Question

\[\frac{dy}{dx} = x^5 \tan^{- 1} \left( x^3 \right)\]

Answer 1

We have,
\[\frac{dy}{dx} = x^5 \tan^{- 1} \left( x^3 \right)\]
\[ \Rightarrow dy = \left\{ x^5 \tan^{- 1} \left( x^3 \right) \right\}dx\]
Integrating both sides, we get
\[\int dy = \int x^5 \tan^{- 1} \left( x^3 \right)dx\]
\[ \Rightarrow y = \int x^5 \tan^{- 1} \left( x^3 \right)dx\]
\[\text{Putting }t = x^3 ,\text{ we get }\]
\[dt = 3 x^2 dx\]
\[ \therefore y = \frac{1}{3}\int t \tan^{- 1} t dt\]

\[ = \frac{1}{3}\left[ \tan^{- 1} t\int t dt - \int\left\{ \frac{d}{dt}\left( \tan^{- 1} t \right)\int t dx \right\}dt \right]\]
\[ = \frac{1}{3} \times \frac{t^2 \tan^{- 1} t}{2} - \frac{1}{6}\int\frac{t^2}{\left( 1 + t^2 \right)}dt\]
\[ = \frac{t^2 \tan^{- 1} t}{6} - \frac{1}{6}\int\frac{t^2 + 1 - 1}{\left( 1 + t^2 \right)}dt\]
\[ = \frac{t^2 \tan^{- 1} t}{6} - \frac{1}{6}\int dt + \frac{1}{6}\int\frac{1}{1 + t^2}dt\]
\[ = \frac{t^2 \tan^{- 1} t}{6} - \frac{1}{6}t + \frac{\tan^{- 1} t}{6} + C\]
\[ = \frac{x^6 \tan^{- 1} x^3}{6} - \frac{1}{6} x^3 + \frac{\tan^{- 1} x^3}{6} + C\]
\[ = \frac{1}{6}\left( x^6 \tan^{- 1} x^3 - x^3 + \tan^{- 1} x^3 \right) + C\]
\[\text{Hence, }y = \frac{1}{6}\left( x^6 \tan^{- 1} x^3 - x^3 + \tan^{- 1} x^3 \right) +\text{C is the solution to the given differential equation.}\]