What is the solution for y^{''}-y=(e^x)/(e^x) ?

The solution for y^{''}-y=(e^x)/(e^x) is y=c_{1}e^x+c_{2}e^{-x}-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-y=(e^x)/(e^x)

y^{^{''}} - y = \frac{e ^{x}}{e ^{x}}

y = c_{1} e^{x} + c_{2} e^{- x} - 1

Solution steps

y^{''} (x) - y = \frac{e ^{x}}{e ^{x}}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} - 1

y = c_{1} e^{x} + c_{2} e^{- x} - 1

What is the solution for y^{''}-y=(e^x)/(e^x) ?
The solution for y^{''}-y=(e^x)/(e^x) is y=c_{1}e^x+c_{2}e^{-x}-1