Verified Solution • 2025–26 Session moderate

QJEEM23Maafb8: JEE Main 2023 Mathematics - Class 12 JEEM23Maafb8 Solution: Let be the solution of the differential equation . Then equals :

Let αx=exp⁡(xβyγ)\alpha x=\exp \left(x^{\beta} y^{\gamma}\right) be the solution of the differential equation 2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2} . Then α+β−γ\alpha+\beta-\gamma equals :

Multiple Choice Options

  • A 1
  • B 0
  • C 3
  • D −1-1

Step-by-Step Verified Solution

amp;2x2ydy−(1−xy2)dx=0amp;⇒2x2ydydx−1+xy2=0amp;⇒2ydydx−1x2+y2x=0amp;⇒2ydydx+y2x=1x2 (L.D.E) amp;Let,y2=t⇒2ydydx=dtdxamp;dtdx+tx=1x2amp; I.F =e∫1xdx=x\begin{aligned} & 2 x^2 y d y-\left(1-x y^2\right) d x=0 \\\\ & \Rightarrow 2 x^2 y \frac{d y}{d x}-1+x y^2=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}-\frac{1}{x^2}+\frac{y^2}{x}=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}+\frac{y^2}{x}=\frac{1}{x^2} \text { (L.D.E) } \\\\ & Let, y^2=t \Rightarrow 2 y \frac{d y}{d x}=\frac{d t}{d x} \\\\ & \frac{d t}{d x}+\frac{t}{x}=\frac{1}{x^2} \\\\ & \text { I.F }=e^{\int \frac{1}{x} d x}=x\end{aligned}

So, the general solution is

amp;⇒t×x=∫x⋅1x2dxamp;⇒y2⋅x=ln⁡x+camp; Also,y(2) =log⁡e2amp;log⁡e22=log⁡e2+c⇒c=log⁡e2amp;⇒y2x=ln⁡x+ln⁡2amp;⇒y2x=ln⁡2xamp;2x=exp⁡(x1y2)\begin{aligned}& \Rightarrow t \times x=\int x \cdot \frac{1}{x^2} d x \\\\& \Rightarrow y^2 \cdot x=\ln x+c \\\\& \text { Also,y(2) }=\sqrt{\log _e 2} \\\\& \log _e^2 2=\log _e 2+c \Rightarrow c=\log _e 2 \\\\& \Rightarrow y^2 x=\ln x+\ln 2 \\\\& \Rightarrow y^2 x=\ln 2 x \\\\& 2 x=\exp \left(x^1 y^2\right)\end{aligned}

By compare it with given solution we get,

amp;α=2,β=1,γ=2amp;⇒α+β−γ=2+1−2=1\begin{aligned}& \alpha=2,\beta=1, \gamma=2 \\\\& \Rightarrow \alpha+\beta-\gamma=2+1-2=1\end{aligned}

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