The general solution of the differential equation $\frac{dy}{dx} = \frac{2x-3y+4}{3x+2y-7}$ is

  • A
    $x^2+y^2=3xy+y+C$
  • B
    $(2x-3y)^2+(3x+2y)^2=C$
  • C
    $x^2-y^2-3xy+4x+7y+C=0$
  • D
    $x^2-3xy-y^2+4x+7y+C=0$

Explore More

Similar Questions

Every curve represented by the general solution of $\frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$ cuts every curve represented by the general solution of $\frac{dy}{dx} + \frac{y^3 e^{y^2-5}}{x \log x} = 0$ at an angle $\theta$. Then,$4\theta - \frac{\pi}{2} =$

Find the family of curves such that the angle between the tangent at any point $(x, y)$ and the tangent to the curve $xy = c^2$ at the point of intersection is $\frac{\pi}{4}$.

Let $b$ be a nonzero real number. Suppose $f: R \rightarrow R$ is a differentiable function such that $f(0)=1$. If the derivative $f^{\prime}$ of $f$ satisfies the equation $f^{\prime}(x) = \frac{f(x)}{b^2+x^2}$ for all $x \in R$,then which of the following statements is/are $TRUE$?
$(A)$ If $b>0$,then $f$ is an increasing function
$(B)$ If $b < 0$,then $f$ is a decreasing function
$(C)$ $f(x)f(-x)=1$ for all $x \in R$
$(D)$ $f(x)-f(-x)=0$ for all $x \in R$

Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$, where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to . . . . . . .

If $\frac{d^2y}{dx^2} + \sin x = 0$,then the solution of the differential equation 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