Question

Find the particular solution of the differential equation \[\frac{dx}{dy} + x \cot y = 2y + y^2 \cot y, y ≠ 0\] given that x = 0 when \[y = \frac{\pi}{2}\].

Answer 1

We have,
\[\frac{dx}{dy} + x \cot y = 2y + y^2 \cot y . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dx}{dy} + Px = Q\]
\[\text{ where }P = \cot y\text{ and }Q = 2y + y^2 \cot y\]
\[ \therefore I . F . = e^{\int P\ dy} \]
\[ = e^{\int\cot y\ dy} \]
\[ = e^{log\left| \sin y \right|} = \sin y\]
\[\text{Multiplying both sides of } \left( 1 \right)\text{ by }I.F. = \sin y,\text{ we get }\]
\[\sin y\left( \frac{dx}{dy} + x \cot y \right) = \sin y\left( y^2 \cot y + 2y \right)\]
\[ \Rightarrow \sin y\frac{dx}{dy} + x \cos y = y^2 \cos y + 2y \sin y\]
Integrating both sides with respect to y, we get

\[ \Rightarrow x \sin y = y^2 \int\cos y\ dy - \int\left[ \frac{d}{dy}\left( y^2 \right)\int\cos y\ dy \right]dy + \int2y \sin y\ dy + C\]
\[ \Rightarrow x \sin y = y^2 \sin y - \int2y \sin y\ dy + \int2y\sin y\ dy + C\]
\[ \Rightarrow x \sin y = y^2 \sin y + C\]
Now, 
\[y = \frac{\pi}{2}\text{ at }x = 0\]
\[ \therefore 0 \times \sin \frac{\pi}{2} = \frac{\pi}{4}^2 \sin \frac{\pi}{2} + C\]
\[ \Rightarrow C = - \frac{\pi}{4}^2 \]
Putting the value of C, we get
\[x \sin y = y^2 \sin y - \frac{\pi}{4}^2 \]
\[\text{ Hence, }x \sin y = y^2 \sin y - \frac{\pi}{4}^2\text{ is the required solution.}\]