Question

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

Answer 1

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
\(\displaystyle{a}{e}^{{-\frac{{x}^{{2}}}{{2}}}}+{e}^{{-\frac{{x}^{{2}}}{{2}}}}\int\frac{{{e}^{{x}}}}{{{e}^{{-\frac{{x}^{{2}}}{{2}}}}}}{\left.{d}{x}\right.}\)