What is the solution for y^'=y(1-0.0004y) ?

The solution for y^'=y(1-0.0004y) is y=-(e^{t+c_{1}})/(1-0.0004e^{t+c_{1)}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=y(1-0.0004y)

y^{^{'}} = y (1 - 0.0004 y)

y = - \frac{e ^{t + c_{1}}}{1 - 0.0004 e ^{t + c_{1}}}

Solution steps

y^{'} (t) = y (1 - 0.0004 y)

Solve separable ODE:

y = - \frac{e ^{t + c_{1}}}{1 - 0.0004 e ^{t + c_{1}}}

y = - \frac{e ^{t + c_{1}}}{1 - 0.0004 e ^{t + c_{1}}}

x = 4 y^{2}, x = 8 + 2 y^{2}

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

t^{2} y^{^{'}} + 2 t y = 3

\int \frac{3}{x ( x ^{2} + 25 )} d x

6 x^{3} + \frac{7}{x}

What is the solution for y^'=y(1-0.0004y) ?
The solution for y^'=y(1-0.0004y) is y=-(e^{t+c_{1}})/(1-0.0004e^{t+c_{1)}}