 # 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}}} geduiwelh 2020-12-28 Answered
Solve the linear equations by considering y as a function of x, that is
$y=y\left(x\right)$
${y}^{\prime }-2xy={e}^{{x}^{2}}$
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Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
${y}^{\prime }=2xy⇔\frac{dy}{dx}=2xy⇔\frac{dy}{y}=2xdx$
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
$\int \frac{dy}{y}=2\int xdx$
Which is
$\mathrm{ln}|y|={x}^{2}+c$
By taking exponents, we obtain
$|y|={e}^{{x}^{2}+c}={e}^{{x}^{2}}\cdot {e}^{c}$
Hence,we obtain
$y=C{e}^{{x}^{2}}$
where $C\phantom{\rule{0.222em}{0ex}}=±{e}^{c}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{y}_{c}={e}^{{x}^{2}}$ is the complementary 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
${\left(u{y}_{c}\right)}^{\prime }+2xu{y}_{c}={e}^{{x}^{2}}$
${u}^{\prime }{y}_{c}+u{y}_{c}^{\prime }+2xu{y}_{c}={e}^{{x}^{2}}$

since ${y}_{c}$ is a solution
Therefore,
${u}^{\prime }{y}_{c}={e}^{{x}^{2}}⇒{u}^{\prime }=\frac{{e}^{{x}^{2}}}{{y}_{c}}$
which gives
$u=\int \frac{{e}^{{x}^{2}}}{{y}_{c}}dx$
Now, we can find the function u:
$u=\int \left(\frac{{e}^{{x}^{2}}}{{e}^{{x}^{2}}}\right)dx$
$=\int dx$
$=x+c$
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 ${y}_{p}=u{y}_{c}$
Therefore
${y}_{p}=x{e}^{{x}^{2}}$
The general solution is
$y=C{y}_{c}+{y}_{p}$
$=C{e}^{{x}^{2}}+x{e}^{{x}^{2}}$
$={e}^{{x}^{2}}\left(x+C\right)$
Integration Factor technique
This equation is linear with $P\left(x\right)=-2x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Q\left(x\right)={e}^{{x}^{2}}$
Hence,

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