What is the solution for y^'=(2x+y)/(2x+y+3) ?

The solution for y^'=(2x+y)/(2x+y+3) is y=\W(1/(3e^{-3x-c_{1)-2}})-2-2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(2x+y)/(2x+y+3)

y^{^{'}} = \frac{2 x + y}{2 x + y + 3}

y = W (\frac{1}{3 e ^{- 3 x - c_{1} - 2}}) - 2 - 2 x

Solution steps

y^{'} (x) = \frac{2 x + y}{2 x + y + 3}

Solve using substitution:

y = W (\frac{1}{3 e ^{- 3 x - c_{1} - 2}}) - 2 - 2 x

y = W (\frac{1}{3 e ^{- 3 x - c_{1} - 2}}) - 2 - 2 x

\int sin^{2} (5 x) d x

\frac{\partial}{\partial x} (5 x^{2} + y^{2})

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (5) = 5

\frac{d}{d x} ((3 x - 2)^{5})

What is the solution for y^'=(2x+y)/(2x+y+3) ?
The solution for y^'=(2x+y)/(2x+y+3) is y=\W(1/(3e^{-3x-c_{1)-2}})-2-2x