Question

Solve the following differential equation:- \[\left( \cot^{- 1} y + x \right) dy = \left( 1 + y^2 \right) dx\]

Answer 1

The given differential equation is \[\left( \cot^{- 1} y + x \right) dy = \left( 1 + y^2 \right) dx\]
This differential equation can be written as\[\frac{dx}{dy} = \frac{\cot^{- 1} y + x}{1 + y^2} \]
\[ \Rightarrow \frac{dx}{dy} + \left( - \frac{1}{1 + y^2} \right)x = \frac{\cot^{- 1} y}{1 + y^2}\]
This is a linear differential equation with
\[P = - \frac{1}{1 + y^2}\text{ and }Q = \frac{\cot^{- 1} y}{1 + y^2}\]
\[I.F.=e^{- \int\frac{1}{1 + y^2}dy} = e^{cot^{- 1} y}\]
Multiply the differential equation by integration factor (I.F.), we get
\[\frac{dx}{dy} e^{cot^{- 1} y} - \frac{x}{\left( 1 + y^2 \right)} e^{cot^{- 1} y} = \frac{\cot^{- 1} y}{\left( 1 + y^2 \right)} e^{cot^{- 1} y} \]
\[\Rightarrow \frac{d}{dy}\left( x\ e^{cot^{- 1} y} \right) = \frac{\cot^{- 1} y}{\left( 1 + y^2 \right)} e^{cot^{- 1} y}\]
Integrating both sides with respect y, we get
\[x e^{cot^{- 1} y} = \int\frac{\cot^{- 1} y}{\left( 1 + y^2 \right)} e^{cot^{- 1} y} dy + C\]
Putting \[t = \cot^{- 1} y\]
\[dt = - \frac{1}{1 + y^2}dy\], we get

\[x e^{cot^{- 1} y} = - \int t e^t dt + C\]
\[ \Rightarrow x e^{cot^{- 1} y} = - e^t \left( t - 1 \right) + C\]
\[ \Rightarrow x e^{cot^{- 1} y} = e^{cot^{- 1}} y \left( 1 - \cot^{- 1} y \right) + C\]