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Popular >

y^'=y^2(t)

Solution

y^{^{'}} = y^{2} (t)

Solution

y = - \frac{2}{t ^{2} + c _{1}}

Solution steps

y^{'} (t) = y^{2} (t)

Solve separable ODE:

y = - \frac{2}{t ^{2} + c _{1}}

y = - \frac{2}{t ^{2} + c _{1}}

Graph

Popular Examples

(dy)/(dx)x= 2/3 (y-1)^{3/2}

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

(x-13)(x-3)y^'=y

(x - 13) (x - 3) y^{^{'}} = y

(dx)/(dy)=3e^{2x-3y}

\frac{d x}{d y} = 3 e^{2 x - 3 y}

x d y - (y + 4) d x = 0

(dy)/(dx)=14e^{5x-y},y(0)=0

\frac{d y}{d x} = 14 e^{5 x - y}, y (0) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^'=y^2(t) ?
The solution for y^'=y^2(t) is y=-2/(t^2+c_{1)}