If y=yx is the solution of the differential equation dy/dx+tanxy=sinx,0≤x≤(π)/3, with y0=0, then y(π)/4 is equal to

Correct answer: (B) 1/(2√2)log_e2. dy/dx+tanxy=sinx; 0≤x≤(π)/3 I.F.=e^(∫tanxdx)=e^lnsecx=secx ysecx=∫tanxdx+C ysecx=lnsecx+C x=0, y=0 ⇒ C=0 ysecx=lnsecx y=cosx·lnsecx y_(x=π4)=(1√2·ln√2y_(x=π4)=1)/(2√2)log_e2

Q37 / 90 · JEE Main 2021 (16 Mar Shift 2) · Mathematics

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (\tan x) y = \sin x, 0 \leq x \leq \frac{π}{3},$ with $y (0) = 0,$ then $y (\frac{π}{4})$ is equal to

A
$\frac{1}{4} \log_{e} 2$
Correct answer
B
$(\frac{1}{2 \sqrt{2}}) \log_{e} 2$
C
$\log_{e} 2$
D
$\frac{1}{2} \log_{e} 2$

Solution

$\frac{d y}{d x} + (\tan x) y = \sin x; 0 \leq x \leq \frac{π}{3}$

$I . F . = e^{\int \tan x d x} = e^{\ln \sec x} = \sec x$

$y \sec x = \int \tan x d x + C$

$y \sec x = \ln |\sec x| + C$

$x = 0, y = 0 \Rightarrow C = 0$

$y \sec x = \ln |\sec x|$

$y = \cos x \cdot \ln |\sec x|$

$y_{x = \frac{π}{4}} = (\frac{1}{\sqrt{2}}) \cdot \ln \sqrt{2}$

$y_{x = \frac{π}{4}} = \frac{1}{2 \sqrt{2}} \log_{e} 2$

Hence, the correct answer is option: 'B'.

If y=yx is the solution of the differential equation dydx+tanxy=sinx,0≤x≤π3, with y0=0, then yπ4 is equal to

Solution

Relevant Questions

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (\tan x) y = \sin x, 0 \leq x \leq \frac{π}{3},$ with $y (0) = 0,$ then $y (\frac{π}{4})$ is equal to