2nd Order Differential Equation ElectroMagnetics

Question

I am trying to solve the 2nd differential equation shown in the picture.

I do not understand how the general solution has been formulated with cosine ore sine. Could any one please help? Thanks

Answer 1

The given differential equation is a linear homogeneous constant-coefficient differential equation. Solutions to $ax’’+bx’+cx=0$ are found by obtaining the roots of the characteristic polynomial $ar^2+br+c=0$. Let $r_1$ and $r_2$ be the roots of this polynomial, then the solution to the differential equation is $x=c_1e^{r_1t}+c_2e^{r_2t}$. If your roots are complex, then using Euler’s formula $e^{it}=\cos t+i\sin t$ you can transform the complex exponentials into sines and cosines.