Answer
y = $cos(\sqrt 3x) + \sqrt 3sin(\sqrt 3x)$
Work Step by Step
Question: y"+3y=0 y(0)=1 y'(0)=3
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(r^{2}+3)e^{rx}$=0
The corresponding characteristic equation is:
$r^{2}+3=0$
Solving the equation with complex numbers gives:
$r^{2}=-3$
$r=\frac{+}{-}\sqrt 3 i$
Adding the constants gives the following general solution:
y = $C_{1}cos(\sqrt 3x)$ + $C_{2}sin(\sqrt 3x)$
To solve the initial-value problem, we need to differentiate the general solution.
y'=$-\sqrt 3C_{1}sin(\sqrt 3x)$ + $\sqrt 3C_{2}cos(\sqrt 3x)$
Fill in the initial value problems in these general solutions.
$1 = C_{1}cos(\sqrt 3*0) + C_{2}sin(\sqrt 3*0)$ --> $C_{1}=1$
$3=-\sqrt 3C_{1}sin(\sqrt 3x) + \sqrt 3C_{2}cos(\sqrt 3x)$ --> $C_{2}=\sqrt 3$
The specific solution is:
y = $cos(\sqrt 3x) + \sqrt 3sin(\sqrt 3x)$
