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Popular >

y^'=-y^3,y(0)=2

Solution

y^{^{'}} = - y^{3}, y (0) = 2

Solution

y = \frac{2}{8 t + 1}

Solution steps

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

Solve separable ODE:

y = \frac{2}{8 t + 1}

y = \frac{2}{8 t + 1}

Graph

Popular Examples

(\partial)/(\partial x)(7xcos(xy))

\frac{\partial}{\partial x} (7 x cos (x y))

(dx)/(dt)=1-t+x-tx

\frac{d x}{d t} = 1 - t + x - t x

7 y^{^{''}} + y = 0

integral of e^{4x} 1/5

\int e^{4 x} \frac{1}{5} d x

(dy)/(dt)=(5+sin(t))*(y/5)

\frac{d y}{d t} = (5 + sin (t)) \cdot (\frac{y}{5})

Frequently Asked Questions (FAQ)

What is the solution for y^'=-y^3,y(0)=2 ?
The solution for y^'=-y^3,y(0)=2 is y= 2/(sqrt(8t+1))