Question

Find general solution:

$\frac{dy}{dx}-y=(x-1)e^x$

Answer 1

Given,

$\frac{dy}{dx}-y=(x-1)e^x$

$\frac{dy}{dx}+Py=Q$

$P=-1\to I.F=e^{\int Pdx}=e^{\int (-1)dx}=e^{-x}$

$y\times I.F=\int Q\times I.Fdx$

$y\times e^{-x}=\int (x-1)e^x\times e^{-x}dx$

$y{e}^{-x}=\int (x-1)dx$

$y{e}^{-x}=\frac{x^2}{2}-x+c$