GATE 2020 EE Q.16: solution and answer

GATE 2020 · EEQ.16 · 1 mark · NAT

Consider the initial value problem below. The value of y at $x = ln 2$ , (rounded off to 3 decimal places) is ________.

\frac{d y}{d x} = 2 x - y, y (0) = 1

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Answer and solution

ANSWER

$y (ln 2) = 2 ln 2 - 0.5 \approx 0.886$ .

Official answer: 0.8774 to 0.8952Source: Official answer key, GATE 2020 EE (IIT Delhi)

Rewrite as a first-order linear ODE in standard form $\frac{d y}{d x} + P (x) y = Q (x)$ : here $\frac{d y}{d x} + y = 2 x$ , so $P = 1$ , $Q = 2 x$ .
Integrating factor: $IF = e^{\int P d x} = e^{x}$ .
Solution rule $y \cdot IF = \int Q \cdot IF d x + C$ : $y e^{x} = \int 2 x e^{x} d x + C = 2 (x - 1) e^{x} + C$ (integration by parts: $\int x e^{x} d x = x e^{x} - e^{x}$ ).
Divide by $e^{x}$ to get the general solution: $y = 2 x - 2 + C e^{- x}$ .
Apply the initial condition $y (0) = 1$ : $1 = 0 - 2 + C \Rightarrow C = 3$ , so $y = 2 x - 2 + 3 e^{- x}$ . Check: $y^{'} = 2 - 3 e^{- x}$ and $2 x - y = 2 - 3 e^{- x}$ , so the ODE holds.
At $x = ln 2$ : $e^{- l n 2} = \frac{1}{2}$ , so $y (ln 2) = 2 ln 2 - 2 + \frac{3}{2} = 2 ln 2 - 0.5 = 1.386294 - 0.5 = 0.886294$
Rounded to 3 decimal places: $y (ln 2) \approx 0.886$ .

KEY IDEA

$\frac{d y}{d x} + y = 2 x$ is a first-order linear ODE. Use the integrating factor $e^{x}$ to get $y = 2 x - 2 + C e^{- x}$ . Then $y (0) = 1$ gives $C = 3$ , and $e^{- l n 2} = \frac{1}{2}$ makes the evaluation exact before you round.

WATCH OUT FOR

Using the initial condition before adding the constant $C$ , or dropping $C$ after integrating by parts. Getting the sign of the $- e^{x}$ term wrong in $\int x e^{x} d x = (x - 1) e^{x}$ . Writing $e^{- l n 2} = 2$ instead of $\frac{1}{2}$ gives $2 ln 2 + 4 \approx 5.386$ .

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GATE 2020 EE Q.16: solution and answer

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