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

Answer: (B)

$e^{-\pi/4}$

Answer 2

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