Find the particular solution of the following differential equation.
$\cos y \, dx + (1 + 2e^{-x}) \sin y \, dy = 0; \quad y(0) = \frac{\pi}{4}$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $e^x + 2 = 3\sqrt{2} \cos y$ ##
$\cos y \, dx + (1 + 2e^{-x}) \sin y \, dy = 0$
$\Rightarrow \int \frac{dx}{1 + 2e^{-x}} = \int \frac{-\sin y}{\cos y} dy$
$\Rightarrow \int \frac{e^x}{2 + e^x} dx = \int \frac{-\sin y}{\cos y} dy$
$\log (e^x + 2) = \log |\cos y| + \log C$
$\log (e^x + 2) = \log |\cos y \cdot C|$
$\Rightarrow e^x + 2 = C |\cos y|$
$\Rightarrow e^x + 2 = \pm C \cos y \Rightarrow e^x + 2 = k \cos y \quad \dots(i)$
Substituting $x = 0, y = \frac{\pi}{4}$ in $(i)$, we get
$1 + 2 = k \cos \frac{\pi}{4}$
$\Rightarrow k = 3\sqrt{2}$
$∴e^x + 2 = 3\sqrt{2} \cos y$
is the particular solution.