The equation of the curve passing through the origin and satisfying the differential equation $(1+x^2) \frac{dy}{dx} + 2xy = 4x^2$ is:

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

Explore More

Similar Questions

If $\cos x \frac{dy}{dx} - y \sin x = 6x$ for $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{6})$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$, where $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $y(0) = -\frac{7}{4}$. Then $y(\frac{\pi}{6})$ is equal to:

If $y=y(x)$ is the solution of $\frac{dy}{dx}=\frac{x-y \cos x}{1+\sin x}$ with $y\left(\frac{\pi}{2}\right)=\frac{\pi^2}{8}$,then $y(\pi)=$

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 the solution curve $y = y(x)$ of the differential equation $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} y = 2x \exp \left( \frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} \right)$ pass through the origin. Then $y(1)$ 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