What is the solution for y^{'''}-2y^{''}-y^'+2y=e^{4x} ?

The solution for y^{'''}-2y^{''}-y^'+2y=e^{4x} is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+1/30 e^{4x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{'''}-2y^{''}-y^'+2y=e^{4x}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{4 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{30} e^{4 x}

Solution steps

y^{'''} (x) - 2 y^{''} (x) - y^{'} (x) + 2 y = e^{4 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{30} e^{4 x}

y = c_{1} e^{2 x} + c_{2} e^{- x} + c_{3} e^{x} + \frac{1}{30} e^{4 x}

What is the solution for y^{'''}-2y^{''}-y^'+2y=e^{4x} ?
The solution for y^{'''}-2y^{''}-y^'+2y=e^{4x} is y=c_{1}e^{2x}+c_{2}e^{-x}+c_{3}e^x+1/30 e^{4x}