SHM can be described as $$x''= -\omega^2x$$
If initial conditions are $x(0) = x_0,\ x'(0)=y_0$ I need to solve this equation by
a) Taking laplace transforms directly
b) Defining the function y(t) = x'(t) and hence obtaining an equivalent pair of coupled, first-order differential equations
I can do part a) where I got $x(t)=\frac{y_0}{\omega}\sin{(\omega t)} + x_0\cos{(\omega t)}$ but I am confused as to what to do for part b. I don't understand what it wants me to find.
Hint: If we call $x'=y_1$ and $x''=y_1'=y_2$, then we can rewrite the ODE as a system:
$y_1'=y_2$
$y_2'=-\omega^2 y_1$
Can you cast this into a matrix form? Then you will need to find the eigenvalues $\lambda_i$ and eigenvectors $v_i$ of this matrix. The solution of the system is then given by (if the eigenvalues are destinct) $$Y=c_1v_1e^{\lambda_1 t}+c_2v_2e^{\lambda_2 t}.$$
Note, that $Y=[y_1,y_2]^T$ is a vector.