JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q11

← JEE Main 2022 · 28 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 11 of 90

Let $y=y(x)$ be the solution of the differential equation $x\left(1-x^2\right)\dfrac{dy}{dx}+\left(3x^2y-y-4x^3\right)=0$ , $x>1$ , with $y(2)=-2$ . Then $y(3)$ is equal to

-18

-12

-6

-3

Answer and worked solution

Official answer: A

Linear form. $y'+\dfrac{3x^2-1}{x(1-x^2)}y=\dfrac{4x^2}{1-x^2}$ ; the integrating factor is $\dfrac1{x(1-x^2)}$ .

Integrate. $\dfrac y{x(1-x^2)}=\displaystyle\int\frac{4x}{(1-x^2)^2}dx=\frac2{1-x^2}+C$ .

Constants. $x=2$ , $y=-2$ : $\frac13=-\frac23+C$ , $C=1$ . At $x=3$ : $\frac y{-24}=-\frac14+1$ , $y=-18$ .

(A)

Topics:Linear first-order equations Partial fractions

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 11