1. ## Find constant

I have the equation y'' = -4y <=> y'' + 4y = 0. This means, I suppose, that I can only find a homogenous solution. Or rather that the particular solution and the homogenous one are the same.

r^2 + 4 = 0 <=> r = +- 2i. So y = Acos2x + Bsin2x. I know from the problem that y(0) = 10. So A*1 + B*0 = 10 <=> A = 10.

Now the purpose is to find out when y = 0 for the first time. To do this I need to find out B. How can I do that? (B=0, but how do I know this?)

Also I'm not studying math in English so I'm not sure if I got the name for everything right, please tell me if anythings wrong!

2. Originally Posted by tnsten
I have the equation y'' = -4y <=> y'' + 4y = 0. This means, I suppose, that I can only find a homogenous solution. Or rather that the particular solution and the homogenous one are the same.

r^2 + 4 = 0 <=> r = +- 2i. So y = Acos2x + Bsin2x. I know from the problem that y(0) = 10. So A*1 + B*0 = 10 <=> A = 10.

Now the purpose is to find out when y = 0 for the first time. To do this I need to find out B. How can I do that? (B=0, but how do I know this?)

Also I'm not studying math in English so I'm not sure if I got the name for everything right, please tell me if anythings wrong!

You'll need two conditions to find both constants. You gave one $y(0)=10$. To find tha $B = 0$, there has to be another condition for that. Now you have the solution

$
y = 10 \cos 2x
$

To find out when $y = 0$, you'll need to find the values of x such that $\cos 2x = 0$ or when $2x = \frac{\pi}{2} + n \pi,\;\; n \in \mathbb{Z}$

3. Aww...yes I just read the problem wrong. It's an equation for a moving particle, and apparently at t = 0 it's also at rest in y = 10. Could that mean y'(0) = 0 ? This condition makes B = 0.

But from what we've learned in physics velocity has nothing to do with rest, Galileo was wrong etc. Instead rest only implies Fnet = 0, so the object does not accelerate. So y''(0) = 0 <=> -40cos2x - 4Bsin2x = 0 and that equation has no solutions...