 # Solve the differential equation. x\frac{dy}{dx}-4y=2x^{4}e^{x} Jason Farmer 2021-09-16 Answered
Solve the differential equation.
$x\frac{dy}{dx}-4y=2{x}^{4}{e}^{x}$
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Step 1
According to the given information, it is required to solve the differential equation.
$x\frac{dy}{dx}-4y=2{x}^{4}{e}^{x}$
Step 2
First divide the whole differential equation by x to get the linear differential form.
$\frac{dy}{dx}-\frac{4y}{x}=2{x}^{3}{e}^{x}$
Step 3
Now, the general linear differential equation and its solution is:
$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
solution of differential equation is:
$y\left(x\right)×IF=\int Q\left(x\right).IFdx+c$
where $IF={e}^{\int P\left(x\right)dx}$
Step 4
Now, solve the given using the above definition.
In the given differential equation is:
$P\left(x\right)=\frac{-4}{x},Q\left(x\right)=2{x}^{3}{e}^{x}$
$IF={e}^{Pdx}={e}^{\int -\frac{4}{x}dx}={e}^{-4\mathrm{log}\left(x\right)}={e}^{\mathrm{log}\left({x}^{-4}\right)}={x}^{-4}$
$IF=\frac{1}{{x}^{4}}$
the solution of the given differential equation is:
$y\left(\frac{1}{{x}^{4}}\right)=\int 2{x}^{3}{e}^{x}\left(\frac{1}{{x}^{4}}\right)dx+c$
$\frac{y}{{x}^{4}}=2\int \frac{{e}^{x}}{x}dx+c$
$\frac{y}{{x}^{4}}=2{E}_{i}\left(x\right)+c\left(\int \frac{{e}^{x}}{x}dx={E}_{i}\left(x\right)+c\right)$
$y\left(x\right)=2{x}^{4}{E}_{i}\left(x\right)+c{x}^{4}$