Differential Equation: Find solution of initial value problem with equation y''+4y=0 with initial conditions y(0) = 0 and y'(0) = 1

Question

Differential Equation: Find solution of initial value problem with equation y''+4y=0  with initial conditions y(0) = 0   and y'(0) = 1

anonymous · Answer

have you learned laplace transforms yet?

anonymous · Answer

I got the general solution to be $$y=c _{1} \cos2t + c _{2} sint 2t$$

anonymous · Answer

no

zepdrix · Answer

Hmm yah your general solution looks good. :D having trouble figuring out c1 and c2?

anonymous · Answer

yep

anonymous · Answer

i know that you have to plug it in but im not sure when to and if it is before or after driving the solution

anonymous · Answer

deriving*

anonymous · Answer

you know y(0)=0

so plug that in

you know y'(0)=1

so find the derivative of y and plug that information in, then solve out for c1 and c2

zepdrix · Answer

$$y(0)=0=c_1\cos(0)+c_2\sin(0)$$
Yah you plug in after you have found your general solution.
You will have a system of equations :) the first one looking like this.

anonymous · Answer

ok got it thx guys