Step 1
According to the given information, it is required to solve the differential equation.
\(\displaystyle{x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{4}{y}={2}{x}^{{{4}}}{e}^{{{x}}}\)
Step 2
First divide the whole differential equation by x to get the linear differential form.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{\frac{{{4}{y}}}{{{x}}}}={2}{x}^{{{3}}}{e}^{{{x}}}\)
Step 3
Now, the general linear differential equation and its solution is:
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{\left({x}\right)}{y}={Q}{\left({x}\right)}\)
solution of differential equation is:
\(\displaystyle{y}{\left({x}\right)}\times{I}{F}=\int{Q}{\left({x}\right)}.{I}{F}{\left.{d}{x}\right.}+{c}\)
where \(\displaystyle{I}{F}={e}^{{\int{P}{\left({x}\right)}{\left.{d}{x}\right.}}}\)
Step 4
Now, solve the given using the above definition.
In the given differential equation is:
\(\displaystyle{P}{\left({x}\right)}={\frac{{-{4}}}{{{x}}}},{Q}{\left({x}\right)}={2}{x}^{{{3}}}{e}^{{{x}}}\)
\(\displaystyle{I}{F}={e}^{{{P}{\left.{d}{x}\right.}}}={e}^{{\int-{\frac{{{4}}}{{{x}}}}{\left.{d}{x}\right.}}}={e}^{{-{4}{\log{{\left({x}\right)}}}}}={e}^{{{\log{{\left({x}^{{-{4}}}\right)}}}}}={x}^{{-{4}}}\)
\(\displaystyle{I}{F}={\frac{{{1}}}{{{x}^{{{4}}}}}}\)
the solution of the given differential equation is:
\(\displaystyle{y}{\left({\frac{{{1}}}{{{x}^{{{4}}}}}}\right)}=\int{2}{x}^{{{3}}}{e}^{{{x}}}{\left({\frac{{{1}}}{{{x}^{{{4}}}}}}\right)}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{\frac{{{y}}}{{{x}^{{{4}}}}}}={2}\int{\frac{{{e}^{{{x}}}}}{{{x}}}}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{\frac{{{y}}}{{{x}^{{{4}}}}}}={2}{E}_{{{i}}}{\left({x}\right)}+{c}{\left(\int{\frac{{{e}^{{{x}}}}}{{{x}}}}{\left.{d}{x}\right.}={E}_{{{i}}}{\left({x}\right)}+{c}\right)}\)
\(\displaystyle{y}{\left({x}\right)}={2}{x}^{{{4}}}{E}_{{{i}}}{\left({x}\right)}+{c}{x}^{{{4}}}\)
According to the given information, it is required to solve the differential equation.
\(\displaystyle{x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{4}{y}={2}{x}^{{{4}}}{e}^{{{x}}}\)
Step 2
First divide the whole differential equation by x to get the linear differential form.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{\frac{{{4}{y}}}{{{x}}}}={2}{x}^{{{3}}}{e}^{{{x}}}\)
Step 3
Now, the general linear differential equation and its solution is:
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{\left({x}\right)}{y}={Q}{\left({x}\right)}\)
solution of differential equation is:
\(\displaystyle{y}{\left({x}\right)}\times{I}{F}=\int{Q}{\left({x}\right)}.{I}{F}{\left.{d}{x}\right.}+{c}\)
where \(\displaystyle{I}{F}={e}^{{\int{P}{\left({x}\right)}{\left.{d}{x}\right.}}}\)
Step 4
Now, solve the given using the above definition.
In the given differential equation is:
\(\displaystyle{P}{\left({x}\right)}={\frac{{-{4}}}{{{x}}}},{Q}{\left({x}\right)}={2}{x}^{{{3}}}{e}^{{{x}}}\)
\(\displaystyle{I}{F}={e}^{{{P}{\left.{d}{x}\right.}}}={e}^{{\int-{\frac{{{4}}}{{{x}}}}{\left.{d}{x}\right.}}}={e}^{{-{4}{\log{{\left({x}\right)}}}}}={e}^{{{\log{{\left({x}^{{-{4}}}\right)}}}}}={x}^{{-{4}}}\)
\(\displaystyle{I}{F}={\frac{{{1}}}{{{x}^{{{4}}}}}}\)
the solution of the given differential equation is:
\(\displaystyle{y}{\left({\frac{{{1}}}{{{x}^{{{4}}}}}}\right)}=\int{2}{x}^{{{3}}}{e}^{{{x}}}{\left({\frac{{{1}}}{{{x}^{{{4}}}}}}\right)}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{\frac{{{y}}}{{{x}^{{{4}}}}}}={2}\int{\frac{{{e}^{{{x}}}}}{{{x}}}}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{\frac{{{y}}}{{{x}^{{{4}}}}}}={2}{E}_{{{i}}}{\left({x}\right)}+{c}{\left(\int{\frac{{{e}^{{{x}}}}}{{{x}}}}{\left.{d}{x}\right.}={E}_{{{i}}}{\left({x}\right)}+{c}\right)}\)
\(\displaystyle{y}{\left({x}\right)}={2}{x}^{{{4}}}{E}_{{{i}}}{\left({x}\right)}+{c}{x}^{{{4}}}\)
