GATE PH 2022 • Q19
MCQ • 1 Mark
Section 1: Mathematical Physics
Differential equations
Subtopic: First and Second Order ODEs
Question Statement:
Consider the ordinary differential equation $y'' - 2x y' + 4y = 0$ and its solution $y(x) = a + bx + cx^2$. Then
Options:
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(A) $a = 0, \; c = -2b \neq 0$
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(B) $c = -2a \neq 0, \; b = 0$
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(C) $b = -2a \neq 0, \; c = 0$
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(D) $c = 2a \neq 0, \; b = 0$
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.19)**
**1. Ordinary Differential Equation Formulation**:
- Given ODE:
$y'' - 2x y' + 4y = 0$
This is Hermite's differential equation $y'' - 2xy' + 2ny = 0$ with $n = 2$.
- Proposed polynomial solution:
$y(x) = a + bx + cx^2$
**2. Step-by-Step Substitution & Coefficient Matching**:
- Compute derivatives of $y(x)$:
$y'(x) = b + 2cx$
$y''(x) = 2c$
- Substitute into the ODE:
$2c - 2x(b + 2cx) + 4(a + bx + cx^2) = 0$
$2c - 2bx - 4cx^2 + 4a + 4bx + 4cx^2 = 0$
- Group by powers of $x$:
$(2c + 4a) + (-2b + 4b)x + (-4c + 4c)x^2 = 0$
$(2c + 4a) + 2bx + 0 = 0$
- For this identity to hold for all $x$:
- Linear term: $2b = 0 \implies b = 0$.
- Constant term: $2c + 4a = 0 \implies 2c = -4a \implies c = -2a \ne 0$.
- Thus, $c = -2a \ne 0$ and $b = 0$.
**3. Official Verification**:
The verified official IIT Master Answer is **Option (B)** (`c = -2a
e 0, b = 0`).
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