The solution of $(1+x^2) \frac{dy}{dx} + 2xy - 4x^2 = 0$ is:

  • A
    $3x(1+y^2) = 4y^3 + c$
  • B
    $3y(1+x^2) = 4x^3 + c$
  • C
    $3x(1-y^2) = 4y^3 + c$
  • D
    $3y(1+y^2) = 4x^3 + c$

Explore More

Similar Questions

Find the general solution of the differential equation:
$\frac{dy}{dx} + (\sec x)y = \tan x$,where $0 \le x \le \frac{\pi}{2}$.

Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}$ for $-1 < x < 1$ such that $f(0)=0$. If $6 \int_{-1/2}^{1/2} f(x) dx = 2\pi - \alpha$,then $\alpha^2$ is equal to . . . . . .

Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in R$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right)$ and $y(0)=0$,then $y(1)$ is:

Find the general solution of the differential equation: $y dx + (x - y^2) dy = 0$

Difficult
View Solution

Let $y=y(x)$ be the solution of the differential equation $\sin x \frac{dy}{dx}+y \cos x=4x, x \in(0, \pi)$. If $y\left(\frac{\pi}{2}\right)=0$,then $y\left(\frac{\pi}{6}\right)$ is equal to

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