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QJEEM21Ma7ef9: JEE Main 2021 Mathematics - Class 12 JEEM21Ma7ef9 Solution: Let C 1 be the curve obtained by the solution of differential…

Let C1 be the curve obtained by the solution of differential equation

2xy{{dy} \over {dx}} = {y^2} - {x^2},x > 0 . Let the curve C2 be the

solution of 2xyx2−y2=dydx{{2xy} \over {{x^2} - {y^2}}} = {{dy} \over {dx}} . If both the curves pass through (1, 1), then the area enclosed by the curves C1 and C2 is equal to :

Multiple Choice Options

  • A π4{\pi \over 4} + 1
  • B π\pi + 1
  • C π\pi −- 1
  • D π2{\pi \over 2} −- 1

Step-by-Step Verified Solution

dydx=y2−x22xy{{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}

Put y=vxy = vx

v+xdvdx=v2x2−x22vx2=v2−12vv + x{{dv} \over {dx}} = {{{v^2}{x^2} - {x^2}} \over {2v{x^2}}} = {{{v^2} - 1} \over {2v}}

xdvdx=v2−1−2v22v=−(v2+1)2vx{{dv} \over {dx}} = {{{v^2} - 1 - 2{v^2}} \over {2v}} = - {{({v^2} + 1)} \over {2v}}

⇒2vv2+1dv=−dxx\Rightarrow {{2v} \over {{v^2} + 1}}dv = - {{dx} \over x}

ln⁡(v2+1)=−ln⁡x+ln⁡c⇒v2+1=cx\ln ({v^2} + 1) = - \ln x + \ln c \Rightarrow {v^2} + 1 = {c \over x}

⇒y2x2+1=cx⇒x2+y2=cx\Rightarrow {{{y^2}} \over {{x^2}}} + 1 = {c \over x} \Rightarrow {x^2} + {y^2} = cx

It pass through (1, 1)

∴\therefore x2+y2−2x=0{x^2} + {y^2} - 2x = 0

Similarly for second differential equation dydx={{dy} \over {dx}} = 2xyx2−y2{{2xy} \over {{x^2} - {y^2}}}

Equation of curve is x2 + y2 −- 2y = 0

Now required area is

JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Differential Equations Question 147 English Explanation

=2∫01(2x−x2−x)dx= 2\int\limits_0^1 {\left( {\sqrt {2x - {x^2}} - x} \right)} dx

=(π2−1)= ({\pi \over 2} - 1) sq. units

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