JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q28

← JEE Main 2022 · 27 Jun · Shift 2

MathematicsInteger+4 / -1Question 28 of 90

Let $y=y(x)$ be the solution of the differential equation $\left(1-x^2\right)dy=\left(xy+\left(x^3+2\right)\sqrt{1-x^2}\right)dx$ , $-1<x<1$ , and $y(0)=0$ . If $\displaystyle\int_{-1/2}^{1/2}\sqrt{1-x^2}\,y(x)\,dx=k$ , then $k^{-1}$ is equal to ______.

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Answer and worked solution

Official answer: 320

Linear equation. $y'-\dfrac x{1-x^2}y=\dfrac{x^3+2}{\sqrt{1-x^2}}$ , integrating factor $\sqrt{1-x^2}$ .

Solve. $y\sqrt{1-x^2}=\dfrac{x^4}4+2x$ (using $y(0)=0$ ).

Integral. $k=\displaystyle\int_{-1/2}^{1/2}\left(\frac{x^4}4+2x\right)dx=\frac1{320}$ , so $k^{-1}=320$ .

Answer: 320

Topics:Linear first-order equations Properties of definite integrals

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JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 28