What is the solution for (dy)/(dx)=kae^{-kx}-by ?

The solution for (dy)/(dx)=kae^{-kx}-by is y=(ka)/(e^{kx)(-k+b)}+(c_{1})/(e^{bx)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=kae^{-kx}-by

\frac{d y}{d x} = ka e^{- k x} - b y

y = \frac{ka}{e ^{k x} ( - k + b )} + \frac{c _{1}}{e ^{b x}}

Solution steps

\frac{d y}{d x} = ka e^{- k x} - b y (x)

Solve linear ODE:

y = \frac{ka}{e ^{k x} ( - k + b )} + \frac{c _{1}}{e ^{b x}}

y = \frac{ka}{e ^{k x} ( - k + b )} + \frac{c _{1}}{e ^{b x}}

\frac{s}{( s ^{2} - 1 ^{2} ) ^{2}}

n = 1 \sum \infty \frac{1}{n ^{3} - n}

y = \frac{3 x}{e ^{x}}

e^{- 3 x}, 2

\int \frac{3 x + 7}{( x + 4 ) ^{2}} d x

What is the solution for (dy)/(dx)=kae^{-kx}-by ?
The solution for (dy)/(dx)=kae^{-kx}-by is y=(ka)/(e^{kx)(-k+b)}+(c_{1})/(e^{bx)}