WBJEE Linear DE — practice questions

Explore 2 free Linear DE questions in WBJEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Linear DE questions and exercises with answers

Showing questions 1–2 of 2.

The general solution of $\dfrac{dy}{dx} + y = e^{-x}$ is:Read the answer and worked solution →The particular solution of $\dfrac{dy}{dx} + y\tan x = \sec x$ given $y(0) = 1$ is:Read the answer and worked solution →

Worked practice questions

The general solution of $\dfrac{dy}{dx} + y = e^{-x}$ is:

$y \cdot e^x = x + c$
$y = e^{-x}(x + c)$
Both A and B are equivalent
$y = x e^x + c$

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Correct answer: C. Both A and B are equivalent

Linear DE with $P = 1, Q = e^{-x}$. IF = $e^x$. $y \cdot e^x = \int e^{-x}\cdot e^x dx + c = x + c$. **So $y \cdot e^x = x + c$, equivalent to $y = e^{-x}(x + c)$.**

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The particular solution of $\dfrac{dy}{dx} + y\tan x = \sec x$ given $y(0) = 1$ is:

$y\sec x = \tan x + 1$
$y\sec x = \sec x + 1$
$y\sec x = \tan x - 1$
$y = \sin x + \cos x$

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Correct answer: A. $y\sec x = \tan x + 1$

**Linear:** $P = \tan x$, IF = $e^{\int \tan x\,dx} = e^{\log\sec x} = \sec x$. So $y\sec x = \int \sec x \cdot \sec x\,dx + c = \tan x + c$. At $x=0, y=1$: $1\cdot 1 = 0 + c$ ⇒ $c = 1$. **Hence $y\sec x = \tan x + 1$.**