Question

Find the particular solution of the differential equation \( \frac{dy}{dx} = e^{2y} \cos x \), when \( x = \frac{\pi}{6} \), \( y = 0 \).

Hint:

Separate variables for exponential-trigonometric equations; apply initial conditions to find constants.

Answer 1

Separate variables: \[ e^{-2y} \, dy = \cos x \, dx. \] Integrate: \[ \int e^{-2y} \, dy = \int \cos x \, dx. \] \[ -\frac{1}{2} e^{-2y} = \sin x + c. \] \[ e^{-2y} = -2 \sin x - 2c = -2 \sin x + C, \quad y = -\frac{1}{2} \ln (-2 \sin x + C). \] Apply condition \( x = \frac{\pi}{6} \), \( y = 0 \): \[ \sin \frac{\pi}{6} = \frac{1}{2}, \quad e^{-2 \cdot 0} = 1 = -2 \cdot \frac{1}{2} + C \Rightarrow 1 = -1 + C \Rightarrow C = 2. \] \[ e^{-2y} = -2 \sin x + 2 = 2 (1 - \sin x), \quad y = -\frac{1}{2} \ln (2 (1 - \sin x)). \] Answer: \( y = -\frac{1}{2} \ln (2 (1 - \sin x)) \).