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

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

Answers (1)

2021-03-12
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 \(\displaystyle{y}={C}{e}^{{-{2}{x}}}+{2}\)
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