Solving differential equation using ode45 with three variables

Hi I'm trying to solve a simple second-order differential equation y''-2y'+y=0, with initial conditions where y'=0 and y=0.

I have successfully made a function of

function dydt=ode5( t, y)

dydt=zeros(2,1);

dydt(1)=y(2);

dydt(2)=2*y(2)-y(1);

end

but I want to make a function that can substitute dydt=p or some other symbol to simply put it as

function dydx=ode9(x,p,y)

p=zeros(2,1);

dydx=p;

dpdx=-p+2*y;

end

but I keep getting errors. Help!

[Torsten]

It's not possible to set a boundary condition for the second derivative for a differential equation of order 2. Only boundary conditions for the function itself and its first derivazive are allowed.

[Jay Kim]

typo fixed! but I'm still stuck with the substitution though

[Torsten]

I don't understand the sense of your code.

1. p is an input parameter, but you reset it to null with p = zeros(2,1).

2. You define dpdx, but the return variable of ode9 is dydx. So dpdx is not used.

Which differential equation do you intend to define in ode9 ?

[Jay Kim]

Thanks for trying to help me by the way.

What I'm trying to understand is, the code works when I put dydx as y(2) and y as y(1) so this time, I'm trying to put dydx as p (or some other symbol) and y as y and make it work

[Jan]

@Jay Kim: It is not clear to me, what you are asking for.

[Torsten]

The only meaningful I can think of is

[X,Y]=ode45(@(x,y)ode9(x,y(1),y(2)),tspan,y0)

function derivatives = ode9(x,y,yp)
derivatives = zeros(2,1)
dydx = yp;
dypdx = 2*yp-y;
derivatives = [dydx;dypdx]
end

[Walter Roberson]

Do I understand correctly that you would like to pass in as p some indication of which y entry to use?

Would the code always be as simple as dydx being assigned y indexed at some fixed value?

Answer 1

[Walter Roberson]

p = 1;
[X,Y]=ode45(@(x,y)ode9(x, p, y), tspan, y0);

function derivatives = ode9(x, p, y)
  derivatives = zeros(2,1);
  derivatives(1) = y(p);
  derivatives(2) = 2*y(p)- y(3-p);
end

Here I had to guess what you wanted. You subtract y for the second of the two derivatives, but y is a vector of two elements. I had to guess that you wanted the other y entry, the one not selected by p. When you have a p that is either 1 or 2, then 3-p would be 2 or 1, thereby selecting the other entry.

Answer 2

[goutham baburaj]

thanks