b=fix(10*rand())-5
c=fix(10*rand())-5
d=fix(10*rand())-5
end

%determining Landa 1
lambda1=(-(-a-d)+sqrt((-a-d)^2-4*(a*d-b*c)))/2

%determining Landa 2
lambda2=(-(-a-d)-sqrt((-a-d)^2-4*(a*d-b*c)))/2

xx=-1:.1:1;
% invariant line 1
p1=(xx*(lambda1-a))/b;
% invariant line 2
p2=(xx*(lambda2-a))/b;

[x,y]=meshgrid(-1:1/10:1,-1:1/10:1);
u=a*x+b*y;
v=c*x+d*y;
w=sqrt(u.^2 +v.^2);quiver(x,y,u./w,v./w,.7);
% if g is one then plot the variant lines
if (g)
hold on
if (isreal(lambda1)==1) %tests to see if lambda1 is a real number
plot(xx,p1,’r')
end
if (isreal(lambda2)==1) %tests to see if lambda2 is a real number
plot(xx,p2,’r')
end
hold off
end
end

______________________________________________________________________________________

Now that we have that setup lets plot the vector field for the linear function . Using our Function we need to say Block5_1(a,b,c,d,g).  Where:

= 1        =1     = 1      =0

Since we don’t want the variant line to show up yet, we want = 0

The following was put into MATLAB and what is spit out (input was in yellow output gray):

______________________________________________________________________________________

Block5_1(1,1,1,0,0)

lambda1 =

1.6180

lambda2 =

-0.6180

______________________________________________________________________________________

As a result from the following command we got:

In this plot of vectors, they all seem to be pointing away from the coordinate (0,0). Also you can see that the vectors seem to split into four different sections. The top left corner seem to be going south-east and south-west. The top right corner is going from south-west to north west and north to northwest. The bottom left corner is going south to south-east and north-east to south-east. Finally the bottom right corner is going north-east and north-west. With these four sections, you can almost see the invariant lines will be. One should be from about (-0.8,1) to (0.8,0) and the other about (-1.1,-0.7) to (1,0.6). The only way to know for sure is graphing them with our predefined Function.

Here is what was put into MATLAB and what is spit out (input was in yellow output gray)::

______________________________________________________________________________________

Block5_1(1,1,1,0,1)

lambda1 =

1.6180

lambda2 =

-0.6180

______________________________________________________________________________________

As a result from the following command we got:

As predicted the variant lines are exactly where we said they would be. Using our equations we got from solving invariant lines algebraically and what MATLAB spit out for our values, our invariant line equations are:

AND          

Now the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow are as follows:

Now we are going to repeat this for the linear functions for several choices of integers chosen randomly in the range . Since our MATLAB Function is already capable of selecting random values from -5 to 5 we will just continue using this. As a refresher all we have to do for our function is input Block5_1(0,0,0,0,1) into MATLAB to get random plots. So lets do so.

The following was put into MATLAB and what is spit out (input was in yellow output gray):

______________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

-4

b =

-1

c =

-1

d =

1

lambda1 =

1.1926

lambda2 =

-4.1926

______________________________________________________________________________________

Before we show the graph lets analyze what MATLAB gave us for random Numbers. It gave us:

= -4        =-1     = -1      =1

And from this we get our linear function

Now that we have established our Linear Function our vector field plot looks like so:

In this plot of vectors, they still seem to be pointing away from or pointing around the coordinate (0,0). Once again the vectors seem to split into four different sections. The top left corner seem to be going east to north-east almost all north. The top right corner is going from west to north-west almost all north. The bottom left corner is going east to south-east almost all south. Finally the bottom right corner is going west to south-west almost all south. With these four sections, you can almost see the invariant lines will be. One should be from about (-1.8,1) to (2.1,-1) and the other about (-1,-0.1) to (1,0.1). The only way to know for sure is graphing them with our predefined Function.

The following was put into MATLAB and what is spit out (input was in yellow output gray):

______________________________________________________________________________________

Block5_1(-4,-1,-1,1,1)

lambda1 =

1.1926

lambda2 =

-4.1926

______________________________________________________________________________________

As a result from the following command we got:

As predicted the variant lines are exactly where we said they would be. The invariant lines seem to be where the vectors start to point towards each other. Using our equations we got from solving invariant lines algebraically and what MATLAB spit out for our values, our invariant line equations are:

AND          

Now the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow are as follows:

Lets now continue and get another random example.

The following was put into MATLAB and what is spit out (input was in yellow output gray):

______________________________________________________________________________________

Block5_1(0,0,0,0,1)

a =

2

b =

2

c =

-3

d =

1

lambda1 =

1.5000 + 2.3979i

lambda2 =

1.5000 – 2.3979i

______________________________________________________________________________________

Lets analyze what MATLAB gave us for random Numbers. It gave us:

= 2        =2     = -3      =1

And from this we get our linear function

Now that we have established our Linear Function our vector field plot looks like so:

In this plot of vectors, they still seem to be pointing away from coordinate (0,0). This linear Function seems to be one of those special cases where there are no defined variant lines. There are no invariant lines because is equal to two different complex numbers. As you can see from above =1.5000 + 2.3979i and 1.5000 – 2.3979i. Since both results are complex numbers they can’t be graphed.

Now the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow are as follows:

Lets now continue and get another random example.

The following was put into MATLAB and what is spit out (input was in yellow output gray):

______________________________________________________________________________________

Block5_1(0,0,0,0,1)

a =

     2

b =

    -1

c =

     1

d =

     4

lambda1 =

     3

lambda2 =

     3
______________________________________________________________________________________

Before we show the graph lets analyze what MATLAB gave us for random Numbers. It gave us:

= 2        =-1     = 1      =4

And from this we get our linear function

Now that we have established our Linear Function our vector field plot looks like so:

In this plot of vectors, they still seem to be pointing away from the coordinate (0,0). This time the vectors seem to split into two different sections. There seems to be two sets of vectors that line up across from each other pointing in the opposite direction. Since is only has one solution this should be the only variant line. The line we think goes from (-1,1) to (1,-1)  The only way to know for sure is graphing them with our predefined Function.

The following was put into MATLAB and what is spit out (input was in yellow output gray):

______________________________________________________________________________________

Block5_1(2,-1,1,4,1)

lambda1 =

     3

lambda2 =

     3
______________________________________________________________________________________
As a result from the following command we got:

As predicted the variant line are exactly where we said it would be. Using our equations we got from solving invariant lines algebraically and what MATLAB spit out for our value, our invariant line equation are:

Now the differential equations corresponding to a flow for which the vectors in the vector field are tangent to the flow are as follows:

PART 2

For this part of the lab my partner and I will be going go through all the cases for the roots of the characteristic equation, describing the geometry of the solutions of the differential equations. Instead of using the examples from the book though my partner will be using our pre-made function and keep randomly going through numbers until we do every possibility. In each case we will also be describing the exact solutions to the differential equations obtained from MATLAB, and relate the form of those solutions to the geometry of the flow.

Our first Possible solution is when you have two positive roots. For this our Function had spit out the following (input was in yellow output gray):

________________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

2

b =

-3

c =

0

d =

3

lambda1 =

3

lambda2 =

2
________________________________________________________________________________________

This random gave us the random numbers we need to get two positive roots. The roots are 3 and 2. The linear equation is . Now from this we get the two differential equations:

The plot for these differential equations is as follows:

One of the first things that my partner and I noticed about this plot is that all the vectors seem to be pointing away from the origin and away from the variant line (-1,0) to (1,0). Before making any conclusions though we are now going to try another example where we get two negative roots.

(input was in yellow output gray):

________________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

1

b =

-2

c =

4

d =

-5

lambda1 =

-1

lambda2 =

-5

________________________________________________________________________________________

This random gave us the random numbers we need to get two negative roots. The roots were -1 and -5. The linear equation is . Now from this we get the two differential equations:

The plot for these differential equations is as follows:

On this plot we noticed  that all the vectors seem to be pointing towards the origin or toward the invariant line (-1,-1) to (1,1). Also  Before making any conclusions though we are now going to try another example where we get two negative roots.

(input was in yellow output gray):

________________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

-4

b =

0

c =

-5

d =

2

lambda1 =

2

lambda2 =

-4

________________________________________________________________________________________

This random gave us the random numbers we need to get two negative roots. The roots were 2 and -4. The linear equation is . Now from this we get the two differential equations:

The plot for these differential equations is as follows:

This is the final circumstance when getting two roots. What we have been saying all along seems to still be true. If you split the plot into four sections, there are two sections that head inward but not pointing at the origin (bottom left and top right) and two sections that head away but not pointing directly away from the origin (bottom right and top left). In this case the origin is known as a saddle point. One of the differences between this and the two positive roots is that all vectors seem to be pointing toward the invariant line (0,-1) to (0,1) in one way or another. Now we will explore the circumstance of getting one root.  Using our function we found the following for a problem in which you get one root.

(input was in yellow output gray):

________________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

3

b =

0

c =

0

d =

3

lambda1 =

3

lambda2 =

3

________________________________________________________________________________________

This random example gave us the random numbers to get one positive root. The root is 3. The linear equation is .Now from this we get the two differential equations:

The plot for these differential equations is as follows:

Much like the Two positive roots, the one positive root goes outward from the origin also. One difference is this one goes perfectly outward. The two root example seemed to have turned a little bit. This one looks as if there has been an explosion in the origin and everythings just shooting outward. This one also doesn’t seem to have that just one clear variant line where all the vectors point away from. Now to see what the one negative root would does.

(input was in yellow output gray):

________________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

-4

b =

0

c =

-3

d =

-4

lambda1 =

-4

lambda2 =

-4

________________________________________________________________________________________

This random example gave us the random numbers to get one negative root. The root is -4. The linear equation is .Now from this we get the two differential equations:

The plot for these differential equations is as follows:

Much like the two negative roots, the one negative root goes inward towards the origin. This one doesn’t seem to have that just one clear variant line where all the vectors point towards to. This plot kind of makes a sink towards the origin. Now that we see one negative roots goes towards the origin and one positive root goes away, we are going to examine the imaginary roots to see if we get the same results as the two root examples.

(input was in yellow output gray):

________________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

2

b =

-5

c =

1

d =

0

lambda1 =

1.0000 + 2.0000i

lambda2 =

1.0000 – 2.0000i

________________________________________________________________________________________

This random example gave us the random numbers to get positive complex roots. The root are 1.0000 + 2.0000i and 1.0000 – 2.0000i. The linear equation is .Now from this we get the two differential equations:

The plot for these differential equations is as follows:

This plot isn’t really like any of the other plots. There dosn’t seem to be any defined variant line and the vectors seem to be going counter clockwise in like a vortex shape. They still seem to be going away from the origin though. This as been true for all examples so far. Now lets find a negative complex root.

(input was in yellow output gray):

________________________________________________________________________________________

Block5_1(0,0,0,0,0)

a =

-5

b =

-4

c =

4

d =

0

lambda1 =

-2.5000 + 3.1225i

lambda2 =

-2.5000 – 3.1225i

________________________________________________________________________________________

This random example gave us the random numbers to get negative complex roots. The root are -2.5000 + 3.1225i and -2.5000 – 3.1225i. The linear equation is .Now from this we get the two differential equations:

The plot for these differential equations is as follows:

This plot isn’t really like any of the other plots. There dosn’t seem to be any defined variant line and the vectors seem to be going clockwise in like a vortex shape. They still seem to be going towards the origin though. This as been true for all examples so far. So this seems positive = outward and negative = inward. This concludes the examples for this block.

Conclusion

This was a quite extensive block. It took many examples to understand geometrically what was happening. We feel though that we have a good understanding in Linear Systems of Differential Equations. We first did our function to help make this block a little bit easier and then went on to solve one example for all possible outcomes.

]]> http://joshuamoranmth212s09.wordpress.com/2009/04/03/block-5/feed/ 0 jmoran1 matlab_vector_field2 block5_1_v block5_11 block5_1_ran_v1 block5_1_ran block5_1_ran2_v block5_1_ran31 block5_1_ran3_v2 block5_2_pp_r2 block5_2_nn_r1 block5_2_np_r block5_2_pp_one block5_2_nn_one block5_2_pp_c block5_2_nn_c http://joshuamoranmth212s09.wordpress.com/2009/03/29/fourth-block/ http://joshuamoranmth212s09.wordpress.com/2009/03/29/fourth-block/#comments Sun, 29 Mar 2009 17:18:58 +0000 jmoran1 http://joshuamoranmth212s09.wordpress.com/?p=193 ]]> Variables Separable Differential Equations

This is a short block that was designed to address a simple class of differential equations that clearly have exact solutions. These are the variables separable differential equations.

A Separable differential equation explanation can be found here.

For this Block My partner (Mike Bosse) and I have two tasks.

1. Choose 6 examples (total) from pp. 24, 25, 26, problems 12-16 pp. 76-77 of the text and carry out the integrations in the variables separable method to obtain exact solutions for these differential equations.

2. In each case, compare your exact solution with that obtained using dsolve in MATLAB

Here are the 6 problems that we chose to analyze:

1,)

4.)    

5.)

6.)

9.)

11.)

PROBLEM 1

Now let’s integrate both sides with respect to x:

=

So once we have this we can simplify the problem by canceling out the on the left hand side and then performing a U-Substitution:

Lets let for the right side of the problem so:

Now that we see this we can represent the sides by e and then solve for y:

Now all we need to do is show our results for dsolve. From previous blocks, we know that the dsolve method should give us an exact solution of what the problem should come out to be. Here is what we got when we typed the equation into MATLAB:

_______________________________________________________

dsolve(’Dy=(-4*x*y)/(x^2+1)’,’x’)
ans =
C10/(x^2 + 1)^2

_______________________________________________________

As you can see they are the same answer. The answers don’t look exactly the same because of C10. But that is just a basic constant that is replaced by C in our equation.

PROBLEM 2

Now we have to integrate both side by x to move on to the next step of the equation:

A U-Substitution is needed in order to simplify the problem so lets let so if we do this we get

As you can see we let the sides again be the base e to simplify the problem to the final answer above where C again is represented as a constant.

Now that we have our final answer we can move onto the Dsolve method.  Here is the code that we entered into MATLAB in order to get the exact solution:

_______________________________________________________

dsolve(’Dy=(-2*y*cos(x))/(sin(x))’,’x’)
ans =
C14/sin(x)^2

_______________________________________________________

As you can see they were the same exact solution except for the same constant value representation. Overall the U-Substition method has worked well in solving most of these integrations.

PROBLEM 5

Just as before, we are going to integrate both sides with respect to x in order to get from the left side and so we can have our seperable variables to solve:

Now we can rewrite the equation for a possible U-Substition:

For this we used the U-Substitution method where we can represnt . Then we can solve for and since , we can solve for what x is going to be. () Once we do all this, we get the following equation:

Now all we have to do is simplify this into an equation where we can get rid of the natural log in order to produce the explicit solution:

Now that we have solved this problem using the U-Substition method we can now check our answer with the Dsolve method used in MATLAB. Here is what we typed in and what we received as an explicit solution.

_______________________________________________________

dsolve(‘Dy=(-x*y)/(x+1)’,'x’)
ans =
C4*exp(log(x + 1) – x)

_______________________________________________________

So as we can see our answers were exactly the same as the dsolve method. With this we still know that the U-Substitution method is still the correct approach to these integrals.

PROBLEM 6

We are going to integrate with respect to x. This will cancel out the on the left hand side which makes it easier to get to the U-Substitution process :

Now we can use the U_Substitution method to get rid of the integrals. We let , so now we can solve for our du so

Now that we have a understanding of what our U-Substition is going to be we can now proceed to solve the problem by plugging in the U-Substitution:

Now you can see here that we have to take the square root of the right side in order to get rid of the so we are going to get a plus – minus situation:

As you can see we now have a solution that looks like it is in it’s simplest form.  We can now move onto the Dsolve method in order to check if our solution is that same as the explicit solution of Dsolve. So here is what we entered into MATLAB and what we got out as a result:

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dsolve(‘Dy=(sqrt(y^2+1))/(x*y)’,'x’)
ans =
-((C17 + log(x))^2 – 1)^(1/2)
((C17 + log(x))^2 – 1)^(1/2)
i
-i

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As you can see both of our solutions that we obtained came out to be the same. Because of this e know that the U-Substitution method was the correct method to use in order to solve for the integration and to reach the explicit solution.

PROBLEM 10

We are going to integrate both sides with respect to t (instead of x) in order to cancel out on the left hand side (which will make it easier to solve for the integration):

Now rewritte in order to make the integral less complicated:

Now we simplify this in order to get to the lowest possible form of the explicit solution:

Our answer varied from MATLAB at first because what it did was make that as so they could factor the problem and make it easier. They used this . We know they did this because the 1 factors out as a constant with the others. So this is what we attained as an aswer:

Now all we have left to do is use the Dsolve function in order to see if the MATLAB gets the same explicit solution.  This is what we typed into MATLAB and what it produced:

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dsolve(‘Dy=(1-t)/(x)’,'t’)
ans =
2^(1/2)*(C21 – (t – 1)^2/2)^(1/2)
-2^(1/2)*(C21 – (t – 1)^2/2)^(1/2

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Overall we can see that both of the explicit solution are the same and the method of integration worked out well the way we performed it.

PROBLEM 16

First we begin to solve the problem by first seperating the variables to each side.

Now our variables are seperated and we can start our integration process:

To solve for the integration we can simplify the problem:

So all we do now is solve for our variable y:

Now all we have left to do is use the Dsolve function in order to see if the MATLAB gets the same explicit solution.  This is what we typed into MATLAB and what it produced:

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dsolve(’ Dy= (y^2-y)/x’,’x’)

ans=

C1*1/1+x

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As you can see we get the same answer just the constant in our factored out when we solved for y.

Conclusion

Overall we can see this was a very good chance to improve our integration skills. We feel more comfortable with the integration process. It made us use a lot of what we learned through calculus. Overall a very good block and glad we did it.

In this Block my partner and I (Mike Bosse) are going to address a major issue in using differential equations to model physical or biological phenomena: the nature of the solution curves to a system of coupled non-linear differential equations. These types of systems of differential equations arise in most serious modeling of physical and biological phenomena. There are very few known exact methods of solution for these systems of differential equations. As a result, people working with these types of systems of differential equations have to resort to numerical methods for obtaining approximations to solution curves.

Edward Lorenz

Edward Lorenz (May 23, 1917 – April 16th, 2008) he was known as one of the most famous mathematicians, and meteorologists  and the founder of his own theory known as the “Chaos Theory”, and from there he discovered his famous butterfly effect. Edward lorenz graduated from Harvard University in Cambridge, Massachusetts. His main love was weather where he worked in the army air corps as a weather forecaster. After his time in the army he became a professor at MIT until he died in his home at the age of 90 due to the deadly fatal disease known as cancer. Once he was in his chaos theory he noticed that his results were varying all over the place and through his work he came up with these three equations and they are known as his system of differential equations.

First thing that we need to do is clearly understand the objective of the block at hand so here I will give you the exact assignment objective that we received:

Objective:

1. Explore how exact solutions to first order differential equations differ Euler approximations.
2. Experiment with Euler approximations to the Lorenz equations for various parameter values and initial conditions.
3. Choose one of the following:

• The Lotka -Volterra predator prey equations  where y is the number of predators,  x is the number of prey,and α, β, γ, δ are paramaters that tune the interaction of the predators and prey.
• Explore solutions of the predator-prey equations for varying paramater values and initial conditions.
• Explore the behavior of the Rossler system of differential equations  , ,   for various paramater values and initial conditions.

First we are going to start off by exploring how exact solutions to first order differential equations differ from Euler approximations. As a starting point for Euler’s Method, we obtained code for MATLAB from Gary Davis. There was no problem with this code, but it just didn’t produce the kind of graphs we were looking for. We wanted one Plot in which we could see both the two dimensional plot and the three dimensional. We obtained this by using a MATLAB function called subplot. What subplot does is takes the amount of plots you want and graphs them in one area. The following is the combination of Gary’s code and subplot that we used.

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Now we are going to use this code towards a system of differential equations. We are going to apply Euler’s method to do the Lorenz equations. We first need to create an m. file document that will be able to process Euler’s method with a simple line of code. For our code we will be using the current values.

Now that we have the values we can make our .m file. We called the .m file lorenz_system. The following is our lorwnz_system:

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With this file and with the following MATLAB code we are able to plot our lorenz_system

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As a result from the following MATLAB commands, we got the following plot:

Now that we have set y values, we are going to try random y values.

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As a result from the following MATLAB commands, we got the following plot:

To see if we get something simialar again, we tried the same MATLAB commands and got the following:

Overall you can see that our two random examples look very alike forming the same vortex shape while our set y values seems to be quite different.

Now that we see how Euler’s plots Lorenz equations, we will now use our next method we learned.This method is known as ODE45.

ODE45

As a result from our last block we saw that the Ode45 method was actually more acurate, where as Euler’s method always seemed to branch off near the end of the plot. So first for Ode45 we copied code from Gary and put it in an m.file  that we called g.m:

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Now that we have this we needed to create another m.file to show how the lorenz system would work. So we created another m.file called lorenz_demo and here is line of code that we created for that:

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Now that we have created this, we are able to run the file in MATLAB. The following is the MATLAB command:

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As a result form the the following command we get the following:

As you can see the graph had formed a shape almost as if it was a butterfly shape. We can really see how the graphs can be so random. We decided to do this procedure one more time with different numbers for our initial conditions. The following is the g.m file we used as our initial conditions:

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Now doing the same command form before (“lorenz_demo(200);“) but using this new g file we get the following plot:

Here we can see we have a completly different plot where we can see this one alost forms the shape of a spiral which is close to the vorex but not in the same manner.

Rossler System of Differential Equations

My partner and I decided to do the rossler system of equations because we saw this as a good example to clearly understand the works of the rossler system. Also we thought we would understand the calculations that involved x,y,and z as components. Here are the formulas for the rossler system of equations:

,,

Now since we have a clear understanding of the equations we can create an m.file to represent the rossler system and one for the lorenz system. These files are known as Rossler_system.m and the other known as Lorenz_system.m. Now that we have this we can create our initial parameters as.

Using this information we can now create our Rossler_system.m file.

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For this part of the Block. We found that ploting both the 2-D and 3-D plots were unneccary. So we edited our code so that our euler_system function would only plot in 3-D. All we did was comment out the 2-D plot by putting a % sign in front of that current line. The following is the MATLAB function we used to plot:

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As a result form the following command we get the following plot:
Now, just like before, we are going to plot two more examples using random y initial values and see if they vary in any way. Here is the formula for the random examples and the plots to follow:

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RANDOM 1:

RANDOM 2:

As you can see we had the graphs vary only by the number of lines in each other and the amount of spacing in between the lines. This is because of the inital conditions that were given in the two random examples. We can now see how the Rossler equation can vary when using random numbers.

Now lets do the Ode45 method for the Rossler system of equations. Here is our new lorenz_demo function that we will use for this part of the Block:

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In order for this function to work we needed to create a .m file with our initial condition values in which we called h.m:

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The follwoing is the MATLAB command that we did that involved these .m files:

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As a result form the following command we get the following plot:

Conclusion

Overall we can see we were able to attain a very good amount of information and understand how Lorenz used his equations in order to show random results. We feel as if we were able to learn something that was very complex and put it into a simple format that was easy to understand. Overall WE thought this block was a success and we feel that we attained a good amount of knowledge.

This block requires two methods of approximations of differential equations. One is Euler’s Method and the other is

Runge-Kutta order method.

Euler’s Method uses the most basic of approximations to the derivative of an unkown function in a differential equation: it approximates the unkown function by a straight line for a very short increment in the independent variable.

Runge-Kutta order method is a method of numerically integrating ordinary differential equations by using a trial step at the midpoint of an interval to cancel out lower-order error terms. and the fourth-order formula is

•    =   
•    =   
•    =   
•    =   
•    =   

MATLAB has built-in procedures for impelmenting Runge-Kutta, and modified Runge-Kutta, methods for numerical approximation of solutions of differential equations. The one we will use for this project is ode45

For this block my partner (Michael Bosse) and I, examined 6 first-order differential equations using MATLAB functions:

1. Euler’s method
2. ode45
3. dsolve

We also will compare and contrast each one to the other.  We had to choose two problems from each of the following sections:

• problems 2-5 on pp. 138-139  of the text
• problems 6-11 on p. 139 of the text
• problems 12-16 on p. 139 of the text

Problem 3

= , 

The first step to examining this example was to edit our Euler’s method Code from our previous block. With our previous block we were entering euler ( function, x_range, y_initial, h ). Also in doing this we had to make the nstep be calculated according to whatever h we had. Then we decided to changed the code so that it would graph red The following is our newest euler method code.

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% Euler’s method for a single differential equation.
% The differential equation is dy/dx = f(x,y)

function [ x, y ] = euler ( f, x_range, y_initial, h )

x(1)=x_range(1);

y(1)=y_initial; 

% The program euler.m will ask for a range of x-values, from x(1) to x(2).x(1) = x_range(1);

% We define dx by dividing this range into an equal number of specified steps

nstep=( x_range(2) - x_range(1) ) / h; 

dx = ( x_range(2) - x_range(1) ) / nstep; 

% We set the initial value of y at the beginning x-value x(1).y(1) = y_initial;  

% We use Euler’s method to update the value of y at new time steps.

% “feval” is used instead of “eval” because we are passing the name of the function f to the program.

for i = 1 : nstep
 x(i+1) = x(i) + dx;
 y(i+1) = y(i) + dx * feval ( f, x(i), y(i) );

end

% The following command plots y as a function of x.

plot(x,y,'r')

 xlabel('x')

 ylabel('y')

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Our next step was to create an example file in which we named “problem_3_blk2″. This file was very similar to the one we did in the first block, but just used the problem three function. please see below:

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Now its time to execute our new euler.m file using our differential equation to create a curve in which we will be able to compare. To do this in MATLAB we said the following command.

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NOTE: We started our boundaries at 0 because the initial value of at = 0 is 1 as the problem states.

As a result from the following command in MATLAB we got the following plot:

This plot looks pretty good but still need to compare it with our other MATLAB functions to see how accurate it really is.

The next Function we looked at was ODE45. ODE45 is a built in MATLAB function that reauires an equation, area of bounds, and an initail point. So we used the same problem_3_blk2.m file and put that into ODE45. To be more specific here is what we typed into MATLAB:

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NOTE: We were limited by our boundaries. We needed to start at zero and also because of the way this function is we couldn’t go past 4.5 without getting an error from ODE45.

As a result from the following command in MATLAB we got the following plot:

With the exception of the boundaries, this plot looks quite similar to euler’s plot. But, before comparing we will graph the exact solution first by using the MATLAB dsolve function.

For this dsolve function we decided to do the function on the computers in class and then would graph them later on on a different computer. For this example we got this when we used the dsolve function

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Now since we didn’t graph and use dsolve on the same computer, we decided to put the answer into a .m file and make it into a function similar to our problem_3_blk2.m file except this function would only require an x input instead of an x and y. We named this file problem_3_sol.m and here is what we put in it:

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function yprime = problem_3_sol(x)

yprime = 1/(sin(x) + 1);
end

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Now that we had our explicit equation file setup we needed a way to plot it. After looking through the help files we found a built in function that would just do that. This function is called fplot. The input for fplot is as follows fplot(function, area of bound, we decided to change the color here to green). Here is what we put into MATLAB

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As a result from the following command in MATLAB we got the following plot:

This plot looks very similar to both our ODE45 and Euler function. The best way to compare all three of these functions would be to plot them all together on one plot. So we decided to make a function that would do just that. In this function we made our EULER PLOT RED, ODE45 PLOT BLUE, and our DSOLVE PLOT GREEN. In doing this we also increased the width of each plot so you could see when the plots over lap easier. We used this function for all our examples and the inputs are exactly the same as our euler’s method inputs. The function also use both our problem_3_blk2.m and our problem_3_sol.m files.  The code can be found below:

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function [ x, y ] = euler_ODE_dsolve ( f, x_range, y_initial, h ) 

x(1)=x_range(1); 

y(1)=y_initial;

nstep=( x_range(2) - x_range(1) ) / h;
 dx = ( x_range(2) - x_range(1) ) / nstep;

for i = 1 : nstep
 x(i+1) = x(i) + dx;
 y(i+1) = y(i) + dx * feval ( f, x(i), y(i) ); end

ode45(f,x_range,y_initial)
 f=strrep(f,'blk2','sol');hold on
 plot(x,y,'r')
 fplot(f,x_range,'g')
 set(findobj(gca,'Type','Line','Color','g'),'LineWidth',4);
 set(findobj(gca,'Type','Line','Color','r'),'LineWidth',3);
 set(findobj(gca,'Type','Line','Color','b'),'LineWidth',2);
 hold off
 xlabel('x')
 ylabel('y')

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To use this function in MATLAB we type said the following command:

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As a result from the following command in MATLAB we got the following plot:

Now that all three functions are on one plot, it is easy to compare them. Eulers method seems to be further and further off as x increases, while ODE45 and DSOLVE seem to be right on. But Remember that ODE45 wouldn’t work after 4.5 so it isn’t perfect. Before making a conclusion we decided to try the next example first.

Problem 5

= , 

The first step we took in examining this problem was to simply plot eulers function since the code was already fixed for the previous problem. Before graphing thought we made our first function file for problem 5 in which we called problem_5_blk2.m. The following was what was put into that file:

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function yprime = problem_5_blk2(x,y)

yprime = y*exp(-x);
end

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Since this file is all set, we next plotted Eulers Function. The following is the command we used to do just that.

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euler(‘problem_5_blk2′,[0,6.8],1,.1)

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NOTE: We started our boundaries at 0 because the initial value of at = 0 is 1 as the problem states.

As a result from the following command in MATLAB we got the following plot:

This plot looks good but still need to compare it with the other functions to see how accurate it is.

The next function that we plotted was ODE45. Since we have already know how to use ODE45 we simply just typed the following command into MATLAB:

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As a result from the following command in MATLAB we got the following plot:

This plot looks quite similar to euler’s plot. But, before comparing we will graph the exact solution first by using the MATLAB dsolve function.

As said before, we decided to do the function on the computers in class and then would plot them later on a different computer. For this example we got this when we used the dsolve function:

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NOTE: exp is

Next we  to put the answer into a .m file and make it into a function similar to our problem_3_sol.m We named this file problem_5_sol.m and here is what we put in it:

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function yprime = problem_5_sol(x)

yprime = exp(1-exp(-x));
end

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Now that we had our explicit equation file setup we plotted it using fpolt. Here is what we put into MATLAB

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As a result from the following command in MATLAB we got the following plot:

This plot looks very similar to both our ODE45 and Euler function. Just like before, the best way to compare is to plot all three at once. To do this we called our function again to accomplish this.

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As a result from the following command in MATLAB we got the following plot:

Now that all three functions are on one plot, it is easy to compare them. Eulers method seems to be further and further off as x increases once again, while ODE45 and DSOLVE seem to be right on. But this time ODE45 seemed to work without any problem. We were starting to get and idea at this point but we wanted to try the rest of the examples first.

Problem 6

= , 

The first step we took in examining this problem was to simply plot eulers function since the code was already fixed for the previous problem. Before graphing thought we made our first function file for problem 6 in which we called problem_6_blk2.m. The following was what was put into that file:

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function yprime = problem_6_blk2(x,y)

yprime = x*y^2;
end

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Since this file is all set, we next plotted Eulers Function. The following is the command we used to do just that.

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euler(‘problem_6_blk2′,[0,1.3],1,.1)

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NOTE: We started our boundaries at 0 because the initial value of at = 0 is 1 as the problem states.

As a result from the following command in MATLAB we got the following plot:

This plot looks good but still need to compare it with the other functions to see how accurate it is.

The next function that we plotted was ODE45. Since we have already know how to use ODE45 we simply just typed the following command into MATLAB:

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NOTE: We were limited by our boundaries. We needed to start at zero and also because of the way this function is we couldn’t go past 1.3 without getting an error from ODE45.

As a result from the following command in MATLAB we got the following plot:

This plot looks quite similar to euler’s plot. But, before comparing we will graph the exact solution first by using the MATLAB dsolve function.

As said before, we decided to do the function on the computers in class and then would plot them later on a different computer. For this example we got this when we used the dsolve function:

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Next we  to put the answer into a .m file and make it into a function similar to our problem_3_sol.m We named this file problem_6_sol.m and here is what we put in it:

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function yprime = problem_6_sol(x)

yprime = -1/(x^2/2-1);
end

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Now that we had our explicit equation file setup we plotted it using fpolt. Here is what we put into MATLAB

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As a result from the following command in MATLAB we got the following plot:

This plot looks very similar to both our ODE45 and Euler function from the range 0 to 1.3.  Seeing this from further away you can see why ODE45 wouldn’t work after 1.3. This the second time seeing this. It seems that ODE45 dosn’t work if the differential equation goes to infinity. Anyways, just like before the best way to compare is to plot all three at once. To do this we called our function again to accomplish this.

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As a result from the following command in MATLAB we got the following plot:

Now that all three functions are on one plot, it is easy to compare them. Eulers method seems to be further and further off as x increases, while ODE45 and DSOLVE seem to be right on. Similar to every example so far. But Remember that ODE45 wouldn’t work after 1.3 so it isn’t perfect.

Problem 10

= , 

The first step we took in examining this problem was to simply plot eulers function since the code was already fixed for the previous problem. Before graphing thought we made our first function file for problem 6 in which we called problem_6_blk2.m. The following was what was put into that file:

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function yprime = problem_10_blk2(x,y)

yprime = x+y;
end

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Since this file is all set, we next plotted Eulers Function. The following is the command we used to do just that.

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euler(‘problem_10_blk2′,[0,3],0,.1)

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NOTE: We started our boundaries at 0 because the initial value of at = 0 is 0 as the problem states.

As a result from the following command in MATLAB we got the following plot:

This plot looks good but still need to compare it with the other functions to see how accurate it is.

The next function that we plotted was ODE45. Since we have already know how to use ODE45 we simply just typed the following command into MATLAB:

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NOTE: We were limited by our boundaries. We needed to start at zero and also because of the way this function is we couldn’t go past 1.3 without getting an error from ODE45.

As a result from the following command in MATLAB we got the following plot:

This plot looks quite similar to euler’s plot. Once again ODE45 wouldn’t work past the 0 to 3ish area. But, before comparing we will graph the exact solution first by using the MATLAB dsolve function.

As said before, we decided to do the function on the computers in class and then would plot them later on a different computer. For this example we got this when we used the dsolve function:

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NOTE: exp is

Next we  to put the answer into a .m file and make it into a function similar to our problem_3_sol.m We named this file problem_10_sol.m and here is what we put in it:

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function yprime = problem_10_sol(x)

yprime = exp(x)-x-1;
end

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Now that we had our explicit equation file setup we plotted it using fpolt. Here is what we put into MATLAB

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As a result from the following command in MATLAB we got the following plot:

This plot looks very similar to both our ODE45 and Euler function from the range 0 to 3.  Seeing this from further away you can see why ODE45 wouldn’t work after 3ish. This the third time seeing this. It seems that ODE45 dosn’t work if the differential equation goes to infinity. Anyways, just like before the best way to compare is to plot all three at once. To do this we called our function again to accomplish this.

__________________________________________________________________________________________

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As a result from the following command in MATLAB we got the following plot:

Now that all three functions are on one plot, it is easy to compare them. Eulers method seems to be further and further off as x increases, while ODE45 and DSOLVE seem to be right on. Similar to every example so far. But Remember that ODE45 isn’t perfect.

Problem 13

= , 

The first step we took in examining this problem was to simply plot eulers function since the code was already fixed for the previous problem. Before graphing thought we made our first function file for problem 13 in which we called problem_13_blk2.m. The following was what was put into that file:

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function yprime = problem_13_blk2(x,y)

yprime = x^3*y-x^2*y^2;
end

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Since this file is all set, we next plotted Eulers Function. The following is the command we used to do just that.

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euler(‘problem_13_blk2′,[-1,5],0,.01)

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NOTE: We started our boundaries at -1 because the initial value of at = -1 is 1 as the problem states.

As a result from the following command in MATLAB we got the following plot:

This plot looks good but still need to compare it with the other functions to see how accurate it is.

The next function that we plotted was ODE45. Since we have already know how to use ODE45 we simply just typed the following command into MATLAB:

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As a result from the following command in MATLAB we got the following plot:

This plot looks quite similar to euler’s plot. We did have a problem with the next part though. Dsolve didn’t seem to work for this problem. We put the equation into dsolve and this is what we got:

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Unfortunately we could not graph this function. So instead we decided to compare ODE45 and euler’s method. What we did was just take out the fplot of our Plot all three one on plot function. This is what the code looked like:

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function [ x, y ] = euler_ODE ( f, x_range, y_initial, h ) 

x(1)=x_range(1); 

y(1)=y_initial;

nstep=( x_range(2) - x_range(1) ) / h;
 dx = ( x_range(2) - x_range(1) ) / nstep;

for i = 1 : nstep
 x(i+1) = x(i) + dx;
 y(i+1) = y(i) + dx * feval ( f, x(i), y(i) ); end

ode45(f,x_range,y_initial)
 f=strrep(f,'blk2','sol');hold on
 plot(x,y,'r')
 set(findobj(gca,'Type','Line','Color','r'),'LineWidth',3);
 set(findobj(gca,'Type','Line','Color','b'),'LineWidth',2);
 hold off
 xlabel('x')
 ylabel('y')

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The following is the command that we used to plot ODE45 and Euler using this function:

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euler_ODE(‘problem_13_blk2′,[-1,5],1,.01)

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As a result from the following command in MATLAB we got the following plot:

Instead of Eulers method being further and further off as x increases, it seems to be right on with ODE45. Part of the reason is for Eulers Method the h had to be changed in order for this function to plot and the lower the h, better the result. Although we couldn’t plot or get an exact dsolve, euler’s method and ODE45 are great ways to get at least an approximation. They seem be most usefull on hard differental equations.

Problem 14

= , 

The first step we took in examining this problem was to simply plot eulers function since the code was already fixed for the previous problem. Before plotting we made a function file for problem 14 in which we called problem_14_blk2.m. The following was what was put into that file:

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function yprime = problem_14_blk2(x,y)

yprime = x^3*exp(y)+3*x^2*sin(y);
end

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Since this file is all set, we next plotted Eulers Function. The following is the command we used to do just that.

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euler(‘problem_14_blk2′,[-.5,5],1,.01)

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NOTE: We started our boundaries at -0.5 because the initial value of at = -1 is 1 as the problem states.

As a result from the following command in MATLAB we got the following plot:

This plot looks good but still need to compare it with the other functions to see how accurate it is.

The next function that we plotted was ODE45. Since we have already know how to use ODE45 we simply just typed the following command into MATLAB:

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As a result from the following command in MATLAB we got the following plot:

This plot looks quite similar to euler’s plot. We did have a problem with the next part though. Dsolve didn’t seem to work for this problem. We put the equation into dsolve and this is what we got:

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So instead we decided to compare ODE45 and euler’s method. Using the same function from the previous problem where we took out fplot, we did the following command:

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euler_ODE(‘problem_14_blk2′,[-0.5,5],1,.01)

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As a result from the following command in MATLAB we got the following plot:

Once again instead of Eulers method being further and further off as x increases, it seems to be right on with ODE45. Part of the reason is for Eulers Method the h had to be changed in order for this function to plot and the lower the h, better the result. Although we couldn’t plot or get an exact dsolve, euler’s method and ODE45 are great ways to get at least an approximation. They seem be most usefull on hard differental equations.

CONCLUSION

Although solving the differental equation and plotting that seems the be the best way to get a result, as we proved sometimes its not possible to do that. Euler’s method isn’t to most accurate all the time  but it does a good job in putting you in the right ballpark. Overall when unable to solve a differenial equation Euler’s Method or ODE45 are two great alteratives to give you that general idea.

The example that my partner (Mike Bosse) and I choose was = .  This example came from page 94 in our textbook A Course In Ordinary Differential Equations by Randall J. Swift and Stephen A. Wirkus.

The first step we took to for this differential equation was plotting the vector field. To do this we used the help of Gary Davis’s code as a starting point and MATLAB. To help make the code more appropriate for our example we changed equal to our function . Then we looked at our solutions and . To accomplish this we changed the input values into the meshgrid function.  meshgrid works like this: meshgrid(x,y), where x is your properties for x-axis on the grid and y is the properties for the y-axis on the grid.

The code that we did in MATLAB for the differental equation = can be found below:

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[X,Y]=meshgrid(-5:0.4:5,-5:0.4:5); % o.4 is the step size for the grid
DY=((Y-1)./(X-1);
DX=ones(size(DY)) ;
DW=sqrt(DX.^2+DY.^2);
quiver(X,Y,DX./DW,DY./DW,0.5,'.');
xlabel('x');
ylabel('y');
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As a result from this code from MATLAB we got the following plot:

From the plot we can see that all the action seems to happen around the area where is equal to one. We obtained this plot by saving it as a jpeg.  If we were to put an imaginary line down = 1, we would see the vector field being split into two regions. The left region points towards the imaginary line while the region on the right points away from the from it. This is because dividing by zero is undefined. When is 1 you get which is equal to . Also since everything happens around = 1 we didn’t feel the need to expand our graph out any further.

Now that we have obtained the vector field for this differential equation, we used Euler’s method to obtain a numerical approximation to a solution curve, first choosing an initial condition of equal to zero.

Euler’s Method

Euler’s Method is the simplest method of obtaining a numerical solution of a differential equation.  Once again, in order to do this method we used code from Gary Davis as a starting point. What we did was copied the code into something called a .m file in MATLAB. In doing this, we made our own function called euler. This way we could keep using this function for future examples. We also created another .m file that the euler function could refer to for the right differential equation. This was done this way to more generalize the Euler function.

In copying the code we ran into a few problems. The problem was on line 17 of the code which was y(i+1)=y(i)+dx*feval(f,x(i),y(i) ); Once we saw this we had no clue what to do until one of the student’s from the class said at the beginning of the function to define and as set variable’s which were equal to zero so = 0 and =0. But, this also didn’t work. So after using the trying a few things and looking at the code, we came up with our own solution. With setting and equal to zero, the function didn’t care what value you passed to it, it set both initials to zero. So after seeing this we set (1) to x_range(1) and (1) to y_inital. This solved the problem.

The following is the code for the euler.m file, or the euler function.

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% Euler’s method for a single differential equation.
% The differential equation is dy/dx = f(x,y)

function [ x, y ] = euler ( f, x_range, y_initial, nstep )

x(1)=x_range(1);

y(1)=y_initial;

% The program euler.m will ask for a range of x-values, from x(1) to x(2).x(1) = x_range(1);

% We define dx by dividing this range into an equal number of specified steps

dx = ( x_range(2) - x_range(1) ) / nstep; % We set the initial value of y at the beginning x-value

x(1).y(1) = y_initial; % We use Euler’s method to update the value of y at new time steps.

% “feval” is used instead of “eval” because we are passing the name of the function f to the program.

for i = 1 : nstep
x(i+1) = x(i) + dx;
y(i+1) = y(i) + dx * feval ( f, x(i), y(i) ); end

% The following command plots y as a function of x.

plot(x,y)
xlabel('x')
ylabel('y')

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For this equation it is asking for four things to be passed to it. (f, x_range, y_initial, nstep) f is the differential equation, x_range is something called and array with asks from where to where on the x-axis do you want to see ploted, y_inital is the inital value of y to start from, and finally nstep is the number of steps to do eulers method. The higher the steps the more accurate your curve will be.

Next, we have the code for the differential equation file in which we called example1.m:

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function yprime = example1(x,y)
yprime = (y-1)./(x-1);

end

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Now executing the euler m-file and calling the specific differential equation that we called example1.m, we will produce a vector of numerical values for x and y. These numerical values to plot an approximation to a solution curve to the differential equation. To do this in MATLAB we said the following command:

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euler(‘example1′,[0,12],.001,60);

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PLEASE NOTE: In order for this to work in MATLAB you MUST have both m-files in the same location on your computer. For example they must be in the same folder. Also make sure example1 is in single quotes

As a result from the following command in MATLAB we got the following Curve:

This curve is pretty consistent with the vector field and for this differential equation all the action is around = 1. So there was no need to widen out this curve. To make sure that the curve and the vector field were consistent we decided to do both on one plot. Combining Gary Davis’s code and both the vector field and curve code, we came up with the following:

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[x,y]=euler('example1',[-1,12],.001,60);
[X,Y]=meshgrid(-1:.5:12,-1:.5:16); %Use CAPITALS to avoid re-assigning x and y
DY=(Y-1)./(X-1);
DX=ones(size(DY));
DW=sqrt(DX.^2+DY.^2);
quiver(X,Y,DX./DW,DY./DW,0.5); % Plot the vector field for the differential equation hold on
plot(x,y) % Plot the numerical solution curve over the vector field hold off
xlabel('x')
ylabel('y')

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After typing this into MATLAB we see the numerically computed solution curve and the vector field tangent to the solution curve. Here is the plot:

EXAMPLE 2

For this example we used the same steps as the first one except the equation was one that Gary Davis had given us which was = .

The first step we took to for this differential equation was plotting the vector field. This time for graphing the vector feild we just copied our previous code from EXAMPLE 1 and edited so that it suited = . This was pretty easy to do since we had already worked out all the bugs thanks to the first example. We looked at our solutions and . After plotting this function a few times we thought that this range got the best result and where you can really see what is happening. The following is the code that was put into MATLAB for this differential equation:

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[T,Y]=meshgrid(-7.6:0.6:1.8,-4:0.6:4); % o.6 is the step size for the grid
DY=Y^2 + T;
DT=ones(size(DY)) ;
DW=sqrt(DT.^2+DY.^2);
quiver(T,Y,DT./DW,DY./DW,0.5,'.');
xlabel('t');
ylabel('y');

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As a result from this code from MATLAB we got the following plot:

From this plot we can see that all of the action happens from = -8 to = 0. Since is the variable that is getting squared from the differential equation = , this plot makes sense.

Now that we have obtained the vector field for this differential equation, we used Euler’s method to obtain a numerical approximation to a solution curve, first choosing an initial condition of equal to zero.

Since we already know that our euler function is generalized all we had to do was create a new m-file that had this differential equation. The Following is what was in that m-file and we decided to call it example2.m:

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function yprime = example2(t,y)
yprime = y.*y + t;

end

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Next we apply example 2 to our already made euler function we get the following that we used in MATLAB:

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euler(‘example2′,[-4,1.8],.001,60);

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REMEMBER: In order for this to work in MATLAB you MUST have both m-files in the same location on your computer. For example they must be in the same folder. Also make sure example1 is in single quotes

As a result from the following command in MATLAB we got the following Curve:

This curve makes sense and is very similar to the vector field. But, to make sure we decided to graph both in the same plot just to make sure. All we had to do was copy the code from example 1 and apply it to this differential equation. In doing that we get the following:

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[x,t]=euler('example2',[-7.6,1.8],.001,60);
[X,T]=meshgrid(-47.6:.5:1.8,-4:.5:4); %Use CAPITALS to avoid re-assigning x and y
DY=Y.*Y+T;
DX=ones(size(DY));
DW=sqrt(DT.^2+DY.^2);
quiver(T,Y,DT./DW,DY./DW,0.5); % Plot the vector field for the differential equation hold on
plot(t,y) % Plot the numerical solution curve over the vector field hold off
xlabel('t')
ylabel('y')

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After typing this into MATLAB we see the numerically computed solution curve and the vector field tangent to the solution curve. Here is the plot:

As you can see the Curve goes in the direction of the vector field.

Overall you can see that my partner and I learned a great deal about MATLAB in the first two week block and even though we ran into a few minor errors and mistakes we were able to make sense of what we were doing and feel confident about it.