Consider the ordinary differential equation:
1
2
dy
dx +
y
x = 1. If y = 2/3 at x = 1,
then the value of y at x = 3 is ______ (rounded off to the nearest integer).
Official Answer Key:
2.0
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.43)**
**1. Intuitive Physical Setup & Given Parameters**:
We are given the first-order linear ordinary differential equation:
$\frac{1}{2}\frac{dy}{dx} + \frac{y}{x} = 1$
with initial condition:
$y(1) = \frac{2}{3}$
We need to find the value of $y$ at $x = 3$ (rounded off to nearest integer).
**2. Step-by-Step Mathematical Derivation**:
1. **Standard Form of Linear ODE**:
Multiply through by 2:
$\frac{dy}{dx} + \frac{2}{x}y = 2$
This is in standard form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \frac{2}{x}$ and $Q(x) = 2$.
2. **Integrating Factor ($I(x)$)**:
$I(x) = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2\ln x} = e^{\ln(x^2)} = x^2$
3. **General Solution**:
$y(x) \cdot I(x) = \int Q(x) I(x) dx + C$
$y \cdot x^2 = \int 2 x^2 dx + C = \frac{2}{3}x^3 + C$
$y(x) = \frac{2}{3}x + \frac{C}{x^2}$
4. **Apply Initial Condition $y(1) = 2/3$**:
$\frac{2}{3} = \frac{2}{3}(1) + \frac{C}{1^2} = \frac{2}{3} + C \implies C = 0$
Therefore, the particular solution is simply:
$y(x) = \frac{2}{3}x$
5. **Evaluate at $x = 3$**:
$y(3) = \frac{2}{3}(3) = 2$
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $y(3) = 2$.
- **Rounding Requirement**: Nearest integer $\implies \mathbf{2}$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **2.0** (Acceptable Range: **2 to 2**)
- **Exam Takeaway**: Integrating factor for $y' + \frac{n}{x}y$ is $x^n$. When $C = 0$, the solution reduces to the linear line $y = \frac{2}{3}x$.
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