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

$x^2(2-x)y^{\text{'}\text{'}}+2x{y}^{\text{'}}-2y=0,u=x$

The given differential equation is $x^2(2-x)y^{\text{'}\text{'}}+2x{y}^{\text{'}}-2y=0$and the solution isrole="math" localid="1664346050732" $u=x$

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

Consider the differential equation.

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

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.

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

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,

$$\begin{gathered} x^2(2-x)\left(x{v}^{\text{'}\text{'}}+2v^{\text{'}}\right)+2x\left(x{v}^{\text{'}}+v\right)-2xv=0 \\ 2x^3{v}^{\text{'}\text{'}}+6x^2{v}^{\text{'}}-x^4{v}^{\text{'}\text{'}}-2x^3{v}^{\text{'}}=0 \\ \left(2x^3-x^4\right)v^{\text{'}\text{'}}=\left(2x^3-6x^2\right)v^{\text{'}} \\ \frac{v^{\text{'}\text{'}}}{v^{\text{'}}}=\frac{2x^2(x-3)}{x^3(2-x)} \end{gathered}$$

Solve further,

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

Solve further,

$$\begin{gathered} \ln v^{\text{'}}=-3\ln x+\ln (2-x)+\ln K \\ {v}^{\text{'}}=\frac{K(2-x)}{x^3} \\ \frac{dv}{dx}=\frac{K(2-x)}{x^3} \\ dv=\left(\frac{K(2-x)}{x^3}\right)dx \end{gathered}$$

Integrate both side of the equation.

$v=K\left(\frac{-1}{x^2}+\frac{1}{x}\right)+C$

Consider$C=0$.

$v=K\left(-\frac{1}{x^2}+\frac{1}{x}\right)$

Thus,

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

So, the second solution of the above equation is,

$y=x^{-1}-1.$

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