JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q73

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MathematicsSingle correct+4 / -1Question 73 of 90

If $y=y(x)$ satisfies the differential equation $x\,dy+\left(x\tan\left(\dfrac yx\right)-y\right)dx=0$ , x > 0, and $y(1)=\dfrac\pi2$ , then a value of $y(\sqrt2)$ is

\dfrac\pi{3\sqrt2}

\dfrac3{2\sqrt2}

\dfrac1{2\sqrt2}

\dfrac{3\pi}{2\sqrt2}

Answer and worked solution

Official answer: D

Homogeneous. With $y=vx$ : $v+xv'=v-\tan v$ , so $\cot v\,dv=-\frac{dx}x$ and $x\sin\frac yx=C$ ; $y(1)=\frac\pi2$ gives $C=1$ .

At $\sqrt2$ : $\sin\frac y{\sqrt2}=\frac1{\sqrt2}$ , so $\frac y{\sqrt2}=\frac\pi4$ or $\frac{3\pi}4$ : one value is $y=\frac{3\pi}{2\sqrt2}$ .

(D)

Topics:Homogeneous equations

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JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 73