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Question
Solve the following differential equation: \[\left( \cot^{- 1} y + x \right) dy = \left( 1 + y^2 \right) dx\] .
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Solution
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
\[x e^{\cot^{- 1}} y = - \int t e^t dt + C\]
\[ \Rightarrow x e^{\cot^{- 1}} y = - e^t \left( t - 1 \right) + C\]
\[ \Rightarrow x e^{\cot^{- 1}} y = e^{\cot^{- 1}} y \left( 1 - \cot^{- 1} y \right) + C\]
