Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If $y(x)$ is the solution of the differential equation $\frac{dy}{dx}+(\frac{2x+1}{x}) y =e^{-2x} , x> 0,$ where $y(1)=\frac{1}{2}e^{-2}$ , then :
Option 1)

$y(\log_{e}2)=\log_{e}4$
Option 2)

$y(x)$ is decreasing in $(\frac{1}{2},1)$
Option 3)

$y(\log_{e}2)=\frac{\log_{e}2}{4}$
Option 4)

$y(x)$ is decreasing in (0,1)

Answers (1)

best_answer

Linear Differential Equation -

$\frac{dy}{dx}+Py= Q$

- wherein

P, Q are functions of x alone.

Linear Differential Equation -

Multiply by $e^{SPdx}$ which is the Integrating factor

- wherein

P is the function of x alone

$\frac{dy}{dx}+\left ( 2+\frac{1}{x} \right )y=e^{-2x}$

$I\cdot F = e^\imath {2+\frac{1}{x}}dx=e^{2x+lnx}=x\varepsilon ^{2x}$

Solution of a differential equation,

$y(1)=\frac{1}{2e^{2}}\Rightarrow c=0$

Hence, $y=\frac{xe^{-2x}}{2}$

$\frac{dy}{dx}=\frac{e^{-2x}}{2}+\frac{x^{2}e^{-2x}(-2)}{2}$

$=e^{-2x}\left [ \frac{1}{2}-x \right ]<0$ if $x>\frac{1}{2}$

$\Rightarrow y(x)$ is decreasing in $(\frac{1}{2},1)$

Option 1)

$y(\log_{e}2)=\log_{e}4$

Option 2)

$y(x)$ is decreasing in $(\frac{1}{2},1)$
Option 3)

$y(\log_{e}2)=\frac{\log_{e}2}{4}$

Option 4)

$y(x)$ is decreasing in (0,1)

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2

anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

Latest Question