Find a general solution. Check your answer by substitution. y''+4y'+(π²+4)y=0

The characteristic equation is

$$\lambda ^2 + 4\lambda + \pi ^2 + 4 = 0.$$

Let's find the roots of the characteristic equation.

$$\begin{align*} \lambda ^2 + 4\lambda + \pi ^2 + 4 = 0 &\iff \lambda _{1/2} = \frac{-4 \pm \sqrt{4^2 -4(\pi ^2 + 4 )}}{2} \\ & \iff \lambda _{1/2} =\frac{-4 \pm 2\sqrt{\cancel{4} -\pi ^2 - \cancel{4} }}{2} = -2 \pm i\pi \\ & \iff \lambda _1 =-2 + i\pi, \quad \lambda _2 = -2 - i\pi \end{align*}$$

So, it has the complex conjugate roots:

$$\lambda _1 = -2 - i\pi (=a + ib)\quad \wedge \quad \lambda _2 = -2 - i\pi (=a-ib)$$

A basis is:

$$\begin{align*} \begin{cases} & \phi _1 (x) = e^{ax} \cos (bx) = e^{-2 x} \cos (\pi x) \\ &\phi _2 (x) =e^{ax} \sin (bx) = e^{-2 x} \sin (\pi x) \end{cases} \\ \end{align*}$$

The general solution is:

$$\begin{align*} y & = c_1 \phi _1 (x) + c_2 \phi _2 (x) \\ & = c_1e^{-2 x} \cos (\pi x) + c_2e^{-2 x} \sin (\pi x) \\ & = \boxed{{\color{#4257b2}{ e^{-2 x} (c_1\cos (\pi x) + c_2\sin (\pi x)) }}} \end{align*}$$

${\color{#c34632} {y''+4y'+(\pi^{2} +4)y=0}}\;\;\;\; \Rightarrow \;\; \textbf{Homogeneous 2nd Order D.E , With Constant Coeff.}$

$\Rightarrow \textbf{Form :}\;\; \boxed{\;ay''+by'+cy=0\;}$

$Where\; : \;a\;,\;b\;,\;c \;\; \Rightarrow Constants$

$In\; order\; to\; solve\; this\; D.E \;, \;we\; need \;to\; get\; the\; "characterestic\; equation\;" by\; replacing$

$\boxed{y'' \;\; \Rightarrow \;\; m^{2}\;\;\;,\;\;\; y'\;\; \Rightarrow \;\; m\;\;\;,\;\;\;y\;\; \Rightarrow \;\; 1}$

$\textbf{Then Characterestic Eq :}\;\; \Rightarrow \;\;{\color{#c34632} {m^{2}+4m+(\pi^{2} +4)=0}}$

$m_{1,2}=\dfrac{-b \;\pm \sqrt{b^{2}-4ac}}{2a}=\dfrac{-4\;\pm \sqrt{16-4\pi^{2} -16}}{2}$

$m_{1,2}=\dfrac{-4\;\pm 2\pi i}{2}$

$m_{1}=-2+\pi i\;\;\;,\;\;\;m_{2}=-2-\pi i\;\;\;\; \Rightarrow \;\; \boxed{\;m_{1,2}=\alpha \pm \beta i\;}$

$\therefore \;\; \textbf{Solution is : }\; \boxed{\color{#c34632} {y=e^{\alpha x}[C_{1}\cos(\beta x)+C_{2}\sin(\beta x)]}}\;\;\; \Rightarrow \;\; \textbf{For Different Complix Roots}$

$$\therefore \boxed{\;\color{#4257b2} {y=e^{-2 x}[C_{1}\cos(\pi x)+C_{2}\sin(\pi x)]}\;}$$