What is the solution for 1/(-0.75y-0.00003)y^'=1 ?

The solution for 1/(-0.75y-0.00003)y^'=1 is y=-(e^{-0.75t-c_{1}})/(0.75)-(0.00003)/(0.75)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/(-0.75y-0.00003)y^'=1

\frac{1}{- 0.75 y - 0.00003} y^{^{'}} = 1

y = - \frac{e ^{- 0.75 t - c_{1}}}{0.75} - \frac{0.00003}{0.75}

Solution steps

\frac{1}{- 0.75 y - 0.00003} y^{'} (t) = 1

Solve separable ODE:

y = - \frac{e ^{- 0.75 t - c_{1}}}{0.75} - \frac{0.00003}{0.75}

y = - \frac{e ^{- 0.75 t - c_{1}}}{0.75} - \frac{0.00003}{0.75}

\int x^{3} x^{2} + 5 d x

\int \frac{9 - 162 x}{49 - 81 x ^{2}} d x

y \to 5 + lim (y - 5)

\int csc^{2} (y) cot (y) d y

What is the solution for 1/(-0.75y-0.00003)y^'=1 ?
The solution for 1/(-0.75y-0.00003)y^'=1 is y=-(e^{-0.75t-c_{1}})/(0.75)-(0.00003)/(0.75)