JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q77

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MathematicsSingle correct+4 / -1Question 77 of 90

Let $y(x)$ be the solution of the differential equation
$(x\log x)\frac{dy}{dx}+y=2x\log x,\quad(x\ge1).$
Then $y(e)$ is equal to :

Answer and worked solution

Official answer: C

$\dfrac{dy}{dx}+\dfrac y{x\log x}=2$ . The integrating factor is $\log x$ :
$y\log x=\int2\log x\,dx=2(x\log x-x)+c.$
At $x=1$ : $0=-2+c$ , so $c=2$ . At $x=e$ : $y=2(e-e)+2=2$ .

Topics:Linear first-order equations

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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 77