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

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

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

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

To verify the given solution$(u=x-1)$we need to find its derivatives:

$$\begin{gathered} d{u}^{\text{'}}=1 \\ {u}^{\text{'}\text{'}}=0 \end{gathered}$$

Substituting this into the equation we get:

$0-(x-1)1+(x-1)=0$

which is consistent and therefore$(u=x-1)$is a solution to this equation. To find the other solution we use method (e) from the chapter. This means we use substitution$y=uv=(x-1)v$where$v\left(x\right)$is a function ofrole="math" localid="1664340304642" $x$. Required derivatives are:

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

Inserting these into the equation we get:

$$\begin{gathered} x(x+1)\left[2v^{\text{'}}+(x-1)v^{\text{'}\text{'}}\right]-(x-1)\left[v+(x-1)v^{\text{'}}\right]+(x-1)v=0 \\ {v}^{\text{'}\text{'}}x(x-1)(x+1)+v^{\text{'}}\left[2x(x+1)-{(x-1)}^2\right]=0 \\ \frac{\text{d}{v}^{\text{'}}}{\text{d}x}=-\frac{x^2+4x-1}{x(x-1)(x+1)}{v}^{\text{'}} \end{gathered}$$

Inserting these into the equation we get:

$$\begin{gathered} x(x+1)\left[2v^{\text{'}}+(x-1)v^{\text{'}\text{'}}\right]-(x-1)\left[v+(x-1)v^{\text{'}}\right]+(x-1)v=0 \\ {v}^{\text{'}\text{'}}x(x-1)(x+1)+v^{\text{'}}\left[2x(x+1)-{(x-1)}^2\right]=0 \\ \frac{\text{d}{v}^{\text{'}}}{\text{d}x}=-\frac{x^2+4x-1}{x(x-1)(x+1)}{v}^{\text{'}} \\ \int \frac{\text{d}{v}^{\text{'}}}{v^{\text{'}}}=-\int \frac{x^2+4x-1}{x(x-1)(x+1)}\text{d}x \\ \ln v^{\text{'}}=-\int \frac{x^2+4x-1}{x(x-1)(x+1)}\text{d}x \end{gathered}$$

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

Using the exponential function on this equation we get:

$\frac{\text{d}v}{\text{d}x}=C_1\frac{{(x+1)}^2}{x{(x-1)}^2}$

$\int \text{d}v=v=C_1\int \frac{{(x+1)}^2}{x{(x-1)}^2}\text{d}x$

$v=C_1\left(\ln x-\frac{4}{x-1}\right)+C_2$

The final solution is therefore:

$y=(x-1)v=C_1\left(\right(x-1)\ln x-4)+C_2(x-1)$

where we recognize the before known solution$(u=x-1)$. The other solution is therefore:

$(x-1)\ln x-4$