Question: Find the general solution to the given equation: $4y''+9y=0$
Answer:
Assume $y=e^{rt}$ is a solution of this equation, then we can write $4r^2+9=0$
or $r^2=\frac{-9}{4}$
so we get $r=\pm\frac{3}{2}i$
Recall that if the roots are complex, we have them in the form of $\lambda\pm i\mu, \mu\neq 0$
in this case $\lambda=0$ and $\mu=\frac{3}{2}$
Then the general equation is given by the following:
$y=c_1e^{\lambda}cos(\mu t)+c_2e^{\lambda}sin(\mu t)$
$y=c_1e^0cos(\frac{3}{2}t)+ c_2e^0sin(\frac{3}{2}t)$
Therefore, the general solution of the given equation is the following:
$y=c_1cos(\frac{3}{2}t)+c_2sin(\frac{3}{2}t)$
