Find the particular solution of the following differential equation $\frac{dy}{dx} + y \cot x = \frac{2}{1 + \sin x}$, given that $y = 0$ when $x = \frac{\pi}{4}$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $x=\frac{\sin y - y \cos y + C}{y \sin y}$ ##
Given, $\frac{dx}{dy} = \frac{y \tan y - x \tan y - xy}{y \tan y}$
$\frac{dx}{dy} + \left( \frac{1}{y} + \frac{1}{\tan y} \right)x = 1$
$I.F. = e^{\int \left(\frac{1}{y} + \cot y\right) dy} = e^{\ln y + \ln \sin y}$
$I.F. = e^{\ln(y \sin y)} = y \sin y$
Solution of the D.E. is:
$x \times I.F. = \int (Q \times I.F.) dy$
$\Rightarrow xy \sin y = \int y \sin y \, dy$
$\Rightarrow xy \sin y = y(-\cos y) - \int (-\cos y) dy$
$\Rightarrow xy \sin y = -y \cos y + \sin y + C$
$\Rightarrow x = \frac{\sin y - y \cos y + C}{y \sin y}$