Question Bank • QB_EM_ADV_005 NAT • 2 Marks

Section 1: Engineering Mathematics

Differential Equations

Subtopic: Higher order linear ODEs with constant coefficients

Question Statement:

For the differential equation $\frac{d^2y}{dx^2} + 4y = \sin(2x)$ with initial conditions $y(0) = 0$ and $y'(0) = 0$, the value of $y\left(\frac{\pi}{4}\right)$ is ________ (round off to two decimal places).

Answer Key:

0.13

Step-by-Step Solution & Explanation

1. Complementary Function (CF): Auxiliary equation: $m^2 + 4 = 0 \implies m = \pm 2i$ $y_c = C_1 \cos(2x) + C_2 \sin(2x)$ 2. Particular Integral (PI): Since $\sin(2x)$ is part of CF, resonance occurs: $y_p = \frac{1}{D^2 + 4} \sin(2x) = x \cdot \frac{1}{2D} \sin(2x) = -\frac{x \cos(2x)}{4}$ 3. General Solution: $y(x) = C_1 \cos(2x) + C_2 \sin(2x) - \frac{x \cos(2x)}{4}$ Applying $y(0) = 0 \implies C_1 = 0$. $y'(x) = 2C_2 \cos(2x) - \frac{\cos(2x)}{4} + \frac{2x \sin(2x)}{4}$ Applying $y'(0) = 0 \implies 2C_2 - \frac{1}{4} = 0 \implies C_2 = \frac{1}{8} = 0.125$. Thus, $y(x) = \frac{1}{8} \sin(2x) - \frac{x}{4} \cos(2x)$. At $x = \frac{\pi}{4}$: $y\left(\frac{\pi}{4}\right) = \frac{1}{8}\sin\left(\frac{\pi}{2}\right) - \frac{\pi/4}{4} \cos\left(\frac{\pi}{2}\right) = \frac{1}{8}(1) - 0 = 0.125 \approx 0.13$
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