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Popular >

y^'=y(y+1)

Solution

y^{^{'}} = y (y + 1)

Solution

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

Solution steps

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

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}}}

Popular Examples

inverse oflaplace 2/(s^2+9)inverse laplace

\frac{2}{s ^{2} + 9}

integral of ln(at+sqrt(1+(at)^2))

\int ln (a t + 1 + (a t)^{2}) d t

integral of (x^3)/((4+x^2)^2)

\int \frac{x ^{3}}{( 4 + x ^{2} ) ^{2}} d x

integral of 1/(x^2-9x+18)

\int \frac{1}{x ^{2} - 9 x + 18} d x

integral of (sqrt(25-x^2))/(x^2)

\int \frac{25 - x ^{2}}{x ^{2}} d x

Frequently Asked Questions (FAQ)

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