What is the solution for (dP)/(dt)=a-bP(t) ?

The solution for (dP)/(dt)=a-bP(t) is P=-(e^{-bt-c_{1}b})/b+a/b

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dP)/(dt)=a-bP(t)

\frac{d P}{d t} = a - b P (t)

P = - \frac{e ^{- b t - c_{1} b}}{b} + \frac{a}{b}

Solution steps

\frac{d P}{d t} = a - b P

Solve separable ODE:

P = - \frac{e ^{- b t - c_{1} b}}{b} + \frac{a}{b}

P = - \frac{e ^{- b t - c_{1} b}}{b} + \frac{a}{b}

\frac{d y}{d x} = \frac{( x y ^{3} )}{1 + x ^{2}}

\frac{1}{x} y^{^{'}} - y - 1 = 0

y^{^{'}} - 2 y = 10, y (0) = 0

y^{2} sec (4 x) \frac{d y}{d x} = 1, y (0) = 0

y^{^{'}} sin (y) + x cos (y) + x = 0

What is the solution for (dP)/(dt)=a-bP(t) ?
The solution for (dP)/(dt)=a-bP(t) is P=-(e^{-bt-c_{1}b})/b+a/b