en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=6y^2+xy^2,y(0)=1

Solution

y^{^{'}} = 6 y^{2} + x y^{2}, y (0) = 1

Solution

y = - \frac{2}{12 x + x ^{2} - 2}

Solution steps

y^{'} (x) = 6 y^{2} + x y^{2}

Solve separable ODE:

y = - \frac{2}{12 x + x ^{2} - 2}

y = - \frac{2}{12 x + x ^{2} - 2}

Popular Examples

(d^2)/(dx^2)((2000t)/(4t+75))

\frac{d ^{2}}{d x ^{2}} (\frac{2000 t}{4 t + 75})

integral of (ycos(x)+2xe^y)

\int (y cos (x) + 2 x e^{y}) d x

implicit xy=2 implicit derivative

x y = 2

derivative of (5sqrt(x)^{ln(x)})

\frac{d}{d x} ((5 x)^{l n (x)})

integral of sqrt(sin(2x))cos(2x)

\int sin (2 x) cos (2 x) d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=6y^2+xy^2,y(0)=1 ?
The solution for y^'=6y^2+xy^2,y(0)=1 is y=-2/(12x+x^2-2)