# Solve differential equationu'-5u=ve^(-5v)

Solve differential equation ${u}^{\prime }-5u=v{e}^{-5v}$

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Dora

$\frac{du}{dv}-5u=v{e}^{-5v}$
$\frac{dy}{dx}+Py=Q$
$I.F.={e}^{\int Pdx}$
$={e}^{-\int 5dv}$
$={e}^{-5\int dv}$
$={e}^{-5v}$
$u\left(I.F.\right)=\int v{e}^{-5v}\left(I.F.\right)dv+c$
$u{e}^{-5v}=\int v{e}^{-5v}{e}^{-5v}dv+c$
$u{e}^{-5v}=\int v{e}^{-10v}dv+c$ (1)
Now find $\int v{e}^{-10v}dv$ (by parts)
$=v\int {e}^{-10v}dv-\int \left[\frac{d}{dv}\left(v\right)\int {e}^{-10v}dv\right]dv$
$=v\frac{{e}^{-10v}}{-10}-\int \left[\frac{{e}^{-10v}}{-10}\right]dv$
$=\frac{v{e}^{-10v}}{-10}+\frac{1}{10}\frac{{e}^{-10v}}{-10}$ (2)
from (1) and (2)
$u{e}^{-5v}=\frac{v{e}^{-10v}}{-10}-\frac{1}{100}{e}^{-10v}+c$
$u=\left(\frac{v{e}^{-10v}}{-10}{e}^{-5v}\right)-\frac{1}{100}\frac{{e}^{-10v}}{{e}^{-5v}}+\frac{c}{{e}^{-5v}}$
$u=\frac{v{e}^{-5v}}{-10}-\frac{1}{100}{e}^{-5v}+c{e}^{5v}$

Jeffrey Jordon