GATE AE 2021 • Q11 MCQ • 1 Mark Engineering Mathematics

Engineering Mathematics

Differential Equations

Subtopic: First Order ODEs

Question Statement:

Consider the differential equation $\frac{d^2 y}{dx^2} + 8\frac{dy}{dx} + 16y = 0$ with boundary conditions $y(0) = 1$ and $\frac{dy}{dx}(0) = 0$. The solution to this equation is:

Options:

  • (A) $y = (1 + 2x)e^{4x}$
  • (B) $y = (1 - 4x)e^{4x}$
  • (C) $y = (1 + 8x)e^{-4x}$
  • (D) $y = (1 + 4x)e^{-4x}$

Official Answer Key:

D

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2021 Q.11)** **1. Intuitive Physical Setup & Given Parameters**: We are given the 2nd-order homogeneous linear ODE with constant coefficients: \frac{d^2 y}{dx^2} + 8\frac{dy}{dx} + 16y = 0 subject to initial conditions $y(0) = 1$ and $y'(0) = 0$. **2. Step-by-Step Mathematical Derivation**: 1. **Characteristic Equation**: Assuming a solution of the form $y = e^{rx}$, the characteristic equation is: r^2 + 8r + 16 = 0 \implies (r + 4)^2 = 0 This yields real, repeated roots: $r_1 = r_2 = -4$. 2. **General Solution for Repeated Roots**: y(x) = (c_1 + c_2 x)e^{-4x} 3. **Apply Initial Conditions**: - At $x = 0$: $y(0) = (c_1 + 0)e^0 = c_1 = 1 \implies c_1 = 1$. - Differentiate to find $y'(x)$: y'(x) = c_2 e^{-4x} - 4(c_1 + c_2 x)e^{-4x} = [c_2 - 4c_1 - 4c_2 x]e^{-4x} - At $x = 0$: $y'(0) = c_2 - 4c_1 = 0 \implies c_2 = 4c_1 = 4(1) = 4$. 4. **Exact Particular Solution**: y(x) = (1 + 4x)e^{-4x} **3. Comprehensive Options & Distractor Analysis**: - **(A) y = (1 + 2x)e^(4x)**: ❌ **INCORRECT**. Wrong sign in exponent ($+4x$ instead of $-4x$) and incorrect coefficient $c_2 = 2$. - **(B) y = (1 - 4x)e^(4x)**: ❌ **INCORRECT**. Positive exponent $+4x$ and negative coefficient $-4$. - **(C) y = (1 + 8x)e^(-4x)**: ❌ **INCORRECT**. Erroneously takes the linear ODE coefficient (8) as $c_2$. - **(D) y = (1 + 4x)e^(-4x)**: ✅ **CORRECT**. Perfectly matches the repeated root characteristic solution and initial conditions. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **D** - **Exam Takeaway**: For repeated roots $r$, the second linearly independent basis solution is $x e^{rx}$. Watch the sign: $r^2 + 8r + 16 = (r+4)^2 = 0 \implies r = -4$.
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