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Popular >

y^'=y(1-y),y(0)=-5

Solution

y^{^{'}} = y (1 - y), y (0) = - 5

Solution

y = \frac{5 e ^{t}}{- 6 + 5 e ^{t}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{5 e ^{t}}{- 6 + 5 e ^{t}}

y = \frac{5 e ^{t}}{- 6 + 5 e ^{t}}

Popular Examples

sum from n=0 to infinity of 1/((1+n)^2)

n = 0 \sum \infty \frac{1}{( 1 + n ) ^{2}}

derivative of 3cos(3x)

\frac{d}{d x} (3 cos (3 x))

derivative of 4x^{5/4}+10x^{3/2}+5x derivative of

4 x^{\frac{5}{4}} + 10 x^{\frac{3}{2}} + 5 x

slope of (1,-2),(3,1)slope

(1, - 2), (3, 1)

y^{''}+36y=-60sin(6t)

y^{^{''}} + 36 y = - 60 sin (6 t)

Frequently Asked Questions (FAQ)

What is the solution for y^'=y(1-y),y(0)=-5 ?
The solution for y^'=y(1-y),y(0)=-5 is y=(5e^t)/(-6+5e^t)