en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Calculus >

y=(2y^4+2x)y^'

Solution

y = (2 y^{4} + 2 x) y^{^{'}}

Solution

y = - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - \frac{c _{1} - c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} - c _{1}^{2} + 4 x}{2}

Solution steps

y = (2 y^{4} + 2 x) y^{'} (x)

Solve exact ODE:

y = - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - \frac{c _{1} - c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} - c _{1}^{2} + 4 x}{2}

y = - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} + c _{1}^{2} + 4 x}{2}, y = - \frac{c _{1} - c _{1}^{2} + 4 x}{2}, y = - - \frac{c _{1} - c _{1}^{2} + 4 x}{2}

Graph

Popular Examples

derivative of (e^{2x})/(sin(3x))

integral of (2x)/3

integral from 0 to 1 of xsqrt(3-x^2)

tangent of f(x)=[1+e^x][ln(x)],\at x=1

integral of te^{-st}sin(t)

Frequently Asked Questions (FAQ)

What is the solution for y=(2y^4+2x)y^' ?
The solution for y=(2y^4+2x)y^' is