What is the solution for y^'+ky=e^{-kx} ?

The solution for y^'+ky=e^{-kx} is y= x/(e^{kx)}+(c_{1})/(e^{kx)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+ky=e^{-kx}

y^{^{'}} + k y = e^{- k x}

y = \frac{x}{e ^{k x}} + \frac{c _{1}}{e ^{k x}}

Solution steps

y^{'} (x) + k y (x) = e^{- k x}

Solve linear ODE:

y = \frac{x}{e ^{k x}} + \frac{c _{1}}{e ^{k x}}

y = \frac{x}{e ^{k x}} + \frac{c _{1}}{e ^{k x}}

x (1 - 5 x)

x \to 0 lim (1 - cos (2 x))

\int \frac{8}{9} x^{- \frac{1}{9}} d x

x \to 3 lim (3 x)

\int \frac{1}{9} x^{2} d x

What is the solution for y^'+ky=e^{-kx} ?
The solution for y^'+ky=e^{-kx} is y= x/(e^{kx)}+(c_{1})/(e^{kx)}