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 \[P = - \frac{1}{1 + y^2} \text { and } Q = \frac{\cot^{- 1} y}{1 + y^2}\].

\[e^{- \int\frac{1}{1 + y^2}dy} = e^{\cot^{- 1}} y\]

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

\[t = \cot^{- 1} y\] and \[dt = - \frac{1}{1 + y^2}dy\] we get

\[x e^{\cot^{- 1}} y = - \int t e^t dt + C\]

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

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