For the following problems, verify the given solution and then, by method (e) above, find a second solution of the given equation.

$\left(x^2+1\right)y^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0$

The given differential equation is$\left(x^2+1\right)y^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0$ and the solution is role="math" localid="1664344939379" $u=x$.

A differential equation is an equation that relates one or more unknown functions and their derivatives.

Consider the differential equation.

$\left(x^2+1\right)y^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0$

The solution of the equation is,

$u=x$

So,

$$\begin{gathered} y=Ax \\ {y}^{\text{'}}=A \\ {y}^{\text{'}\text{'}}=0 \end{gathered}$$

Consider the left-hand side of the given differential equation.

$$\begin{gathered} \left(x^2+1\right)y^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y \\ =\left(x^2+1\right)\left(0\right)-2x\left(A\right)+2Ax \\ =0-2Ax+2Ax \\ =0 \end{gathered}$$

So, $u=x$is the solution of the given differential equation.

Let,

$$\begin{gathered} y=uv \\ =xv \end{gathered}$$

So,

$$\begin{gathered} y^{\text{'}}=x{v}^{\text{'}}+v{y}^{\text{'}\text{'}} \\ =x{v}^{\text{'}\text{'}}+v^{\text{'}}+v^{\text{'}} \\ =x{v}^{\text{'}\text{'}}+2v^{\text{'}} \end{gathered}$$

The differential equation can now be written as,

$$\begin{gathered} \left(x^2+1\right)\left(x{v}^{\text{'}\text{'}}+2v^{\text{'}}\right)-2x\left(x{v}^{\text{'}}+v\right)+2xv=0 \\ {x}^3{v}^{\text{'}\text{'}}+x{v}^{\text{'}\text{'}}+2v^{\text{'}}=0 \\ {v}^{\text{'}\text{'}}\left(x^3+x\right)=-2v^{\text{'}} \\ \frac{v^{\text{'}\text{'}}}{v^{\text{'}}}=\frac{-2}{x\left(x^2+1\right)} \end{gathered}$$

Solve further,

$$\begin{gathered} \frac{d\left(v^{\text{'}}\right)}{v^{\text{'}}}=\frac{-2}{\left(x^3+x\right)} \\ \frac{d\left(v^{\text{'}}\right)}{v^{\text{'}}}=\frac{-2}{x}+\frac{2x}{\left(x^2+1\right)} \\ \int \frac{d{v}^{\text{'}}}{v}=\int \left(\frac{-2}{x}+\frac{2x}{x^2+1}\right)dx \end{gathered}$$

Solve further,

$$\begin{gathered} \ln v^{\text{'}}=-2\ln x+\ln \left(x^2+1\right)+\ln K \\ {v}^{\text{'}}=\frac{K\left(x^2+1\right)}{x^2} \\ \frac{dv}{dx}=\frac{K\left(x^2+1\right)}{x^2} \\ dv=\left(\frac{K\left(x^2+1\right)}{x^2}\right)dx \end{gathered}$$

Integrate both side of the equation.

$v=K\left(x-x^{-1}\right)+C$

Consider $C=0$.

$v=K\left(x-x^{-1}\right)$

Thus,

$$\begin{gathered} y=uv \\ =Kx\left(x-x^{-1}\right) \\ =-K\left(1-x^2\right) \end{gathered}$$

So, the second solution of the above equation is,

$y=1-x^2.$

Therefore, the proof that of the solution of the differential is stated above and the second solution is $y=1-x^2$.