JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q25

← JEE Main 2022 · 26 Jun · Shift 1

MathematicsInteger+4 / -1Question 25 of 90

Let the solution curve $y=y(x)$ of the differential equation $\left(4+x^2\right)dy-2x\left(x^2+3y+4\right)dx=0$ pass through the origin. Then $y(2)$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 12

Linear form. $\dfrac{dy}{dx}-\dfrac{6x}{4+x^2}y=2x$ , integrating factor $(4+x^2)^{-3}$ .

Integrate. $\dfrac y{(4+x^2)^3}=\displaystyle\int\frac{2x\,dx}{(4+x^2)^3}=-\frac1{2(4+x^2)^2}+C$ ; $y(0)=0$ gives $C=\frac1{32}$ .

At $x=2$ . $y=8^3\left(\frac1{32}-\frac1{128}\right)=512\cdot\frac3{128}=12$ .

Answer: 12

Topics:Linear first-order equations

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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 25