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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

integral of 1/(x^2)+x

\int \frac{1}{x ^{2}} + x d x

derivative of x^{{z}(x})

\frac{d}{d x} (x^{z (x)})

limit as x approaches 2-of 3x+6

x \to 2 - lim (3 x + 6)

(\partial)/(\partial y)(xcos(xy))

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

(\partial)/(\partial y)(x^2+xy-25)

\frac{\partial}{\partial y} (x^{2} + x y - 25)

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= 1/(sqrt(2t+1))