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QJEEM20Mab3e9: JEE Main 2020 Mathematics - Class 12 JEEM20Mab3e9 Solution: The general solution of the differential equation + xy = 0 is : …

The general solution of the differential equation

1+x2+y2+x2y2\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} + xy dydx{{dy} \over {dx}} = 0 is :

(where C is a constant of integration)

Multiple Choice Options

  • A 1+y2+1+x2=12log⁡e(1+x2−11+x2+1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C
  • B 1+y2−1+x2=12log⁡e(1+x2−11+x2+1)+C\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C
  • C 1+y2+1+x2=12log⁡e(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C
  • D 1+y2−1+x2=12log⁡e(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C

Step-by-Step Verified Solution

1+x2+y2+x2y2\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} + xy dydx{{dy} \over {dx}} = 0

⇒\Rightarrow (1+x2)(1+y2)\sqrt {\left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)} + xy dydx{{dy} \over {dx}} = 0

⇒\Rightarrow (1+x2)(1+y2)\sqrt {\left( {1 + {x^2}} \right)} \sqrt {\left( {1 + {y^2}} \right)} = -xy dydx{{dy} \over {dx}}

⇒\Rightarrow ∫ydy1+y2=−∫1+x2xdx\int {{{ydy} \over {\sqrt {1 + {y^2}} }}} = - \int {{{\sqrt {1 + {x^2}} } \over x}} dx .....(1)

Now put 1 + x2 = u2 and 1 + y2 = v2

2xdx = 2udu and 2ydy = 2vdv

⇒\Rightarrow xdx = udu and ydy = vdv

Substitute these values in equation (1)

∫vdvv=−∫u2duu2−1\int {{{vdv} \over v}} = - \int {{{{u^2}du} \over {{u^2} - 1}}}

⇒\Rightarrow ∫dv=−∫u2−1+1u2−1du\int {dv} = - \int {{{{u^2} - 1 + 1} \over {{u^2} - 1}}} du

⇒\Rightarrow v = −∫(1+1u2−1)du- \int {\left( {1 + {1 \over {{u^2} - 1}}} \right)} du

⇒\Rightarrow v = -u - 12log⁡e∣u−1u+1∣{1 \over 2}{\log _e}\left| {{{u - 1} \over {u + 1}}} \right| + C

⇒\Rightarrow 1+y2\sqrt {1 + {y^2}} = −1+x2- \sqrt {1 + {x^2}} + 12log⁡e∣1+x2+11+x2−1∣{1 \over 2}{\log _e}\left| {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right| + C

⇒\Rightarrow 1+y2+1+x2=12log⁡e(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C

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