Verified Solution • 2025–26 Session moderate

QJEEM20Ma413c: JEE Main 2020 Mathematics - Class 12 JEEM20Ma413c Solution: If x 3 dy + xy dx = x 2 dy + 2y dx; y(2) = e and x > 1, then…

If x3dy + xy dx = x2dy + 2y dx; y(2) = e and
x > 1, then y(4) is equal to :

Multiple Choice Options

  • A e2{{\sqrt e } \over 2}
  • B 12+e{1 \over 2} + \sqrt e
  • C 32+e{3 \over 2} + \sqrt e
  • D 32e{3 \over 2}\sqrt e

Step-by-Step Verified Solution

x3dy+xy dx=x2dy+2y dx{x^3}dy + xy\,dx = {x^2}dy + 2y\,dx

⇒dy(x3−x2)=dx(2y−xy)\Rightarrow dy({x^3} - {x^2}) = dx(2y - xy)

⇒\Rightarrow −∫1ydx- \int {{1 \over y}dx} = ∫x−2x2(x−1)dx\int {{{x - 2} \over {{x^2}(x - 1)}}dx}

⇒\Rightarrow −ln⁡y=∫(Ax+Bx2+C(x−1))dx- \ln y = \int {\left( {{A \over x} + {B \over {{x^2}}} + {C \over {(x - 1)}}} \right)dx}

Where A = 1, B = +2, C = −- 1

⇒−ln⁡y=ln⁡x−2x−ln⁡(x−1)+λ\Rightarrow - \ln y = \ln x - {2 \over x} - \ln (x - 1) + \lambda

As y(2)=ey(2) = e

⇒−1=ln⁡2−1−0+λ\Rightarrow - 1 = \ln 2 - 1 - 0 + \lambda

∴\therefore λ=−ln⁡2\lambda = - \ln 2

⇒ln⁡y=−ln⁡x+2x+ln⁡(x−1)+ln⁡2\Rightarrow \ln y = - \ln x + {2 \over x} + \ln (x - 1) + \ln 2

Now put x = 4 in equation

⇒ln⁡y=−ln⁡4+12+ln⁡3+ln⁡2\Rightarrow \ln y = - \ln 4 + {1 \over 2} + \ln 3 + \ln 2

⇒lny=ln(32)+12ln⁡e\Rightarrow {\mathop{\rm lny}\nolimits} = ln\left( {{3 \over 2}} \right) + {1 \over 2}\ln e

⇒y=32e\Rightarrow y = {3 \over 2}\sqrt e

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