First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives

\(\displaystyle{y}'=-{2}{y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=-{2}{y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{y}}=-{2}{\left.{d}{x}\right.}\)

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

\(\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=-{2}\int{\left.{d}{x}\right.}\)

Which is

\(\displaystyle \ln{{\left|{{y}}\right|}}=-{2}{x}+{c}\)

By taking exponents, we obtain

\(\displaystyle{\left|{{y}}\right|}={e}^{{-{2}{x}+{c}}}={e}^{{-{2}{x}}}\cdot{e}^{c}\)

Hence,we obtain

\(\displaystyle{y}={C}{e}^{{-{2}{x}}}\)

where \(\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}={e}^{{-{2}{x}}}\) is the complementary solution .

Next, we need to find the particular solution \(y_p\).

Therefore, we consider \(uy_c\), and try to find u, a function of x, that will make this work.

Let's assume that \(uy_c\) is a solution of the given equation. Hence, it satisfies the given equation.

Substituting \(uy_c\), and its derivative in the equation gives

\(\displaystyle{\left({u}{y}_{{c}}\right)}'+{2}{u}{y}_{{c}}={4}\)

\(\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}+{2}{u}{y}_{{c}}={4}\)

\(\displaystyle{u}'{y}_{{c}}+\underbrace{{{u}{\left({y}'_{{c}}+{2}{y}_{{c}}\right)}}}_{{\text{=0 }\ }}={4}\)

since \(y_c\) is a solution

Therefore,

\(\displaystyle{u}'{y}_{{c}}={4}\Rightarrow{u}'=\frac{4}{{{y}_{{c}}}}\)

which gives

\(\displaystyle{u}=\int\frac{4}{{{y}_{{c}}}}{\left.{d}{x}\right.}\)

Now, we can find the function u:

\(\displaystyle{u}=\int{\left(\frac{4}{{{e}^{{-{2}{x}}}}}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle={4}\int{e}^{{{2}{x}}}{\left.{d}{x}\right.}\)

\(\displaystyle={4}\cdot\frac{1}{{2}}{e}^{{{2}{x}}}+{c}_{{1}}\)

\(\displaystyle={2}{e}^{{{2}{x}}}+{c}_{{1}}\)

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,

\(\displaystyle{u}={2}{e}^{{{2}{x}}}\)

Recall that \(y_p=uy_c\)

Therefore

\(\displaystyle{y}_{{p}}={e}^{{-{2}{x}}}\cdot{2}{e}^{{{2}{x}}}\)

=2 The general solution is

\(y=Cy_c+y_p\)

\(\displaystyle={C}{e}^{{-{2}{x}}}+{2}\)

Integration Factor technique

This equation is linear with \(\displaystyle{P}{\left({x}\right)}={2}{\quad\text{and}\quad}{Q}{\left({x}\right)}={e}^{x}\)

Hence,

\(\displaystyle{h}=\int{2}{\left.{d}{x}\right.}={2}{x}\)

So, an integrating factor is

\(\displaystyle{e}^{h}={e}^{{{2}{x}}}\)

and the general solution is

\(\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}\)

\(\displaystyle={e}^{{-{2}{x}}}{\left({c}+\int{4}{e}^{{{2}{x}}}{\left.{d}{x}\right.}\right)}\)

\(\displaystyle={e}^{{-{2}{x}}}{\left({c}+{4}\cdot\frac{1}{{2}}{e}^{{{2}{x}}}\right)}\)

\(\displaystyle={c}{e}^{{-{2}{x}}}+{2}\)

Answer \(\displaystyle{y}={C}{e}^{{-{2}{x}}}+{2}\)