Let $y=y(x)$ be the solution of the differential equation $\cos x(3 \sin x+\cos x+3) dy = (1+y \sin x(3 \sin x+\cos x+3)) dx$; $0 \leq x \leq \frac{\pi}{2}, y(0)=0$. Then,$y\left(\frac{\pi}{3}\right)$ is equal to ..... .

  • A
    $2 \log _{e}\left(\frac{2 \sqrt{3}+9}{6}\right)$
  • B
    $2 \log _{e}\left(\frac{2 \sqrt{3}+10}{11}\right)$
  • C
    $2 \log _{e}\left(\frac{\sqrt{3}+7}{2}\right)$
  • D
    $2 \log _{e}\left(\frac{3 \sqrt{3}-8}{4}\right)$

Explore More

Similar Questions

The curve satisfying the differential equation $y \, dx - (x + 3y^2) \, dy = 0$ and passing through the point $(1, 1)$ also passes through the point

The integrating factor of the differential equation $(1-x^2) \frac{dy}{dx} + xy = kx$ for $(-1 < x < 1)$ is . . . . . . .

The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........

Solve the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x$.

Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the tangent line $Y-y=Y^{\prime}(x)(X-x)$ and the coordinate axes,where $(x, y)$ is any point on the curve,is always $\frac{-y^2}{2 Y^{\prime}(x)}+1$,where $Y^{\prime}(x) \neq 0$. If $Y(1)=1$,then $12 Y(2)$ equals

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo