The general solution of $\frac{dy}{dx} = x + \sin x \cos y + x \cos y + \sin x$ is

  • A
    $\tan \frac{x}{2} = \frac{y^2}{2} - \cos x + C$
  • B
    $\tan \frac{y}{2} = \frac{x^2}{2} - \cos x + C$
  • C
    $\sec^2 \frac{y}{2} = \frac{x^2}{2} - \cos x + C$
  • D
    $\tan \frac{y}{2} = \frac{x^2}{2} + \cos x + C$

Explore More

Similar Questions

The solution of the differential equation $\frac{dy}{dx} + \sin^2 y = 0$ is

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x$,when $x = 0, y = 1$ is .....

Let $x=x(t)$ and $y=y(t)$ be solutions of the differential equations $\frac{dx}{dt}+ax=0$ and $\frac{dy}{dt}+by=0$ respectively,where $a, b \in R$. Given that $x(0)=2$,$y(0)=1$,and $3y(1)=2x(1)$,the value of $t$ for which $x(t)=y(t)$ is:

If $y = f(x)$ is a monotonically increasing function such that $(\frac{dy}{dx})^2 = 6 - \frac{dy}{dx}$ and $y(0) = 5$, then $y(3) = ...$

The solution of the differential equation $\left( {1 + {e^{2y}}} \right){e^{{{\tan }^{ - 1}}x}}dx - \left( {1 + {x^2}} \right)\left( {{e^y} + {{\left( {{e^y} - 1} \right)}^2}} \right)dy = 0$ is

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