GATE AG 2025 • Q43
NAT • 2 Marks
Section 1: Engineering Mathematics
Differential Equations
Subtopic: Linear and non-linear first order Ordinary Differential Equations (ODE)
Question Statement:
Find $y$ at $x = 0.3$, given $y(0) = 0$, for $\frac{dy}{dx} = 3y + 2e^x$. (Rounded to 3 decimal places)
Official Answer Key:
1.106 to 1.120
Step-by-Step Solution & Explanation
• **Governing Differential Equation:** $\frac{dy}{dx} - 3y = 2e^x$, which is a standard first-order linear ordinary differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = -3$ and $Q(x) = 2e^x$.
• **Integrating Factor (I.F.):** $\text{I.F.} = e^{\int -3\,dx} = e^{-3x}$.
• **General Solution:**
$y \cdot e^{-3x} = \int 2e^x \cdot e^{-3x}\,dx = 2 \int e^{-2x}\,dx = -e^{-2x} + C$
Multiplying both sides by $e^{3x}$ gives:
$y(x) = -e^x + C e^{3x}$
• **Boundary Condition:** $y(0) = 0 \implies 0 = -e^0 + C e^0 = -1 + C \implies C = 1$.
• **Particular Solution:** $y(x) = e^{3x} - e^x$.
• **Evaluation at $x = 0.3$:**
$y(0.3) = e^{3(0.3)} - e^{0.3} = e^{0.9} - e^{0.3} = 2.459603 - 1.349859 = 1.109744 \approx 1.110$
• **Conclusion:** Rounded to 3 decimal places, $y(0.3) = 1.110$ (Official range: 1.106 to 1.120).
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.
Launch CBT Mock Test