What is the solution for y^'=y^3-6x^2y^3,y(1)=-1/2 ?

The solution for y^'=y^3-6x^2y^3,y(1)=-1/2 is y=-sqrt(-1/(2(x-2x^3-1)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=y^3-6x^2y^3,y(1)=-1/2

y^{^{'}} = y^{3} - 6 x^{2} y^{3}, y (1) = - \frac{1}{2}

y = - - \frac{1}{2 ( x - 2 x ^{3} - 1 )}

Solution steps

y^{'} (x) = y^{3} - 6 x^{2} y^{3}

Solve separable ODE:

y = - - \frac{1}{2 ( x - 2 x ^{3} - 1 )}

y = - - \frac{1}{2 ( x - 2 x ^{3} - 1 )}

What is the solution for y^'=y^3-6x^2y^3,y(1)=-1/2 ?
The solution for y^'=y^3-6x^2y^3,y(1)=-1/2 is y=-sqrt(-1/(2(x-2x^3-1)))