en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=y^3,y(0)=1

Solution

y^{^{'}} = y^{3}, y (0) = 1

Solution

y = - \frac{1}{2 t - 1}

Solution steps

y^{'} (t) = y^{3}

Solve separable ODE:

y = - \frac{1}{2 t - 1}

y = - \frac{1}{2 t - 1}

Popular Examples

(dy)/(dt)+1/2 y=17sin(2t)

\frac{d y}{d t} + \frac{1}{2} y = 17 sin (2 t)

integral of cos(sqrt(2x))

\int cos (2 x) d x

integral of (tan(x))^4

\int (tan (x))^{4} d x

derivative of (x^2-5x+4/(|x-1|))

\frac{d}{d x} (\frac{x ^{2} - 5 x + 4}{∣ x - 1 ∣})

derivative of x^{1-a}

\frac{d}{d x} (x^{1 - a})

Frequently Asked Questions (FAQ)

What is the solution for y^'=y^3,y(0)=1 ?
The solution for y^'=y^3,y(0)=1 is y=sqrt(-1/(2t-1))