Trigonometry, Stationary Points and the Second Derivative

I'm having trouble getting the answer to this particular question out;

Given that the $\frac{d^2y}{dx^2}$ is equal to $9\sin3x$ ;
Find y if there is a stationary point at $(\frac{\pi}{2}, 1)$
Show that $\frac{d^2y}{dx^2} + 9y = 0$

What you have here

Quote:

Originally Posted by Flay

I'm having trouble getting the answer to this particular question out;

Given that the $\frac{d^2y}{dx^2}$ is equal to $9\sin3x$ ;

Find y if there is a stationary point at $(\frac{\pi}{2}, 1)$
Show that $\frac{d^2y}{dx^2} + 9y = 0$

is a differntial equation...here is what you do...since $\frac{d^2y}{dx^2}=9sin(3x)$ ......ok now seperate to get $d^2y=9sin(3x)dx$ then integrate to get $y'=-3cos(3x)+C$ ...now stop before we get to y we must find that c...we know there is a stationary point at c which means the slope is 0 at $\frac{\pi}{2{$ which means that y'=0...so thefore $0=-3cos\bigg(3\cdot\frac{\pi}{2}\bigg)+C$ solve and you see c=0...next integrate again to get y and you see that $y=-sin(3x)+C$ now using that condition again we know that $1=-sin\bigg(3\cdot\frac{\pi}{2}\bigg)+C_1$ solving we get x=0..therfore $y=-sin(3x)$

for part two just add nine times what we just found to what $\frac{d^2y}{dx^2}$ is to verify its zero

Got it. I was making yet another stupid mistake.