particular solution of the second-order linear equation I'm trying to find particular solution of the second-order linear equation but I can't find $y_{1}$ and $y_{2}$ according to $y = c_{1}y_{1} + c_{2}y_{2}$
$$x^{2}y^{''}-2xy^{'}+2y=0,   y(1) = 3, y'(1) = 1$$
If $r$ is used, $x^{2}r^{2}-2xr=-2$ then $xr(xr - 2) = -2$, I can't go on from there to find $y_{1}$ and $y_{2}$
 A: Assume $y=x^r$
then 
if $y=x^r$ then 
$y'=rx^{r-1}$ 
and 
$y''=r(r-1)x^{r-2}$. 
Substitute in 
$$x^{2}y^{''}-2xy^{'}+2y=0$$
to get
$$r(r-1)x^r -2r x^r+2x^r=0$$
divide by $y$ to get
$$r(r-1) -2r +2=(r-1)(r-2)=0$$
therefore 
$$y=ax + b x^2$$
Solve
$y(1) =a+b=3$
$y'(1) = a+2b=1$
to get
$$y=5x -2 x^2$$
A: $$y''(x)x^2+2y'(x)x+2y(x)=0\Longleftrightarrow$$

Assume a solution to this Euler-Cauchy equation will be proportional to $e^{\lambda}$ for some constant $\lambda$.
Substitute $y=x^{\lambda}$ into the differential equation:

$$x^2\frac{\text{d}^2}{\text{d}x^2}\left(x^{\lambda}\right)+2x\frac{\text{d}}{\text{d}x}\left(x^{\lambda}\right)+2x^{\lambda}=0\Longleftrightarrow$$

Substitute $\frac{\text{d}^2}{\text{d}x^2}\left(x^{\lambda}\right)=(\lambda-1)\lambda x^{\lambda-2}$ and $\frac{\text{d}}{\text{d}x}\left(x^{\lambda}\right)=\lambda x^{\lambda-1}$:

$$\lambda^2x^{\lambda}+\lambda x^{\lambda}+2x^{\lambda}=0\Longleftrightarrow$$
$$x^{\lambda}\left(\lambda^2+\lambda+2\right)=0\Longleftrightarrow$$

Assuming $x\ne0$, the zeros must come from the polynomial:

$$\lambda^2+\lambda+2=0\Longleftrightarrow$$
$$\lambda=-\frac{1}{2}\pm\frac{i\sqrt{7}}{2}$$

The roots $\lambda=-\frac{1}{2}\pm\frac{i\sqrt{7}}{2}$ give $y_1(x)=\text{C}_1x^{-\frac{1}{2}+\frac{i\sqrt{7}}{2}}$, $y_2(x)=\text{C}_2x^{-\frac{1}{2}-\frac{i\sqrt{7}}{2}}$ as solutions, 
where $\text{C}_1$ and $\text{C}_2$ are arbitrary constants.
The general solution is the sum of the above solutions:

$$y(x)=y_1(x)+y_2(x)=\text{C}_1x^{-\frac{1}{2}+\frac{i\sqrt{7}}{2}}+\text{C}_2x^{-\frac{1}{2}-\frac{i\sqrt{7}}{2}}$$

Using $x^{\lambda}=e^{\lambda\ln(x)}$, apply Euler's identity $e^{a+bi}=e^a\cos(b)+ie^a\sin(b)$:

$$y(x)=\frac{\left(\text{C}_1+\text{C}_2\right)\cos\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}+\frac{i\left(\text{C}_1-\text{C}_2\right)\sin\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}\Longleftrightarrow$$
$$y(x)=\frac{\text{C}_1\cos\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}+\frac{\text{C}_2\sin\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{\sqrt{x}}\Longleftrightarrow$$

Now using some algebra to find $\text{C}_1$ and $\text{C}_2$ with $y(1)=3$ and $y'(1)=1$:

$$y(x)=\frac{21\cos\left(\frac{1}{2}\sqrt{7}\ln(x)\right)+5\sqrt{7}\sin\left(\frac{1}{2}\sqrt{7}\ln(x)\right)}{7\sqrt{x}}$$
