# Solve the linear equations by considering y as a function of x, that is, y = y(x). y'+2y=4

Solve the linear equations by considering y as a function of x, that is, $$y = y(x)$$.
$$y'+2y=4$$

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Szeteib
Variation of parameters
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 $$\displaystyle{y}={C}{e}^{{-{2}{x}}}+{2}$$