Solve the PDE $u_{xx}+u=0$

Question

Solve the PDE $u_{xx}+u=0$. The answer is $u(x,y)=f(y)cos(x)+g(y)sin(x)$ but I don't understand where the cos and sin terms come from. I know that the solution to $u_{xx}=o$ is $u(x,y)=f(y)x+g(y)$.

Answer 1

$$u_{xx}+u=0$$ Since you differentiate wrt x twice, consider the ode $$\frac {d^2u}{dx^2}+u=0$$ $$r^2+1=0 \implies r=\pm i$$ $$u(x,y)=K_1(y)\cos(x)+K_2(y)\sin(x)$$