(a) Find Wronskian $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE
$$(x^{2}+1)y''-2xy'+2y=0.$$
Dividing but sides by $(x^{2}+1)$, we get
$$L[y]=y''-\frac{2x}{(x^{2}+1)}y'+\frac{2}{(x^{2}+1)}y=0,$$
where $p(x)=\frac{2x}{(x^{2}+1)}$, and $q(t)=\frac{2}{(x^{2}+1)}$.
By Abel's Theorem,
$$\begin{align}W(y_1,y_2)(x)&=c\exp(\int-{p(x)dx})\\&=c\exp(\int\frac{2x}{(x^{2}+1)}dx)\\&=ce^{\ln(x^{2}+1)}\\&=c(x^{2}+1).\end{align}$$
Let $c=1 \Rightarrow W(y_1,y_2)(x)=x^{2}+1$.
b) Check that $y_1(x)=x$ is a solution and find another linearly independent solution.
Since $y_1(x)=x \Rightarrow y_1 '(x)=1$, and $y_1 ''(x)=0$
Plugging $y_1$, $y_1 '$, and $y_1 ''$ into the ODE, we have
$$\begin{align}(x^{2}+1)\cdot 0-2x\cdot 1+2\cdot x&=0\\{-2x+2x}&={0}\end{align}$$
$y_1(x)$ satisfies the ODE $\Rightarrow$ $y_1(x)$ is a solution.
Given $y_1(x)$, we can find another linearly independent solution.
We know from the definition of the Wronskian that
$$W(y_1,y_2)(x)=y_1y_2 '-y_1 'y_2=xy_2 '-y_2$$
Equating the two expressions for the Wronskian, we get
$$xy_2 '-y_2=x^{2}+1$$
Dividing both sides by $x$, and multiplying by integrating factor $\mu=\frac{1}{x}$,
$$(\frac{1}{x}y_2)'=1+\frac{1}{x^2}$$
$$\frac{1}{x}y_2=\int{(1+\frac{1}{x^2})}dx+C$$
$$y_2(x)=x^{2}-1+Cx$$
$$y_2(x)=x^{2}-1$$
c)Write the general solution. Find solution such that $y(0)=1$, $y'(0)=1$
The general solution to the ODE is
$$y(x)=c_1 x+c_2(x^{2}-1).$$
$\Rightarrow y'(x)=c_1+2c_2 x$
$$1=c_1 \cdot 0+c_2(0^{2}-1)$$
$$1=c_1+2c_2 \cdot 0$$
$$\cases{c_1=1\\c_2=-1}$$
Thus, the solution that satisfies $y(0)=1$, $y'(0)=1$ is
$$y(x)=x-x^{2}+1.$$
