What is the solution for 2y^'+4y=-8y^3 ?

The solution for 2y^'+4y=-8y^3 is y=sqrt(1/(e^{4t-c_{1))-2}},y=-sqrt(1/(e^{4t-c_{1))-2}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^'+4y=-8y^3

2 y^{^{'}} + 4 y = - 8 y^{3}

y = \frac{1}{e ^{4 t - c_{1}} - 2}, y = - \frac{1}{e ^{4 t - c_{1}} - 2}

Solution steps

2 y^{'} (t) + 4 y = - 8 y^{3}

Solve Bernoulli ODE:

y = \frac{1}{e ^{4 t - c_{1}} - 2}, y = - \frac{1}{e ^{4 t - c_{1}} - 2}

y = \frac{1}{e ^{4 t - c_{1}} - 2}, y = - \frac{1}{e ^{4 t - c_{1}} - 2}

y = f (y, x) (g (x))

\frac{d}{d x} (x y z)

x \to - 0.999 lim (\frac{1}{x ^{2}})

\frac{\partial}{\partial z} (y e^{z} sin (x z))

x \to - \infty lim (5 f (x))

What is the solution for 2y^'+4y=-8y^3 ?
The solution for 2y^'+4y=-8y^3 is y=sqrt(1/(e^{4t-c_{1))-2}},y=-sqrt(1/(e^{4t-c_{1))-2}}