Given $y = A\cos2x + B\sin2x$; find the value of $A$ and $B$ if $y = 3$ when $x = \frac{\pi}{2}$ and $\frac{dy}{dx} = 4$ when $x = 0$

Question

There is an additional part to this question where it asks to show that $\frac{d^2y}{dx^2} + 4y =0$. I was able to solve that part but I got stuck where I had to find the values of A and B. In the textbook it says that the values are (-3,2) but I am unable to solve for the answer.

Work: \begin{align} \frac{dy}{dx} &= -2A \sin2x + 2B \cos2x \\ 4 &= -2A \sin(0) + 2B \cos(0) \\ 4 &= 2B \\ B &= 2, \end{align}

If $y = 3$, when $x= \frac{\pi}{2}$, then $3 = A \cos\pi + B \sin\pi$.

I don’t know how to move foward beyond this, please help!

Answer 1

Let $y(x)= A \cos 2x+B \sin 2x$. Then $3=y( \pi /2)=-A$.

We have $y'(x)= -2A \sin 2x+2B \cos 2x$, hence $4=y'(0)=2B$.