The linear ordinary differential equation $y'' + y = 0$ has the family of solutions $y = A \sin(x) + B \cos(x)$ Determine whether $y(0)=2, y'(\pi/2)=3$ is a unique solution. If not, does it have no solutions or infinite?
I found $B=2$ and $A=5x$ giving the general solution $5x\sin(x)+2\cos(x)$
Am I right?
$$y'' + y = 0, y(0) = 2, y'(\pi/2) = 0 \tag 1$$ then $y = a\cos x + b \sin x$ is a solution. suppose we want $y(0) = 2.$ that requires $a = 2.$ now we have $$y = 2\cos x + b \sin x, y' = -2\sin x + b\cos x \implies y'(\pi/2) = -2 \text{ for any } b.$$ this means the boundary value problem $(1)$ has no solutions. that is there is no function $y$ that can satisfy all three requirements.
a different boundary value problem $$y'' + y = 0, y(0) = 2, y'(\pi/2) = -2 \tag 2$$ will have infinitely many solutions $y = 2\cos x + b \sin , b \text{ arbitrary.}$