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

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

The given differential equation is $x^2{y}^{\text{'}\text{'}}+(x+1)y^{\text{'}}-y=0$and the solution is $u=x+1$.

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

Consider the differential equation.

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

The solution of the equation is,

role="math" localid="1664340789299" $u=x+1$

So,

$$\begin{gathered} y=A(x+1) \\ {y}^{\text{'}}=A \\ {y}^{\text{'}\text{'}}=0 \end{gathered}$$

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

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

So, $u=x+1$is the solution of the given differential equation.

Let,

$$\begin{gathered} y=uv \\ =(x+1)v \end{gathered}$$So,

$$\begin{gathered} y^{\text{'}}=(x+1)v^{\text{'}}+v \\ {y}^{\text{'}\text{'}}=v^{\text{'}}+v^{\text{'}}+(x+1) \\ {v}^{\text{'}\text{'}}=(x+1)v^{\text{'}\text{'}}+2v^{\text{'}} \end{gathered}$$

The differential equation can now be written as,$$\begin{gathered} x^2\left((x+1)v^{\text{'}\text{'}}+2v^{\text{'}}\right)+(x+1)\left(v+(x+1)v^{\text{'}}\right)-(x+1)v=0 \\ {v}^{\text{'}}\left(3x^2+1+2x\right)+v^{\text{'}\text{'}}\left(x^3+x^2\right)=0 \\ \frac{v^{\text{'}\text{'}}}{v^{\text{'}}}=\frac{-\left(3x^2+1+2x\right)}{\left(x^3+x^2\right)} \\ \frac{d\left(v^{\text{'}}\right)}{v^{\text{'}}}=\frac{-\left(3x^2+1+2x\right)}{\left(x^3+x^2\right)} \end{gathered}$$

Write the expression in partial fraction form.

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

Integrate the above equation.

$$\begin{gathered} \ln v^{\text{'}}=-3\ln (x+1)+x^{-1}-\ln x+\ln (x+1)+\ln K \\ \ln v^{\text{'}}=\ln \left(K{(x+1)}^2{x}^{-1}\right)+\frac{1}{x} \\ {v}^{\text{'}}=\frac{K{e}^{1/x}}{\left(x{(x+1)}^2\right)} \\ \frac{dv}{dx}=\frac{K{e}^{1/x}}{\left(x{(x+1)}^2\right)} \end{gathered}$$

Integrate both side of the equation.

$v=-\frac{Kx{e}^{1/x}}{x+1}+C$

Consider$C=0$.

$v=\frac{-Kx{e}^{1/x}}{x+1}$

Thus,

$$\begin{gathered} y=uv \\ =-\frac{Kx{e}^{1/x}}{x+1}(x+1) \\ =-Kx{e}^{1/x} \end{gathered}$$

So, the second solution of the above equation is,

$y=x{e}^{1/x}.$

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