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

Correct answer: A. - 3 e aligned & (d y)/(d x)+(y)/(x x)=(3)/(2 x^2) & I.F. =e^ ∫ (1)/(x x) d x =e^ ( (x)) = x & y x=∫ (3 x)/(2 x^2) d x & =(3 x)/(2) ∫ x^ -2 d x-∫((3)/(2 x) · ∫ x^ -2 d x) d x & =(3 x)/(2)(-(1)/(x))-∫ (3)/(2 x)(-(1)/(x)) d x & y. x=(-3 n x)/(2 x)-(3)/(2 x)+C aligned aligned & y(e^ -1 )=0 & 0(-1)= 3 e 2 - 3 e 2 +C C=0 & y= -3 nx 2 x - 3 2 x & y(e)= -3 2 e - 3 2 e = -3 e aligned

JEE Main 20246 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation (2 x _e x)...

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

. Then,

y(e) is equal to

Options

A $-\frac{3}{\mathrm{e}}$
B $-\frac{3}{2 e}$
C $-\frac{2}{3 e}$
D $-\frac{2}{\mathrm{e}}$

Correct answer

A. $-\frac{3}{\mathrm{e}}$

Step-by-step solution

\begin{aligned} & \frac{d y}{d x}+\frac{y}{x \ell \ln x}=\frac{3}{2 x^2} \\ & \therefore \text{ I.F. }=e^{\int \frac{1}{x \ln x} d x}=e^{\ln (\ln (x))}=\ln x \\ & \therefore y \ell \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 \\ & \text{ y. } \ln x=\frac{-3 \ell n x}{2 x}-\frac{3}{2 x}+C\end{aligned}

\begin{aligned} & \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 \mathrm{y}=\frac{-3 \ell \mathrm{nx}}{2 \mathrm{x}}-\frac{3}{2 \mathrm{x}} \\ & \therefore \mathrm{y}(\mathrm{e})=\frac{-3}{2 \mathrm{e}}-\frac{3}{2 \mathrm{e}}=\frac{-3}{\mathrm{e}}\end{aligned}

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