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JEE Main 2019 · Mathematics · Differential Equations · medium

If y=y(x) satisfies dy/dx=y/x+tan y/x for x>0 with y(1)=(π)/(6), then y(√2) equals

Single correct option

If y=y(x)y=y(x) satisfies dydx=yx+tan⁡yx\dfrac{dy}{dx}=\dfrac yx+\tan\dfrac yx for x>0x>0 with y(1)=π6y(1)=\dfrac\pi6, then y(2)y(\sqrt2) equals
  1. A. π4\dfrac{\pi}{4}
  2. B. 2 π2\dfrac{\sqrt2\,\pi}{2}
  3. C. 2 π4\dfrac{\sqrt2\,\pi}{4}
  4. D. π3\dfrac{\pi}{3}
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Answer: (C) 2 π4\dfrac{\sqrt2\,\pi}{4}

The equation is homogeneous. Put y=vxy=vx, so dydx=v+xdvdx\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}. Then v+xv′=v+tan⁡vv+xv'=v+\tan v, i.e. x dvdx=tan⁡vx\,\dfrac{dv}{dx}=\tan v. Separating variables, cot⁡v dv=dxx\cot v\,dv=\dfrac{dx}{x}, so ln⁡∣sin⁡v∣=ln⁡x+ln⁡C\ln|\sin v|=\ln x+\ln C, giving sin⁡yx=Cx\sin\dfrac yx=Cx. Using y(1)=π6y(1)=\dfrac\pi6: sin⁡π6=C\sin\dfrac\pi6=C, so C=12C=\dfrac12. Hence sin⁡yx=x2\sin\dfrac yx=\dfrac x2. At x=2x=\sqrt2: sin⁡y2=22\sin\dfrac{y}{\sqrt2}=\dfrac{\sqrt2}{2}, so y2=π4\dfrac{y}{\sqrt2}=\dfrac\pi4 and y=2 π4y=\dfrac{\sqrt2\,\pi}{4}. Check by direct substitution: with y=xsin⁡−1(x/2)y=x\sin^{-1}(x/2) we get y′=sin⁡−1(x/2)+x/21−x2/4=yx+tan⁡yxy'=\sin^{-1}(x/2)+\dfrac{x/2}{\sqrt{1-x^2/4}}=\dfrac yx+\tan\dfrac yx, and y(1)=sin⁡−1(1/2)=π6y(1)=\sin^{-1}(1/2)=\dfrac\pi6, so the particular solution is valid for 0<x<20<x<2.
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