Verified Solution • 2025–26 Session moderate

QJEEM24Ma3608: JEE Main 2024 Mathematics - Class 12 JEEM24Ma3608 Solution: Let be the solution of the differential equation and . Then, is …

Let y=y(x)y=y(x) be the solution of the differential equation \left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0 and y(e−1)=0y\left(e^{-1}\right)=0 . Then, y(e)y(e) is equal to

Multiple Choice Options

  • A −3e-\frac{3}{\mathrm{e}}
  • B −32e-\frac{3}{2 \mathrm{e}}
  • C −23e-\frac{2}{3 \mathrm{e}}
  • D −2e-\frac{2}{\mathrm{e}}

Step-by-Step Verified Solution

amp;(2xlog⁡x)dydx+2y=3xlog⁡xamp;dydx+yxlog⁡x=32x2amp;IF=e∫1xlog⁡x=elog⁡(log⁡x)=log⁡xamp;y×log⁡x=∫log⁡x×32x2dx+Camp;ylog⁡x=32[−log⁡xx−1x]+Camp;y(e−1)=0amp;⇒0=32[e−e]+Camp;C=0amp;ylog⁡x=−32[log⁡x+1x]amp;y(e)→y(e)×1=−32[1+1e]=−3e\begin{aligned}& (2 x \log x) \frac{d y}{d x}+2 y=\frac{3}{x} \log x \\& \frac{d y}{d x}+\frac{y}{x \log x}=\frac{3}{2 x^2} \\& I F=e^{\int \frac{1}{x \log x}}=e^{\log (\log x)}=\log x \\& y \times \log x=\int \log x \times \frac{3}{2 x^2} d x+C \\& y \log x=\frac{3}{2}\left[\frac{-\log x}{x}-\frac{1}{x}\right]+C \\& y\left(e^{-1}\right)=0 \\& \Rightarrow 0=\frac{3}{2}[e-e]+C \\& C=0 \\& y \log x=\frac{-3}{2}\left[\frac{\log x+1}{x}\right] \\& y(e) \rightarrow y(e) \times 1=\frac{-3}{2}\left[\frac{1+1}{e}\right]=-\frac{3}{e}\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring