x²(x² – 1)\(\frac{dy}{d\mathrm x}\) + x(x² + 1)y = x² – 1

\(\Rightarrow \frac{dy}{d\mathrm x}+\frac{\mathrm x(\mathrm x^2+1)}{\mathrm x^2(\mathrm x^2-1)}y\) \(=\frac{\mathrm x^2-1}{\mathrm x^2(\mathrm x^2-1)}=\frac{1}{\mathrm x^2}\)

Which is a linear differential equation.

where, p \(=\frac{\mathrm x(\mathrm x^2+1)}{\mathrm x^2(\mathrm x^2-1)}\) \(=\frac{\mathrm x^2+1}{\mathrm x(\mathrm x^2-1)}\) \(=\frac{\mathrm x^2-1+2}{\mathrm x(\mathrm x^2-1)}\)

\(=\frac{1}{\mathrm x}+\frac{2}{\mathrm x(\mathrm x^2-1)}\)

\(=\frac{1}{\mathrm x}+\frac{2}{\mathrm{x(x-1)(x+1)}}\)

let \(\frac{2}{\mathrm x(\mathrm x+1)(\mathrm x+1)}\) \(=\frac{A}{\mathrm x}+\frac{B}{\mathrm x+1}+\frac{C}{\mathrm x-1}\)

Then A(x + 1)(x – 1) + Bx(x – 1) + Cx(x + 1) = 2

For x = 0, –A = 2 ⇒ A = –2

For x = 1, 2C = 2 ⇒ C = 1

For x = –1, 2B = 2 ⇒ B = 1

Hence, \(\frac{2}{\mathrm x(\mathrm x-1)(\mathrm x+1)}\) \(=\frac{-2}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\)

\(\therefore p=\frac{1}{\mathrm x}-\frac{2}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\) 

\(=-\frac{1}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\).

Now, I.F \(=e^{\int pd\mathrm x}\) \(=\int\limits_e\left(\frac{-1}{\mathrm x}+\frac{1}{\mathrm x+1}+\frac{1}{\mathrm x-1}\right)d\mathrm x\)

\(=e^{-\log \mathrm x+\log(\mathrm x+1)+\log(\mathrm x-1)}\)

\(=e^{\log\left(\frac{(\mathrm x+1)(\mathrm x-1)}{\mathrm x}\right)}\)

\(=\frac{\mathrm x^2-1}{\mathrm x}\)

\(=\left(\mathrm x-\frac{1}{\mathrm x}\right)\).

Now, complete solution of given linear differential equation is

\(y.\left(\mathrm x-\frac{1}{\mathrm x}\right)\) \(=\int 2\times \left(\mathrm x-\frac{1}{\mathrm x}\right)d\mathrm x\)

\(=\int \frac{1}{\mathrm x^2}\left(\mathrm x-\frac{1}{\mathrm x}\right)d\mathrm x\)

\(=\int \left(\frac{1}{\mathrm x}-\frac{1}{\mathrm x^3}\right)d\mathrm x\)

\(=\log \mathrm x + \frac{1}{2\mathrm x^2}+C\)

Hence, solution of given differential equation is \(y\left(\mathrm x-\frac{1}{\mathrm x}\right)\) \(=\log \mathrm x + \frac{1}{2\mathrm x^2}+C\).