\(\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x\)
\(\frac{d y}{d x} =\frac{\tan ^{-1} y-x}{1+y^{2}}=\frac{\tan ^{-1} y}{1+y^{2}}-\frac{x}{1+y^{2}}\)
\(\frac{d x}{d y}+\frac{1 \cdot x}{1+y^{2}} =\frac{\tan ^{-1} y}{1+y^{2}} ; p=\frac{1}{1+y^{2}}\)
\(\text {If }=e^{\int P d y}=\int_{e} \frac{1}{1+y^{2}} d y=e^{\tan ^{-1} y}\)
समीकरण का हल \(\Rightarrow x \times \mathrm{IF}=\int(\mathrm{Q} \times \mathrm{IF}) d y+c\)
\(x \times e^{\tan ^{-1} y}=\int\left[\frac{\tan ^{-1} y}{1+y^{2}} \times e \tan ^{-1} y\right] d y+c \) .....(i)
माना कि \(\mathrm{I}=\int \frac{\tan ^{-1} y}{1+y^{2}} \times e^{\tan ^{-1} y} d y\)
पुनः माना कि \(\tan ^{-1} y=t \)
\(\Rightarrow \frac{1}{1+y^{2}} d y=d t\)
तो \(\mathrm{I} =\int t e^{t} d t=t e^{t}-\int\left[\frac{d(t)}{d t} \int e^{t} d t\right] d t \)
\(=t e^{t}-e^{t}=e^{t}(t-1)=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)\)
समीकरण (i) से, \(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\tan ^{-1} y-1\right)+c\)
\(x=\left(\tan ^{-1} y-1\right)+c e^{-\tan ^{-1} y}\)
