Question: Find the solution of the given initial value problem.
$$y''+5y'+3y=0, y(0)=1, y'(0)=0$$
Find roots of characteristic equation:
$$r = \frac{-5\pm\sqrt{5^2-4(1)(3)}}{2(1)}
= \frac{-5\pm\sqrt{13}}{2}$$
So the general solution is:
\begin{align}
y &= c_1e^{\frac{-5+\sqrt{13}}{2}}+
c_2e^{\frac{-5-\sqrt{13}}{2}}\notag\\
\implies y' &= \frac{-5+\sqrt{13}}{2}c_1e^{\frac{-5+\sqrt{13}}{2}}
- \frac{5+\sqrt{13}}{2}c_2e^{\frac{-5-\sqrt{13}}{2}}\notag
\end{align}
Plug in the given initial value:
\begin{align}
y(0)=1 &\implies c_1+c_2=1\notag\\
y'(0)=0 &\implies \frac{-5+\sqrt{13}}{2}c_1- \frac{5+\sqrt{13}}{2}c_2=0\notag
\end{align}
Solving this system for $c_1$, $c_2$, we get:
\begin{align}
c_1 &= \frac{5+\sqrt{13}}{2\sqrt{13}}\notag\\
c_2 &= 1-\frac{5+\sqrt{13}}{2\sqrt{13}}\notag
\end{align}
Therefore:
$$y= \frac{5+\sqrt{13}}{2\sqrt{13}}e^{\frac{-5+\sqrt{13}}{2}}+
\left(\frac{\sqrt{13}-5}{2\sqrt{13}} \right) e^{\frac{-5-\sqrt{13}}{2}}$$
