What is the solution for x^'=1-2.2x ?

The solution for x^'=1-2.2x is x=-(e^{-2.19999…t-c_{1}})/(2.2)+1/(2.2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^'=1-2.2x

x^{^{'}} = 1 - 2.2 x

x = - \frac{e ^{- 2.19999 \dots t - c_{1}}}{2.2} + \frac{1}{2.2}

Solution steps

x^{'} (t) = 1 - 2.2 x

Solve separable ODE:

x = - \frac{e ^{- 2.19999 \dots t - c_{1}}}{2.2} + \frac{1}{2.2}

x = - \frac{e ^{- 2.19999 \dots t - c_{1}}}{2.2} + \frac{1}{2.2}

y^{^{'}} = 4 x^{3} y^{4}

\frac{d y}{d x} = 6 x^{4} y, y > 0

\frac{d y}{d x} - 2 y^{2} cos^{2} (x) = 3 x^{2} y^{2}

\frac{d y}{d x} y = 3 x^{5} + 4 x^{2} + x^{2} + 1000

y^{^{'}} - 2 y^{^{'}} - 3 y = 0

What is the solution for x^'=1-2.2x ?
The solution for x^'=1-2.2x is x=-(e^{-2.19999…t-c_{1}})/(2.2)+1/(2.2)