## Related questions with answers

Find a general solution. Check your answer by substitution. y''+2πy'+π²y=0

Solutions

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

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

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

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

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

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

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

$m_{1,2}=\dfrac{-2\pi \;\pm 0}{2}$

$\boxed{\;m_{1,2}=-\pi\;}$

$\therefore \;\; \textbf{Solution is : }\; \boxed{\color{#c34632} {y=C_{1}e^{mx}+C_{2}xe^{mx}}}\;\;\; \Rightarrow \;\; \textbf{For Repeated Roots}$

$\therefore \boxed{\;\color{#4257b2} {y=C_{1}e^{-\pi x}+C_{2}xe^{-\pi x}}\;}$

The characteristic equation is

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

Let's find the roots of the characteristic equation.

$\lambda ^2 + 2\pi \lambda + \pi ^2 = 0 \iff (\lambda + \pi)^2 = 0$

So, it has the real double root:

$\lambda = -\pi$

A basis is:

$\begin{align*} \begin{cases} & \phi _1 (x) = e^{\lambda x} = e^{-\pi x} \\ & \phi _2 (x)= xe^{\lambda x} = xe^{-\pi x} \end{cases} \end{align*}$

The general solution is:

$\begin{align*} y & = c_1 \phi _1 (x) + c_2 \phi _2 (x) \\ & = c_1e^{-\pi x} + c_2xe^{-\pi x} \\ & = \boxed{{\color{#4257b2}{ e^{-\pi x} (c_1+ c_2x) }}} \end{align*}$

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