Solve this equation pls y'+xy=e^x y(0)=1

Question

Solve this equation pls

y^{'} + x y = e^{x}

y (0) = 1

Answer 1

First, let us find the solution of the homogeneous equation
$y^{'} + x y = 0$
We get,
$y^{'} + x y = 0$
$\Rightarrow y^{'} = - x y$
$\Rightarrow \frac{d y}{d x} = - x y$
$\Rightarrow \frac{d y}{y} = (- x) d x$
By integrating both sides, we get,
$\int \frac{d y}{y} = \int (- x) d x$
$\Rightarrow \ln (y) - \frac{x^{2}}{2} + \ln (C)$
$\Rightarrow \log (y) - \log (C) = - \frac{x^{2}}{2}$
$\Rightarrow (\frac{y}{C}) = - \frac{x^{2}}{2}$
$\Rightarrow y = C e^{\frac{x^{2}}{2}}$
So, the general solution of the given differential equation is given by
$a e^{- \frac{x^{2}}{2}} + e^{- \frac{x^{2}}{2}} \int \frac{e^{x}}{e^{- \frac{x^{2}}{2}}} d x$