ODE and IVP: need to solve $y(0)=2$ and $y'(0)=1,$ I have $y=A\sin(2x)+B\cos(2x)$

Question

Solve $$y''+4y=0 \\ y(0)=2 \;, y'(0)=1$$

I have managed to solve $y(0)=2$ where $B=2,$ but I'm struggling with $y'(0)=1$

Is $y'=2A\cos(2x)-2B\sin(2x) ?$

Answer 1

$$y=A\sin(2x)+B\cos(2x)$$ $$y(0)=2 \implies B=2$$ Yes when you differentiate you get: $$y'=2A\cos(2x)-2B\sin(2x) $$ Then initial condtion gives: $$y'(0)=2A=1 \implies A=\dfrac 12$$ So that the solution is : $$y(x)=\dfrac 12\sin(2x)+2\cos(2x)$$