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Solving two differential equations using ode45

Asked by Nagarajan on 3 Jun 2019
Latest activity Commented on by Nagarajan on 4 Jun 2019
I have two differential equations in the form
-y"-5y'-y=3x"+4x'^2+2x+6t
-y"-2.3y'-2y=2x"+3x'^2+x
I would like to solve these equations using ODE's
the inital conditions are x(0), y(0), x'(0) and y'(0) are known
Thanks in advance

  5 Comments

I need asssitance in converting the above equations into the ODE solver form
for Ex:
fun=@(t,y)[y(2);......];
Do you have access to the Symbolic Math Toolbox?
Yeah i do have access
I need assistance while converting the variables
fun=@(t,y)[.....]; % change of variables..
Y0=[x0; x0prime; y0; y0prime]; % initial conditions
tspan=[0 2];
[T,Y]=ode45(fun,tspan,Y0);

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2 Answers

Answer by Stephan
on 3 Jun 2019
Edited by Stephan
on 3 Jun 2019
 Accepted Answer

This will help:
Follow the given examples and adapt them to your needs - for example:
syms y(t) x(t) t
ode(1) = -(diff(y,t,2))-5*diff(y,t)-y == 3*diff(x,t,2)+4*diff(x,t)^2+2*x+6*t;
ode(2) = -(diff(y,t,2))-2.3*diff(y,t)-2*y == 2*diff(x,t,2)+3*diff(x,t)^2+x;
pretty(ode)
[V,S] = odeToVectorField(ode)
fun = matlabFunction(V,'Vars',{'t','Y'})
x0 = [2 0 3 0.25];
[t, sol] = ode45(fun,[0 2], x0);
subplot(2,2,1)
plot(t,sol(:,1),'-r')
subplot(2,2,2)
plot(t,sol(:,2),'-b')
subplot(2,2,3)
plot(t,sol(:,1),'-g')
subplot(2,2,4)
plot(t,sol(:,2),'-k')

  1 Comment

Thank you stephan, a clear explanation.

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Answer by Steven Lord
on 3 Jun 2019

Let v = [x; x'; y; y']. This means v' = [x'; x''; y'; y''].
Can you express x' in terms of some of the elements of v? Sure, it's just v(2). Expressing y' in terms of elements of v is also easy, it's v(4). So we have half the right-hand side of our system of ODEs. We still have two pieces to fill in; I'll represent them by placeholders in the (pseudo)code below.
dvdt = [v(2);
something;
v(4);
somethingElse];
Now things get a bit trickier. You can't isolate y'' or x'' as a function solely of x, x', y, y', and t. So you're going to need to use a Mass matrix. The ODE you're going to solve is not v' = f(t, y) but M*v' = f(t, y). If we rewrite your first ODE to put the second derivatives together on the left side of the equals sign:
-y''-3*x'' = 5*y'+y+4*x'^2+2*x+6*t
The left side is [0 -3 0 -1]*v (to show this, expand it out: 0*x'-3*x''+0*y'-y''). Let's put that in the second row of our mass matrix. The first and third rows are just the corresponding rows of the identity matrix, since [1 0 0 0]*v == x' and v(2) == x' and similarly for [0 0 1 0]*v and v(4) both representing y'.
M = [1 0 0 0;
0 -3 0 -1;
0 0 1 0;
aDifferentSomethingElse];
dvdt = [v(2);
something;
v(4);
somethingElse];
Fill in the something part of dvdt using the right side of that rewritten first ODE. Do the same type of manipulation (pull the second derivates to the left side of the equals sign) for the second ODE to fill in aDifferentSomethingElse and somethingElse.
Now write a function to evaluate dvdt as a function of t and v. You will need to use odeset to create an options structure containing your Mass matrix M. Pass the function and the options structure into ode45.
Technically you could have pulled both the first and second derivatives to the left side of the equals sign, and had all the elements of that row of the mass matrix be non-zero. But I wanted to keep this a little simple (too late, I know) so I pulled just the second derivatives into the Mass matrix.

  1 Comment

Thank you steven, Your answer gave a clear explanation in solving the ODE's

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