Thread: 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.

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

Differentiate eq(1), and substitute from eq(3),

Substitute for in eq(2) also,

Now differentiate twice and then subtract to eliminate the 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.

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