What is the solution for y^'=(y+2)^2 ?

The solution for y^'=(y+2)^2 is y=-1/(t+c_{1)}-(2t)/(t+c_{1)}-(2c_{1})/(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+2)^2

y^{^{'}} = (y + 2)^{2}

y = - \frac{1}{t + c _{1}} - \frac{2 t}{t + c _{1}} - \frac{2 c _{1}}{t + c _{1}}

Solution steps

y^{'} (t) = (y + 2)^{2}

Solve separable ODE:

y = - \frac{1}{t + c _{1}} - \frac{2 t}{t + c _{1}} - \frac{2 c _{1}}{t + c _{1}}

y = - \frac{1}{t + c _{1}} - \frac{2 t}{t + c _{1}} - \frac{2 c _{1}}{t + c _{1}}

\frac{d y}{d x} = 12 x y - 2 x

y^{^{'}} = (\frac{x}{y}) (\frac{y ^{2} + 1}{x ^{2} - 1})

\frac{d y}{d x} + 9 x e^{y} = 0

\frac{d x}{d t} + \frac{x}{100} = 75, x (0) = 5000

f^{^{'}} (x) = - 2 f (x)

What is the solution for y^'=(y+2)^2 ?
The solution for y^'=(y+2)^2 is y=-1/(t+c_{1)}-(2t)/(t+c_{1)}-(2c_{1})/(t+c_{1)}