# x^2y''+(3x^2+4x)y'+2(x^2+3x+1)y=0 I try to solve this equation by finding \mu such

${x}^{2}y{}^{″}+\left(3{x}^{2}+4x\right){y}^{\prime }+2\left({x}^{2}+3x+1\right)y=0$
I try to solve this equation by finding $\mu$ such that the equation become exact.( I know there are other ways for solving this equation).
${\left(\mu {y}^{\prime }\right)}^{\prime }={\mu }^{\prime }{y}^{\prime }+\mu y{}^{″}=\mu y{}^{″}+\mu \frac{3{x}^{2}+4x}{{x}^{2}}{y}^{\prime }+\mu \frac{2\left({x}^{2}+3x+1\right)}{{x}^{2}}y⇒$
${\mu }^{\prime }{y}^{\prime }=\mu \frac{3{x}^{2}+4x}{{x}^{2}}{y}^{\prime }+\mu \frac{2\left({x}^{2}+3x+1\right)}{{x}^{2}}y⇒\frac{{\mu }^{\prime }}{\mu }=\frac{3{x}^{2}+4x}{{x}^{2}}+\frac{2\left({x}^{2}+3x+1\right)}{{x}^{2}}\frac{y}{{y}^{\prime }}$
How am I supposed to solve it?
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Step 1
There is a method to solve equations like this one that uses a clever substitution. The idea is to use substitution to remove the y' term. Whenever you have an equation in the form of
, you substitute . Divide through by ${x}^{2}$ to get

Then substitute in
Therefore,
.
Step 2
Divide through by $\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}$ to get

Using standard methods in ODEs, we get
$\mu ={c}_{1}{e}^{-\frac{x}{2}}+{c}_{2}{e}^{\frac{x}{2}}$.
Therefore, $y=\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}\mu =\frac{{e}^{-3\frac{x}{2}}}{{x}^{2}}\left({c}_{1}{e}^{-\frac{x}{2}}+{c}_{2}{e}^{\frac{x}{2}}\right)$.

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