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QJEEM22Ma3791: JEE Main 2022 Mathematics - Class 12 JEEM22Ma3791 Solution: If x = x(y) is the solution of the differential equation ; then …

If x = x(y) is the solution of the differential equation

ydxdy=2x+y3(y+1)ey, x(1)=0y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0 ; then x(e) is equal to :

Multiple Choice Options

  • A e3(ee−1){e^3}({e^e} - 1)
  • B ee(e3−1){e^e}({e^3} - 1)
  • C e2(ee+1){e^2}({e^e} + 1)
  • D ee(e2−1){e^e}({e^2} - 1)

Step-by-Step Verified Solution

dxdy−2xy=y2(y+1)ey\frac{d x}{d y}-\frac{2 x}{y}=y^{2}(y+1) e^{y}

 If =e∫−2ydy=e−2ln⁡y=1y2\text { If }=e^{\int-\frac{2}{y} d y}=e^{-2 \ln y}=\frac{1}{y^{2}}

Solution is given by

amp;x⋅1y2=∫y2(y+1)ey⋅1y2dy⇒amp;xy2=∫(y+1)eydy⇒amp;xy2=yey+c\begin{aligned}&x \cdot \frac{1}{y^{2}}=\int y^{2}(y+1) e^{y} \cdot \frac{1}{y^{2}} d y \\\\\Rightarrow & \frac{x}{y^{2}}=\int(y+1) e^{y} d y \\\\\Rightarrow & \frac{x}{y^{2}}=y e^{y}+c\end{aligned}

⇒x=y2(yey+c)\Rightarrow x=y^{2}\left(y e^{y}+c\right) at, y=1,x=0y=1, x=0

⇒0=1(1⋅e1+c)⇒c=−e\Rightarrow 0=1\left(1 \cdot e^{1}+c\right) \Rightarrow c=-e at y=ey=e ,

x=e2(e.ee−e)x=e^{2}\left(e . e^{e}-e\right)

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