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

Question
Differential equations
Solve this equation pls $$\displaystyle{y}'+{x}{y}={e}^{{x}}$$
$$\displaystyle{y}{\left({0}\right)}={1}$$

2021-01-03
First, let us find the solution of the homogeneous equation
$$\displaystyle{y}'+{x}{y}={0}$$
We get,
$$\displaystyle{y}'+{x}{y}={0}$$
$$\displaystyle\Rightarrow{y}'=-{x}{y}$$
$$\displaystyle\Rightarrow\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=-{x}{y}$$
$$\displaystyle\Rightarrow\frac{{\left.{d}{y}\right.}}{{y}}={\left(-{x}\right)}{\left.{d}{x}\right.}$$
By integrating both sides, we get,
$$\displaystyle\int\frac{{\left.{d}{y}\right.}}{{y}}=\int{\left(-{x}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow{\ln{{\left({y}\right)}}}-\frac{{x}^{{2}}}{{2}}+{\ln{{\left({C}\right)}}}$$
$$\displaystyle\Rightarrow{\log{{\left({y}\right)}}}-{\log{{\left({C}\right)}}}=-\frac{{x}^{{2}}}{{2}}$$
$$\displaystyle\Rightarrow{\left(\frac{{y}}{{C}}\right)}=-\frac{{x}^{{2}}}{{2}}$$
$$\displaystyle\Rightarrow{y}={C}{e}^{{\frac{{x}^{{2}}}{{2}}}}$$
So, the general solution of the given differential equation is given by
$$\displaystyle{a}{e}^{{-\frac{{x}^{{2}}}{{2}}}}+{e}^{{-\frac{{x}^{{2}}}{{2}}}}\int\frac{{{e}^{{x}}}}{{{e}^{{-\frac{{x}^{{2}}}{{2}}}}}}{\left.{d}{x}\right.}$$

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