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Popular >

y^'=y(1-y),y(0)=8

Solution

y^{^{'}} = y (1 - y), y (0) = 8

Solution

y = \frac{8 e ^{t}}{- 7 + 8 e ^{t}}

Solution steps

y^{'} (t) = y (1 - y)

Solve Bernoulli ODE:

y = \frac{8 e ^{t}}{- 7 + 8 e ^{t}}

y = \frac{8 e ^{t}}{- 7 + 8 e ^{t}}

Popular Examples

derivative of xcos(x)-sin(x)derivative of

x cos (x) - sin (x)

derivative of 1/6 pir^3 derivative of

\frac{1}{6} π r^{3}

derivative of ln(sinh(9x))

\frac{d}{d x} (ln (sinh (9 x)))

(dy)/(dx)=(y^2-1)/(x^2-1),y(8)=8

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (8) = 8

derivative of (x^2+x^3)

\frac{d}{d x} ((x^{2} + x)^{3})

Frequently Asked Questions (FAQ)

What is the solution for y^'=y(1-y),y(0)=8 ?
The solution for y^'=y(1-y),y(0)=8 is y=(8e^t)/(-7+8e^t)