differential equation problem with Matlab

damightytom

1. Solve the differential equation

y=10, dy/dx=0 when x=0
z=15, dz/dx=0 when x=0

d2y/dx2 = - 2y/3 + (z – y)/3
d2z/dx2 = -(z – y)

plot y,z,dydx,dzdx doe x=0->10

2. Relevant equations

I am having trouble getting my head around the problem.
I need help setting a variable for dy/dx and dz/dx in my function for ODE45 to find the value.


3. The attempt at a solution

So after setting up my script as such
[y z] = ode45('fun',[0 10], vekt);

vekt being the vector for all values at x=0

I need to set up my function so ODE has something to work with.
I know what to do when I have d/dx in the equation since you can set dy/dx = z and then d2y/dx2 = y'. But now I have a value for dydx at x=0 but I don't have that in the equation.

So I'm at a loss.
Anyone can give me any pointers?

tjackson3

You don't even need to use ODE45 for this. There's a simpler way; cf. http://www.math.tamu.edu/REU/comp/matode.pdf, which uses the function dsolve

damightytom

Thanks for the heads up, but the main problems for me is to understand what I should do with dydx and dzdx that can't be found in the equation but have a value at x=0

damightytom

Do I have to find the antiderivitive for d2y/dx2 in order to get dy/dx and then solve it?

Or is there some kind of trick into getting dy/dx in the equation?

SteamKing

Remember that numerical solvers for ODEs work on solving first order ODEs only. How would you transform the second equation to give you first order ODEs only?

damightytom

d2z/dx2 = -(z – y)

z''=-(z-y)

q=[y';y'']

y(1)=z''
y(2)=z'

Can I do something like that so I have ekv=y(2); -(z-y) for the second equation?

I'm out in very deep water, I would kinda know what to do if I saw dy/dx in the equations, but now I don't so I'm pretty much guessing.