I'm kinda stuck
Solve the initial value problem
$$y''+3y=0$$
$$y(0)=1$$
$$y'(0)=3$$
$$r^2+3=0$$
$$r=\sqrt3$$ or
$$r=-\sqrt3$$
$$y=c_1e^{\sqrt3x}+c_2e{-\sqrt3x}$$
$$y(0)=c_1+c_2=1$$
$$y'=\sqrt3c_1e^{\sqrt3x}-\sqrt3c_2e{-\sqrt3x}$$
$$y'(0)=\sqrt3c_1-\sqrt3c_2=3$$

Question

I'm kinda stuck 
Solve the initial value problem 
$$y''+3y=0$$
$$y(0)=1$$
$$y'(0)=3$$
$$r^2+3=0$$
$$r=\sqrt3$$ or 
$$r=-\sqrt3$$
$$y=c_1e^{\sqrt3x}+c_2e{-\sqrt3x}$$
$$y(0)=c_1+c_2=1$$
$$y'=\sqrt3c_1e^{\sqrt3x}-\sqrt3c_2e{-\sqrt3x}$$
$$y'(0)=\sqrt3c_1-\sqrt3c_2=3$$

lgbasallote · Answer

you made a fatal mistake already...

lgbasallote · Answer

$$r^2 + 3 = 0$$
$$\implies r^2 = -3$$

anonymous · Answer

Oh my. It's an imaginary number

lgbasallote · Answer

yes

anonymous · Answer

Ok let's start over. Give me a minute

lgbasallote · Answer

sure. this isnt a pretty one

anonymous · Answer

It's not that ugly if you use Euler's identity :)

lgbasallote · Answer

everything in math is ugly :p lol

anonymous · Answer

$$y=c_1e^{i\sqrt3x}+c_2e^{-i\sqrt3x}$$
$$y(0)=c_1+c_2=0$$
for y' can I just take the derivative normally or do I have to do something special for the imaginary number?

lgbasallote · Answer

uhhh when you have $$r = a + bi$$
aren't you supposed to use
$$e^{ax} (c_1 \sin bx + c_2 \cos bx)$$
or something that looks like that..i can't remember this much

anonymous · Answer

I'm not too familiar with imaginary numbers myself...let's see what google has to say

anonymous · Answer

I think you're right

anonymous · Answer

brb

anonymous · Answer

Ok let's see. 
$$r_1, r_2: \alpha \pm i\beta$$
in our case would this be:
$$r_1=i\sqrt3$$
$$r_2=-i\sqrt3$$
$$y=e^x( c_1 cos\sqrt3x-c_2sin\sqrt3x)$$

anonymous · Answer

Nope. $\alpha=0$, so $e^{\alpha x}=1$.

anonymous · Answer

so $$y=( c_1 cos\sqrt3x-c_2sin\sqrt3x)$$

anonymous · Answer

Yeah, but you don't need (parentheses).

anonymous · Answer

THanks!

anonymous · Answer

yw

anonymous · Answer

How does this look?
$$y(0)=c_1cos\sqrt3(0)-c_2sin\sqrt3(0)=1$$