Verified Solution • 2025–26 Session moderate

QJEEM20Maaf9f: JEE Main 2020 Mathematics - Class 12 JEEM20Maaf9f Solution: If ; y(1) = 1; then a value of x satisfying y(x) = e is :

If dydx=xyx2+y2{{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}} ; y(1) = 1; then a value of x satisfying y(x) = e is :

Multiple Choice Options

  • A 2e\sqrt 2 e
  • B 123e{1 \over 2}\sqrt 3 e
  • C e2{e \over {\sqrt 2 }}
  • D 3e\sqrt 3 e

Step-by-Step Verified Solution

dydx=xyx2+y2{{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}

Let y = vx

∴\therefore dydx{{dy} \over {dx}} = v + x. dvdx{{dv} \over {dx}}

⇒\Rightarrow v + x. dvdx{{dv} \over {dx}} = x(vx)x2+v2x2{{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}} = v1+v2{v \over {1 + {v^2}}}

⇒\Rightarrow x. dvdx{{dv} \over {dx}} = v1+v2{v \over {1 + {v^2}}} - v = v−v−v31+v2{{v - v - {v^3}} \over {1 + {v^2}}} = −v31+v2- {{{v^3}} \over {1 + {v^2}}}

⇒\Rightarrow ∫1+v2v3dv=−∫dxx\int {{{1 + {v^2}} \over {{v^3}}}} dv = - \int {{{dx} \over x}}

⇒\Rightarrow −12v2+log⁡v- {1 \over {2{v^2}}} + \log v = −log⁡x+C- \log x + C

⇒\Rightarrow −12x2y2+log⁡(yx)- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right) = −log⁡x+C- \log x + C ......(1)

putting x = 1, y = 1 we get

⇒\Rightarrow C = −12- {1 \over 2}

From eq. (1)

−12x2y2+log⁡(yx)- {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right) = −log⁡x−12- \log x - {1 \over 2}

Put y = e

−12x2e2+log⁡(ex)- {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right) = −log⁡x−12- \log x - {1 \over 2}

⇒\Rightarrow x2 = 3e2

⇒\Rightarrow x = ±\pm 3 e\sqrt e

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