Solve differential equation with three variables

Question

I have the following differential equation:

$$z''+y'= \cos (x)$$

$$y''-z= \sin (x)$$

with $z(0)=1$, $z'(0)=-1$, $y(0)=1$ and $y'(0)=0$.

These differential equations involves three variables $x, y, z$. Can someone drop a hint on how to start this.

Answer 1

I'm just going to assume $y$ and $z$ are functions of $x$ since there are only two equations and no initial conditions on $x$ are provided.

Here's a hint. Differentiate the first equation $$ z''' + y'' = -\sin x $$

Subtract this from the second equation to get $$ z'''+ z = -2\sin x $$

This is a non-homogeneous linear ODE that you can solve.