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

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 
${\displaystyle y{ }^{″}+f\left(x\right)y^{\prime }+g\left(x\right)y=0}$, you substitute ${\displaystyle y=\exp (-\int \frac{f\left(x\right)}{2}, dx )\mu }$. Divide through by ${\displaystyle x^2}$ to get
${\displaystyle y{ }^{″}+\underset{=f\left(x\right)}{\underbrace{\frac{3x+4}{x}}}{y}^{\prime }+\underset{=g\left(x\right)}{\underbrace{\frac{2(x^2+3x+1)}{x^2}}}y=0.}$
Then substitute in ${\displaystyle y=\exp (-\int \frac{3x+4}{2x}, dx )\mu =\frac{e^{-3\frac{x}{2}}}{x^2}\mu .}$
Therefore,
${\displaystyle \frac{e^{-3\frac{x}{2}}}{x^2}\mu { }^{″}-\frac{e^{-3\frac{x}{2}}}{4x^2}\mu =0}$.
Step 2
Divide through by $\frac{e^{-3\frac{x}{2}}}{x^2}$ to get
${\displaystyle \mu { }^{″}-\frac{1}{4}\mu =0.}$
Using standard methods in ODEs, we get
${\displaystyle \mu =c_1{e}^{-\frac{x}{2}}+c_2{e}^{\frac{x}{2}}}$.
Therefore, ${\displaystyle y=\frac{e^{-3\frac{x}{2}}}{x^2}\mu =\frac{e^{-3\frac{x}{2}}}{x^2}(c_1{e}^{-\frac{x}{2}}+c_2{e}^{\frac{x}{2}})}$.