Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If y = y(x) is the solution of the differential equation

$\frac{dy}{dx}=(tanx-y)sec^{2}x$ ,

$x\epsilon (\frac{-\pi}{2},\frac{\pi}{2})$ , such that y(0) = 0 , then

$y(\frac{-\pi}{4})$ is equal to :

Option 1)
$e-2$

Option 2)
$\frac{1}{2}-e$

Option 3)
$2+\frac{1}{e}$

Option 4)
$\frac{1}{e}-2$

Answers (1)

Given differential eqn

$\frac{dy}{dx}=(tanx-y)sec^{2}x$

$=>\frac{dy}{dx}+y sec^{2}x+=tanxsec^{2}x$

I.F. = $e^{\int sec^{2}x dx }$ = $e^{tanx}$

Solution of given differential eqn

$y\cdot e^{tanx}=\int tanx \cdot sec^{2}x\cdot e^{tanx} dx$

$=tanx\cdot -\int sec^{2}x\cdot e^{tanx} dx$

$= e^{tanx} (tanx-1)$

Given that $y(0)=0$ => $C=1$

$y\cdot e^{tanx}=e^{tanx}(tanx-1)+1$

put $x=\frac{-\pi}{4}$

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

So, correct option is (1)

Option 1)

$e-2$

Option 2)

$\frac{1}{2}-e$

Option 3)

$2+\frac{1}{e}$

Option 4)

$\frac{1}{e}-2$

Posted by

Vakul

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

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs

anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question