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Popular >

y^'=y-(y)^2

Solution

y^{^{'}} = y - (y)^{2}

Solution

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

Solution steps

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

Solve separable ODE:

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

Graph

Popular Examples

(dy)/(dx)=(x-6)e^{-2y},y(6)=ln(6)

\frac{d y}{d x} = (x - 6) e^{- 2 y}, y (6) = ln (6)

0= 1/10 y+(100*10^{-3})y^'

0 = \frac{1}{10} y + (100 \cdot 1 0^{- 3}) y^{^{'}}

(dQ)/(dt)+2/300 Q=0.8

\frac{d Q}{d t} + \frac{2}{300} Q = 0.8

y^'-(1+tan^2(x))y=0

y^{^{'}} - (1 + tan^{2} (x)) y = 0

2(dy}{dx}-1/y =\frac{2x)/y

2 \frac{d y}{d x} - \frac{1}{y} = \frac{2 x}{y}

Frequently Asked Questions (FAQ)

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