What is the solution for 1500/230 (1-30x^')= x/(0.1)+75x^' ?

The solution for 1500/230 (1-30x^')= x/(0.1)+75x^' is x=(e^{-0.03694…t-c_{1}})/(23)+15/23

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1500/230 (1-30x^')= x/(0.1)+75x^'

\frac{1500}{230} (1 - 30 x^{^{'}}) = \frac{x}{0.1} + 75 x^{^{'}}

x = \frac{e ^{- 0.03694 \dots t - c_{1}}}{23} + \frac{15}{23}

Solution steps

\frac{1500}{230} (1 - 30 x^{'} (t)) = \frac{x}{0.1} + 75 x^{'} (t)

Solve separable ODE:

x = \frac{e ^{- 0.03694 \dots t - c_{1}}}{23} + \frac{15}{23}

x = \frac{e ^{- 0.03694 \dots t - c_{1}}}{23} + \frac{15}{23}

y^{^{'}} + a y + \frac{2 \cdot y}{x} = 0

\frac{d P}{d t} = P - P^{- 2}

y^{^{'}} = \frac{2 y}{7 x}

\frac{d y}{d x} = y^{2} - y + 2

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

What is the solution for 1500/230 (1-30x^')= x/(0.1)+75x^' ?
The solution for 1500/230 (1-30x^')= x/(0.1)+75x^' is x=(e^{-0.03694…t-c_{1}})/(23)+15/23