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

Answer 2

Step 1: Recognize the type of differential equation
The ODE \( \dfrac{dy}{dx} + 2y = \sin(2x) \) is linear, first order, with constant coefficient. Use the integrating factor method with \( p(x)=2 \).

Step 2: Compute the integrating factor
\[ \mu(x)=e^{\int 2\,dx}=e^{2x}. \] Multiplying the ODE by \( e^{2x} \) gives \[ e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{2x}\sin(2x) \;\;\Longrightarrow\;\; \frac{d}{dx}\Big(y\,e^{2x}\Big)=e^{2x}\sin(2x). \]

Step 3: Integrate to obtain \( y \)
\[ y\,e^{2x}=\int e^{2x}\sin(2x)\,dx + C. \] Use the standard integral \[ \int e^{ax}\sin(bx)\,dx=e^{ax}\,\frac{a\sin(bx)-b\cos(bx)}{a^2+b^2}+C, \] with \( a=2,\; b=2 \). Then \[ \int e^{2x}\sin(2x)\,dx =e^{2x}\,\frac{2\sin(2x)-2\cos(2x)}{2^2+2^2} =e^{2x}\,\frac{2\sin(2x)-2\cos(2x)}{8} =e^{2x}\,\frac{\sin(2x)-\cos(2x)}{4}. \] Hence \[ y\,e^{2x}=e^{2x}\,\frac{\sin(2x)-\cos(2x)}{4}+C \;\;\Longrightarrow\;\; y(x)=\frac{\sin(2x)-\cos(2x)}{4}+C\,e^{-2x}. \]

Step 4: Apply the initial condition
Given \( y(0)=\tfrac{3}{4} \). Evaluate: \[ y(0)=\frac{\sin 0-\cos 0}{4}+C\,e^{0} =\frac{0-1}{4}+C =-\frac{1}{4}+C =\frac{3}{4} \;\;\Longrightarrow\;\; C=1. \] Therefore \[ y(x)=\frac{\sin(2x)-\cos(2x)}{4}+e^{-2x}. \]

Step 5: Evaluate at \( x=\tfrac{\pi}{8} \)
At \( x=\tfrac{\pi}{8} \), we have \( 2x=\tfrac{\pi}{4} \), so \( \sin\!\left(\tfrac{\pi}{4}\right)=\cos\!\left(\tfrac{\pi}{4}\right)=\tfrac{\sqrt{2}}{2} \). Hence \[ \frac{\sin(2x)-\cos(2x)}{4}\Bigg|_{x=\pi/8} =\frac{\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}}{4}=0, \] and \[ y\!\left(\tfrac{\pi}{8}\right)=e^{-2(\pi/8)}=e^{-\pi/4}. \]

Final answer

\( e^{-\pi/4} \)