Solve the differential equation. x\frac{dy}{dx}-4y=2x^{4}e^{x}

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