Verified Solution • 2025–26 Session moderate

QJEEM23Maff9e: JEE Main 2023 Mathematics - Class 12 JEEM23Maff9e Solution: If the solution curve of the differential equation , passes…

If the solution curve f(x,y)=0f(x, y)=0 of the differential equation

\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0 ,

passes through the points (1,0)(1,0) and (α,2)(\alpha, 2) , then αα\alpha^{\alpha} is equal to :

Multiple Choice Options

  • A e2e2e^{\sqrt{2} e^{2}}
  • B e2e2e^{2 e^{\sqrt{2}}}
  • C ee2e^{e^{2}}
  • D e2e2e^{2 e^{2}}

Step-by-Step Verified Solution

We have, \left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0

Put xlog⁡ex=tx \log _e x=t

amp;⇒(x⋅1x+log⁡ex)dxdy=dtdyamp;⇒(1+log⁡ex)dxdy=dtdyamp;∴dtdy−t=ey\begin{aligned}& \Rightarrow \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\& \Rightarrow \left(1+\log _e x\right) \frac{d x}{d y} =\frac{d t}{d y} \\\\& \therefore \frac{d t}{d y}-t =e^{y}\end{aligned}

Now, IF =e∫−dy=e−y=e^{\int-d y}=e^{-y}

amp;∴ General solution, t(e−y)=∫(ey⋅e−y)dy+camp;⇒te−y=∫dy+camp;⇒te−y=y+camp;⇒(xlog⁡ex)e−y=y+c..........(i)\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\\\ & \Rightarrow t e^{-y}=\int d y+c \\\\ & \Rightarrow t e^{-y}=y+c \\\\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}

Equation (i), passes through the point (1,0)(1,0)

amp;∴0=0+camp;⇒c=0amp;∴(xlog⁡ex)e−y=yamp;⇒xlog⁡ex=yey\begin{aligned}& \therefore 0=0+c \\\\& \Rightarrow c=0 \\\\& \therefore \left(x \log _e x\right) e^{-y}=y \\\\& \Rightarrow x \log _e x=y e^y\end{aligned}

Which passes through (α,2)(\alpha, 2)

amp;∴αlog⁡eα=2e2amp;⇒log⁡eαα=2e2amp;⇒αα=e2e2\begin{aligned}& \therefore \alpha \log _e \alpha=2 e^2 \\\\& \Rightarrow \log _e \alpha^\alpha=2 e^2 \\\\& \Rightarrow \alpha^\alpha=e^{2 e^2}\end{aligned}

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