Question

Find the particular solution of the differential equation: \[ \frac{dy}{dx} = e^{2y} \cos{x}, \quad \text{when} \quad x = \frac{\pi}{6}, \, y = 0 \]

Hint:

When solving differential equations, separate the variables and integrate both sides. Apply initial conditions to find the constant of integration.

Answer 1

Step 1: Separate the variables in the given differential equation: \[ \frac{dy}{dx} = e^{2y} \cos{x} \] Rearrange to separate the variables: \[ \frac{1}{e^{2y}} \, dy = \cos{x} \, dx \] Step 2: Integrate both sides: \[ \int \frac{1}{e^{2y}} \, dy = \int \cos{x} \, dx \] The left side: \[ \int \frac{1}{e^{2y}} \, dy = \int e^{-2y} \, dy = -\frac{1}{2} e^{-2y} \] The right side: \[ \int \cos{x} \, dx = \sin{x} \] So, we have: \[ -\frac{1}{2} e^{-2y} = \sin{x} + C \] Step 3: Apply the initial condition \( x = \frac{\pi}{6}, y = 0 \): \[ -\frac{1}{2} e^{0} = \sin{\frac{\pi}{6}} + C \] \[ -\frac{1}{2} = \frac{1}{2} + C \] \[ C = -1 \] Step 4: Substitute \( C = -1 \) into the equation: \[ -\frac{1}{2} e^{-2y} = \sin{x} - 1 \] \[ e^{-2y} = 2(1 - \sin{x}) \] Step 5: Take the natural logarithm: \[ -2y = \ln{(2(1 - \sin{x}))} \] \[ y = -\frac{1}{2} \ln{(2(1 - \sin{x}))} \] Thus, the particular solution is: \[ \boxed{y = -\frac{1}{2} \ln{(2(1 - \sin{x}))}} \]