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Popular >

y^'=2y(5-y)

Solution

y^{^{'}} = 2 y (5 - y)

Solution

y = \frac{5}{e ^{- 10 t + c_{1}} + 1}

Solution steps

y^{'} (t) = 2 y (5 - y)

Solve Bernoulli ODE:

y = \frac{5}{e ^{- 10 t + c_{1}} + 1}

y = \frac{5}{e ^{- 10 t + c_{1}} + 1}

Popular Examples

tangent of f(x)=(x^2-2)(x^3-3x),\at x=-2 tangent of

f (x) = (x^{2} - 2) (x^{3} - 3 x), at x = - 2

limit as x approaches 0 of (x^2-5)/(x-2)

x \to 0 lim (\frac{x ^{2} - 5}{x - 2})

(d^2)/(dx^2)xy+2x^2=5

\frac{d ^{2}}{d x ^{2}} x y + 2 x^{2} = 5

integral of sin^3(x)*cos(x)

\int sin^{3} (x) \cdot cos (x) d x

(\partial)/(\partial x)(\sqrt[4]{6x^2+4})

Frequently Asked Questions (FAQ)

What is the solution for y^'=2y(5-y) ?
The solution for y^'=2y(5-y) is y= 5/(e^{-10t+c_{1)}+1}