# Thread: Reduction to single higher-order linear ODE

1. ## Reduction to single higher-order linear ODE

I need to reduce the following system to a single higher-order linear ODE for y

x'(t) - y'(t) + z(t) = 0
x(t) + y'(t) - z'(t) = 1
y(t) + z'(t) = t3- 1

I've tried setting ( z = y' - x' ) & (x = 1 - y' - z')
However on substituting these it ends up as a big mess and I can't go anywhere. Could you someone tell me what substitution or technique I should use to work this problem?

Thanks.

2. ## Re: Reduction to single higher-order linear ODE

Can get you to a cubic, don't know if it's possible to do better

Differentiate eq(1), $\displaystyle x'' - y''+z'=0$ and substitute from eq(3), $\displaystyle x''-y''+(t^{3}-1-y)=0..........(4)$

Substitute for $\displaystyle z'$ in eq(2) also, $\displaystyle x+y'-(t^{3}-1-y)=1...........(5).$

Now differentiate $\displaystyle (5)$ twice and then subtract $\displaystyle (4)$ to eliminate the $\displaystyle x''$ term.

Note that you lose information during this process though, The 1 on the RHS of the second equation differentiates out and so could be any number at all.

3. ## Re: Reduction to single higher-order linear ODE

Yes that works, thanks a lot. I can't see a way of keeping the "1" unless I integrate the first equation and sub x(t) in the second. But I don't think that's possible. I believe this answer is sufficient. Thanks again