Verified Solution • 2025–26 Session moderate

QJEEM19Ma60cb: JEE Main 2019 Mathematics - Class 12 JEEM19Ma60cb Solution: The solution of the differential equation, = (x – y) 2 , when…

The solution of the differential equation,

dydx{{dy} \over {dx}} = (x – y)2, when y(1) = 1, is :

Multiple Choice Options

  • A −- loge ∣1+x−y1−x+y∣\left| {{{1 + x - y} \over {1 - x + y}}} \right| = x + y −- 2
  • B loge ∣2−x2−y∣\left| {{{2 - x} \over {2 - y}}} \right| = x −- y
  • C loge ∣2−y2−x∣\left| {{{2 - y} \over {2 - x}}} \right| = 2(y −- 1)
  • D −- loge ∣1−x+y1+x−y∣\left| {{{1 - x + y} \over {1 + x - y}}} \right| = 2(x −- 1)

Step-by-Step Verified Solution

x −- y = t

⇒\Rightarrow   dydx=1−dtdx{{dy} \over {dx}} = 1 - {{dt} \over {dx}}

⇒\Rightarrow   1 −- dtdx{{dt} \over {dx}} = t2 ⇒\Rightarrow    ∫dt1−t2\int {{{dt} \over {1 - {t^2}}}} = ∫1dx\int {1dx}

⇒\Rightarrow    12ℓn(1+t1−t)=x+λ{1 \over 2}\ell n\left( {{{1 + t} \over {1 - t}}} \right) = x + \lambda

⇒\Rightarrow    12ℓn(1+x−y1−x+y)=x+λ{1 \over 2}\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = x + \lambda given y(1) = 1

⇒\Rightarrow    12ℓn(1)=1+λ⇒λ=−1{1 \over 2}\ell n(1) = 1 + \lambda \Rightarrow \lambda = - 1

⇒\Rightarrow    ℓn(1+x−y1−x+y)\ell n\left( {{{1 + x - y} \over {1 - x + y}}} \right) = 2(x −- 1)

⇒\Rightarrow    −ℓn(1−x+y1+x−y)- \ell n\left( {{{1 - x + y} \over {1 + x - y}}} \right) = 2(x −- 1)

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