The general solution of the differential equation 1 + x ^ 2 + y ^ 2 + x ^ 2 y ^ 2 + xy(dy)/(dx) = 0 (where C is a constant of integration)

Correct answer: C. 1 + y ^ 2 + 1 + x ^ 2 = (1)/(2) _ e ( 1 + x ^ 2 + 1 1 + x ^ 2 - 1 ) + C (1 + x ^ 2 )(1 + y ^ 2 ) + xy(dy)/(dx) = 0 Integrating, ⇒ ∫ ( 2y 2 1 + y ^ 2 )dy = - ∫ ( 1 + x ^ 2 x 1 + x ^ 2 )dx ⇒ 1 + y ^ 2 = - ∫ x 1 + x ^ 2 dx - ∫ x x ^ 2 1 + x ^ 2 dx Put 1 + x ^ 2 = t ⇒ x ^ 2 = t ^ 2 - 1 to solve RHS 2nd integration, ⇒ 1 + y ^ 2 = - 1 + x ^ 2 - ∫ t ( t ^ 2 - 1)t dt ⇒ 1 + y ^ 2 + 1 + x ^ 2 = - ∫ 1 t ^ 2 - 1 dt ⇒ 1 + y ^ 2 + 1 + x ^ 2 = (1)/(2) [ 1 + x ^ 2 + 1 1 + x ^ 2 - 1 ] + C

JEE Main 20206 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 6 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2020 · 6 September · Morning Shift · Mathematics · Differential Equations

The general solution of the differential equation 1 + x ^ 2 + y ^ 2...

The general solution of the differential equation

\sqrt{1 + {x}^{2} + {y}^{2} + {x}^{2}{y}^{2}} + xy\dfrac{dy}{dx} = 0

(where C is a constant of integration)

Options

A $\sqrt{1 + {y}^{2}} + \sqrt{1 + {x}^{2}} = \dfrac{1}{2}\log_{e}\left(\dfrac{\sqrt{1 + {x}^{2}} - 1}{\sqrt{1 + {x}^{2}} + 1}\right) + C$
B $\sqrt{1 + {y}^{2}} - \sqrt{1 + {x}^{2}} = \dfrac{1}{2}\log_{e}\left(\dfrac{\sqrt{1 + {x}^{2}} - 1}{\sqrt{1 + {x}^{2}} + 1}\right) + C$
C $\sqrt{1 + {y}^{2}} + \sqrt{1 + {x}^{2}} = \dfrac{1}{2}\log_{e}\left(\dfrac{\sqrt{1 + {x}^{2}} + 1}{\sqrt{1 + {x}^{2}} - 1}\right) + C$
D $\sqrt{1 + {y}^{2}} - \sqrt{1 + {x}^{2}} = \dfrac{1}{2}\log_{e}\left(\dfrac{\sqrt{1 + {x}^{2}} + 1}{\sqrt{1 + {x}^{2}} - 1}\right) + C$

Correct answer

C. $\sqrt{1 + {y}^{2}} + \sqrt{1 + {x}^{2}} = \dfrac{1}{2}\log_{e}\left(\dfrac{\sqrt{1 + {x}^{2}} + 1}{\sqrt{1 + {x}^{2}} - 1}\right) + C$

Step-by-step solution

$\sqrt{\left(1 + {x}^{2}\right)\left(1 + {y}^{2}\right)} + xy\dfrac{dy}{dx} = 0$

Integrating,

$⇒ ∫ \left(\dfrac{2y}{2\sqrt{1 + {y}^{2}}}\right)dy = - ∫ \left(\dfrac{1 + {x}^{2}}{x\sqrt{1 + {x}^{2}}}\right)dx$
$⇒ \sqrt{1 + {y}^{2}} = - ∫ \dfrac{x}{\sqrt{1 + {x}^{2}}}dx - ∫ \dfrac{x}{{x}^{2}\sqrt{1 + {x}^{2}}}dx$

Put $\sqrt{1 + {x}^{2}} = t ⇒ {x}^{2} = {t}^{2} - 1$ to solve RHS 2nd integration,

$⇒ \sqrt{1 + {y}^{2}} = - \sqrt{1 + {x}^{2}} - ∫ \dfrac{t}{\left({t}^{2} - 1\right)t}dt$
$⇒ \sqrt{1 + {y}^{2}} + \sqrt{1 + {x}^{2}} = - ∫ \dfrac{1}{{t}^{2} - 1}dt$
$⇒ \sqrt{1 + {y}^{2}} + \sqrt{1 + {x}^{2}} = \dfrac{1}{2}\ln \left[\dfrac{\sqrt{1 + {x}^{2}} + 1}{\sqrt{1 + {x}^{2}} - 1}\right] + C$

Practice Differential Equations on Quantrex Academy →

More from Differential Equations