$\textrm{Solve the differential equation}$
\begin{align*}\displaystyle
e^x\frac{dy}{dx}+5e^x y&=4, x>0 \\
&=
\end{align*}
confused about the $\frac{dy}{dx}$
one way to go about it is
$e^x \dfrac{dy}{dx} + 5 e^x y = 4$
$\dfrac {dy}{dx} + 5 y = 4e^{-x}$
solving the homogeneous equation
$\dfrac{dy}{dx} = -5y$
$y = c_1 e^{-5x}$
let $y_p = c_2 e^{-x}$
$-e^{-x}+5 c_2 e^{-x} = 4 e^{-x}$
$5c_2 - 1 = 4$
$c_2 = 1$
$y(x) = c_1 e^{-5x} + e^{-x}$
When the equation is first order linear, it's usually expected that the student should use an integrating factor.
$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} + 5\,y &= 4\,\mathrm{e}^{-x} \end{align*}$
The integrating factor is $\displaystyle \begin{align*} \mathrm{e}^{\int{5\,\mathrm{d}x}} = \mathrm{e}^{5\,x} \end{align*}$, so multiplying both sides by the integrating factor gives
$\displaystyle \begin{align*} \mathrm{e}^{5\,x}\,\frac{\mathrm{d}y}{\mathrm{d}x} + 5\,\mathrm{e}^{5\,x}\,y &= \mathrm{e}^{4\,x} \\ \frac{\mathrm{d}}{\mathrm{d}x}\,\left( \mathrm{e}^{5\,x}\,y \right) &= \mathrm{e}^{4\,x} \\ \mathrm{e}^{5\,x}\,y &= \int{ \mathrm{e}^{4\,x} \,\mathrm{d}x} \\ \mathrm{e}^{5\,x}\,y &= \frac{1}{4}\,\mathrm{e}^{4\,x} + C \\ y &= \frac{1}{4}\,\mathrm{e}^{-x} + C\,\mathrm{e}^{-5\,x} \end{align*}$
Yes, it is.
The first answer is correct, the second is not. Prove It is correct that the "integrating factor" is $\displaystyle e^{5x}$ but he lost a "4" when he multiplied both sides of the equation by that.