What is the solution for y^'=(y-y^2)/x ?

The solution for y^'=(y-y^2)/x is y=-(e^{c_{1}}x)/(1-e^{c_{1)}x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(y-y^2)/x

y^{^{'}} = \frac{y - y ^{2}}{x}

y = - \frac{e ^{c_{1}} x}{1 - e ^{c_{1}} x}

Solution steps

y^{'} (x) = \frac{y - y ^{2}}{x}

Solve separable ODE:

y = - \frac{e ^{c_{1}} x}{1 - e ^{c_{1}} x}

y = - \frac{e ^{c_{1}} x}{1 - e ^{c_{1}} x}

x^{\frac{2}{5}}

9 - t

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 3 x}{x + 1})

n = 1 \sum \infty 6 e^{- 5 n}

\int \frac{12 x ^{3} - 24 x ^{2} + 5}{x ^{2} - 2 x} d x

What is the solution for y^'=(y-y^2)/x ?
The solution for y^'=(y-y^2)/x is y=-(e^{c_{1}}x)/(1-e^{c_{1)}x}