Question Bank • QB_EM_ADV_053
NAT • 2 Marks
Section 1: Engineering Mathematics
Differential Equations
Subtopic: Linear and non-linear first order Ordinary Differential Equations (ODE)
Question Statement:
Solve the initial value problem:
$\frac{dy}{dx} + 2x y = 2x, \quad y(0) = 4$
The value of $y(1)$ is ________ (round off to three decimal places).
Step-by-Step Solution & Explanation
**Method 1: Integrating Factor Method**
The equation is linear in standard form $\frac{dy}{dx} + P(x) y = Q(x)$ with $P(x) = 2x$ and $Q(x) = 2x$.
$\text{I.F.} = e^{\int 2x \, dx} = e^{x^2}$
Multiplying through by the integrating factor:
$\frac{d}{dx} \left( y e^{x^2} \right) = 2x e^{x^2}$
Integrating both sides:
$y e^{x^2} = \int 2x e^{x^2} \, dx = e^{x^2} + C \implies y(x) = 1 + C e^{-x^2}$
Applying the initial condition $y(0) = 4$:
$1 + C e^0 = 4 \implies C = 3$
Thus $y(x) = 1 + 3 e^{-x^2}$. At $x = 1$:
$y(1) = 1 + 3 e^{-1} = 1 + \frac{3}{2.71828} \approx 1 + 1.1036 = 2.1036 \approx 2.104$
**Method 2: Separation of Variables**
$\frac{dy}{dx} = 2x(1 - y) \implies \frac{dy}{1 - y} = 2x \, dx \implies -\ln|1 - y| = x^2 + K$
$1 - y = A e^{-x^2} \implies y = 1 - A e^{-x^2}$
With $y(0) = 4 \implies A = -3 \implies y(x) = 1 + 3e^{-x^2}$. Both methods yield $2.104$.
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