Solve the linear equations by considering y as a function of x, that is y=y(x) displaystyle{y}'-{2}{x}{y}={e}^{{{x}^{2}}}

Solve the linear equations by considering y as a function of x, that is y=y(x) displaystyle{y}'-{2}{x}{y}={e}^{{{x}^{2}}}

Question
Solve the linear equations by considering y as a function of x, that is
\(y=y(x)\)
\(\displaystyle{y}'-{2}{x}{y}={e}^{{{x}^{2}}}\)

Answers (1)

2020-12-29
Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
\(\displaystyle{y}'={2}{x}{y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={2}{x}{y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{y}}={2}{x}{\left.{d}{x}\right.}\)
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
\(\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}={2}\int{x}{\left.{d}{x}\right.}\)
Which is
\(\displaystyle \ln{{\left|{{y}}\right|}}={x}^{2}+{c}\)
By taking exponents, we obtain
\(\displaystyle{\left|{{y}}\right|}={e}^{{{x}^{2}+{c}}}={e}^{{{x}^{2}}}\cdot{e}^{c}\)
Hence,we obtain
\(\displaystyle{y}={C}{e}^{{{x}^{2}}}\)
where \(\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}={e}^{{{x}^{2}}}\) 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
\(\displaystyle{\left({u}{y}_{{c}}\right)}'+{2}{x}{u}{y}_{{c}}={e}^{{{x}^{2}}}\)
\(\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}+{2}{x}{u}{y}_{{c}}={e}^{{{x}^{2}}}\)
\(\displaystyle{u}'{y}_{{c}}+\underbrace{{{u}{\left({y}'_{{c}}+{2}{x}{y}_{{c}}\right)}}}_{{\text{=0 }\ }}={e}^{{{x}^{2}}}\)
since \(y_c\) is a solution
Therefore,
\(\displaystyle{u}'{y}_{{c}}={e}^{{{x}^{2}}}\Rightarrow{u}'=\frac{{{e}^{{{x}^{2}}}}}{{{y}_{{c}}}}\)
which gives
\(\displaystyle{u}=\int\frac{{{e}^{{{x}^{2}}}}}{{{y}_{{c}}}}{\left.{d}{x}\right.}\)
Now, we can find the function u:
\(\displaystyle{u}=\int{\left(\frac{{{e}^{{{x}^{2}}}}}{{{e}^{{{x}^{2}}}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\int{\left.{d}{x}\right.}\)
\)=x+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=x
Recall that \(y_p=uy_c\)
Therefore
\(\displaystyle{y}_{{p}}={x}{e}^{{{x}^{2}}}\)
The general solution is
\(y=Cy_c+y_p\)
\(\displaystyle={C}{e}^{{{x}^{2}}}+{x}{e}^{{{x}^{2}}}\)
\(\displaystyle={e}^{{{x}^{2}}}{\left({x}+{C}\right)}\)
Integration Factor technique
This equation is linear with \(\displaystyle{P}{\left({x}\right)}=-{2}{x}{\quad\text{and}\quad}{Q}{\left({x}\right)}={e}^{{{x}^{2}}}\)
Hence,
\(\displaystyle{h}=-{2}\int{x}{\left.{d}{x}\right.}=-{x}^{2}\)
So, an integrating factor is
\(\displaystyle{e}^{h}={e}^{{-{x}^{2}}}\)
and the general solution is
\(\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{{{x}^{2}}}{\left({c}+\int{e}^{{{x}^{2}}}\cdot{e}^{{-{x}^{2}}}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{{{x}^{2}}}{\left({c}+\int{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{{{x}^{2}}}{\left({c}+{x}\right)}\)
Answer \(\displaystyle{y}={e}^{{{x}^{2}}}{\left({x}+{C}\right)}\)
0

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