$\left(1+{\mathit{x}}^2\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ + y = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ x
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ + $\frac{\mathit{y}}{1+{\mathit{x}}^2}$ = $\frac{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}{1+{\mathit{x}}^2}$ ... (i)
Given equation is linear with
So, I.F. = ${\mathit{e}}^{\int \frac{1}{1+{\mathit{x}}^2}\mathit{d}\mathit{x}}$ = ${\mathit{e}}^{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}$
Solution of (i)
$\mathit{y}{\mathit{e}}^{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}$ = ∫ ${\mathit{e}}^{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}\left(\frac{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}{1+{\mathit{x}}^2}\right)$ dx ... (ii)
For R.H.S,let $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ x = t ⇒ $\frac{1}{1+\mathit{t}{\mathit{x}}^2}$ dx = dt
By substituting in equation(ii)
$\mathit{y}{\mathit{e}}^{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}$ = ∫ ${\mathit{e}}^{\mathit{t}}$ . tdt
⇒ $\mathit{y}{\mathit{e}}^{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}$ = $\left[\mathit{t}{\mathit{e}}^{\mathit{t}}-{\mathit{e}}^{\mathit{t}}\right]$ + C
⇒ $\mathit{y}{\mathit{e}}^{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}$ = ${\mathit{e}}^{\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}$ $\left(\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}-1\right)$ + C
⇒ y = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}-1+\mathit{C}{\mathit{e}}^{-\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{x}}$