Question

If \( y = y(x) \) is the solution of the differential equation \(\frac{dy}{dx} + 2y = \sin(2x), \quad y(0) = \frac{3}{4},\)then \(y\left(\frac{\pi}{8}\right)\) is equal to:

• A. \( e^{-\pi/8} \)
• B. \( e^{-\pi/4} \)
• C. \( e^{\pi/4} \)
• D. \( e^{\pi/8} \)

Answer 1

The Correct Option is B

Given differential equation:

\[ \frac{dy}{dx} + 2y = \sin(2x), \quad y(0) = \frac{3}{4} \]

The integrating factor (I.F) is:

\[ \text{I.F} = e^{\int 2dx} = e^{2x} \]

Multiplying through by the integrating factor:

\[ ye^{2x} = \int e^{2x} \sin(2x) \, dx \]

To solve the integral, we use integration by parts:

\[ ye^{2x} = e^{2x} \left( \frac{2 \sin 2x - 2 \cos 2x}{4 + 4} \right) + C \]

\[ ye^{2x} = e^{2x} \left( \frac{\sin 2x - \cos 2x}{4} \right) + C \]

Using the initial condition \( y(0) = \frac{3}{4} \):

\[ \frac{3}{4} = \left( \frac{1}{4} (0 - 2) \right) + C \]

\[ \frac{3}{4} = -\frac{1}{4} + C \implies C = 1 \]

Thus, the solution is:

\[ y = \frac{\sin 2x - \cos 2x}{8} + e^{-2x} \]

To find \( y\left(\frac{\pi}{8}\right) \):

\[ y\left(\frac{\pi}{8}\right) = \frac{1}{8} \left( 2 \sin \frac{\pi}{4} - 2 \cos \frac{\pi}{4} \right) + e^{-\pi/4} \]

Since \( \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \): \[ y\left(\frac{\pi}{8}\right) = 0 + e^{-\pi/4} = e^{-\pi/4} \]