Let y=y(x) be the solution of the differential equation (2 x log e x ) d y/d x+2 y=3/x log e x, x 0 and y ( e -1 )=0 . Then, y ( e ) is equal to

Mathematics · Differential Equations

JEE Main 2024 — 6 April, Shift 1 — Question 18

Let $y=y(x)$ 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 $\mathrm{y}\left(\mathrm{e}^{-1}\right)=0$ . Then, $\mathrm{y}(\mathrm{e})$ is equal to

$\frac{d y}{d x}+\frac{y}{x \ln x}=\frac{3}{2 x^{2}}$

$\therefore$ I.F. $=\mathrm{e}^{\int \frac{1}{\mathrm{x} \ln \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\ln (\ln (\mathrm{x}))}=\ln \mathrm{x}$

$\therefore y \ln x=\int \frac{3 \ln x}{2 x^{2}} d x$

$=\frac{3 \ln x}{2} \int x^{-2} d x-\int\left(\frac{3}{2 x} \cdot \int x^{-2} d x\right) d x$

$=\frac{3 \ln x}{2}\left(-\frac{1}{x}\right)-\int \frac{3}{2 x}\left(-\frac{1}{x}\right) d x$

$y \cdot \ln x=\frac{-3 \ln x}{2 x}-\frac{3}{2 x}+C$

$\because \mathrm{y}\left(\mathrm{e}^{-1}\right)=0$

$\therefore 0(-1)=\frac{3 \mathrm{e}}{2}-\frac{3 \mathrm{e}}{2}+\mathrm{C} \Rightarrow \mathrm{C}=0$

$\therefore y=\frac{-3 \ln x}{2 x}-\frac{3}{2 x}$

$\therefore \mathrm{y}(\mathrm{e})=\frac{-3}{2 \mathrm{e}}-\frac{3}{2 \mathrm{e}}=\frac{-3}{\mathrm{e}}$

Answer key and solution verified before publishing.

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