What is the solution for (dy)/(dt)+ay=be^{-bt} ?

The solution for (dy)/(dt)+ay=be^{-bt} is y= b/(e^{bt)(-b+a)}+(c_{1})/(e^{at)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(dy)/(dt)+ay=be^{-bt}

\frac{d y}{d t} + a y = b e^{- b t}

y = \frac{b}{e ^{b t} ( - b + a )} + \frac{c _{1}}{e ^{a t}}

Solution steps

\frac{d y}{d t} + a y = b e^{- b t}

Solve linear ODE:

y = \frac{b}{e ^{b t} ( - b + a )} + \frac{c _{1}}{e ^{a t}}

y = \frac{b}{e ^{b t} ( - b + a )} + \frac{c _{1}}{e ^{a t}}

What is the solution for (dy)/(dt)+ay=be^{-bt} ?
The solution for (dy)/(dt)+ay=be^{-bt} is y= b/(e^{bt)(-b+a)}+(c_{1})/(e^{at)}