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Popular >

y^'=y-y^2,y(0)=10

Solution

y^{^{'}} = y - y^{2}, y (0) = 10

Solution

y = \frac{10 e ^{t}}{- 9 + 10 e ^{t}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{10 e ^{t}}{- 9 + 10 e ^{t}}

y = \frac{10 e ^{t}}{- 9 + 10 e ^{t}}

Popular Examples

derivative of 2x^{9/2}

\frac{d}{d x} (2 x^{\frac{9}{2}})

derivative of sin(xx^{-2})

\frac{d}{d x} (sin (x) x^{- 2})

integral of e^{-nx}

\int e^{- n x} d x

(x(dy)/(dx)+2y)=3

(x \frac{d y}{d x} + 2 y) = 3

tangent of f(x)=sqrt(9-x),\at x=2 tangent of

f (x) = 9 - x, at x = 2

Frequently Asked Questions (FAQ)

What is the solution for y^'=y-y^2,y(0)=10 ?
The solution for y^'=y-y^2,y(0)=10 is y=(10e^t)/(-9+10e^t)