OlmekinjP

2021-01-13

Solve differential equation $dy+5ydx={e}^{-5x}dx$

d2saint0

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ (1)
where P(x) or Q(x) are constants or function of x alone
Integrating factor of (1) is
$I.F.={e}^{\int P\left(x\right)dx}$
Required solution is
$y\left(I.F.\right)=\int Q\left(x\right)\left(I.F\right)dx+C$
$dy+\left(5y\right)dx={e}^{-5x}dx$
$⇒\frac{dy}{dx}+5y={e}^{-5x}\left(\ast \right)$
Since this equation is in the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ so it is linear ODE comparing $\left(\ast \right)$ with $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ we get $P\left(x\right)=5$, $Q\left(x\right)={e}^{-5x}$
So integrating factor is
$I.F.={e}^{\int P\left(x\right)dx}$
$={e}^{\int 5dx}$
$={e}^{5x}$
$y\left(I.F.\right)=\int Q\left(x\right)\left(I.F.\right)dx+c$
$⇒y\left({e}^{5x}\right)=\int \left({e}^{5x}\right)\left({e}^{5x}\right)dx+C$
$⇒y\left({e}^{5x}\right)=\int dx+C$
$=y\left({e}^{5x}\right)=x+C$
$⇒y=\frac{x+C}{{e}^{5x}}$

Jeffrey Jordon