Solve the linear equations by considering y as a function of x, that is, y = y(x). displaystylefrac{{{left.{d}{y}right.}}}{{{left.{d}{x}right.}}}-{y}={4}{e}^{x},{y}{left({0}right)}={4}

Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
${\displaystyle \frac{dy}{dx}=y\iff \frac{dy}{y}=dx}$
Now, the variables are separated, x appears only on the right side, and y only on the left.
Integrate the left side in relation to y, and the right side in relation to x
${\displaystyle \int \frac{dy}{y}=\int dx}$
which is${\displaystyle \ln \left|y\right|=x+c}$
By taking exponents, we obtain
${\displaystyle \left|y\right|=e^{x+c}=e^x\cdot e^c}$
Hence,we obtain
${\displaystyle y=C{e}^x}$
where ${\displaystyle C=\pm e^c\text{and}{y}_c=e^x}$ is the compleentary solution
Next, we need to find the particular solution $y_p$.
Therefore, we consider $u{y}_c$, and try to find u, a function of x, that will make this work.
Let's assume that $u{y}_c$. is a solution of the given equation. Hence, it satisfies the given equation. Substituting $u{y}_c$, and its derivative in the equation gives
${\displaystyle {\left(u{y}_c\right)}^{\prime }-u{y}_c=4e^x}$
${\displaystyle u^{\prime }{y}_c+u{y}_c^{\prime }-u{y}_c=4e^x}$
${\displaystyle u^{\prime }{y}_c+\underset{\text{=0 }}{\underbrace{u(y_c^{\prime }-y_c)}}=4e^x}$
since $y_c$ is a solution
Therefore
${\displaystyle u^{\prime }{y}_c=4e^x\Rightarrow u^{\prime }=\frac{4e^x}{y_c}}$
which gives
${\displaystyle u=\int \frac{e^x}{y_c}dx}$
Now, we can find the function u :
${\displaystyle u=\int \frac{4e^x}{e^x}dx}$
${\displaystyle =4\int dx}$
$=4x+c$
Since we need to find only one function that will male this work, we don’t need to introduce the constant of integration c. Hence,
$u=4x$
Recall that $y_p=u{y}_c$. Therefore,
${\displaystyle y_p=4x{e}^x}$
The general solution is
$y=C{y}_c+y_p$
${\displaystyle =e^x(C+4x)}$
Integrating Factor technique
This equation is linear with ${\displaystyle P\left(x\right)=-1\text{and}Q\left(x\right)=4e^x}$
Hence,
${\displaystyle h=\int Pdx=-\int dx=-x}$
So, an integrating factor is
${\displaystyle e^h=e^{-x}}$
and the general solution is
${\displaystyle y\left(x\right)=e^{-h}(c+\int Q{e}^{h}dx)}$