How to solve this differential equation 2

Question

\begin{align} \ x\frac{dy}{dx}-4y &= x^6e^x \\\\ \end{align} By dividing on x we get :

\begin{align} \ \frac{dy}{dx}-4\frac{y}{x}=x^5e^x \end{align} Now I got the integrating factor"by e^integralOf(4/x) that's going to be : \begin{align} \ x^{-4} \end{align}

By multiply the I.F in every side then Integral I got :

\begin{align} \ \frac{-x^{-3}}{3} +4\frac{x^{-6}}{6}y=xe^x -x+C \end{align}

That's What I got , but the final answer in the source that I study from shows another final answer . What's the wrong in my solution ?

Note : The final answer in the source is :

\begin{align} \ y=x^4[xe^x-e^x]+C \end{align}

Answer 1

Basically, you multiply the ODE $\frac{dy}{dx} - \frac{4y}{x} = x^5e^x$ by the integrating factor $x^{-4}$ and then we get $$ x^{-4} \frac{dy}{dx} -4yx^{-5} = xe^x$$ which then reduces to $$ \frac{d(yx^{-4})}{dx} = xe^x$$ and so we integrate both sides $$C+yx^{-4} = \int{xe^x}\,dx$$ Then we use integration by parts $$\int{xe^x}\,dx = xe^x - \int{e^x}\,dx = xe^x - e^x + K$$ and so $$y = x^5e^x - x^4e^x +Lx^4$$

Answer 2

$$\frac{\mathrm dy}{\mathrm dx}-\frac{4y}{x}=x^5e^x \\\text{I.F.}=\exp\biggl({-\int 4\frac{\mathrm dx}{x}}\biggr)=\frac{1}{x^4}\\\implies y=x^4\int x e^x\mathrm dx$$

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In your case you made a mistake in step $3$, it should be $(yx^{-4})'=xe^x$. Can you proceed?