What is the solution for y^{'''}-2y^{''}-y^'+2y=e^{5t} ?

The solution for y^{'''}-2y^{''}-y^'+2y=e^{5t} is y=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t+1/72 e^{5t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}-2y^{''}-y^'+2y=e^{5t}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{5 t}

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{72} e^{5 t}

Solution steps

y^{'''} (t) - 2 y^{''} (t) - y^{'} (t) + 2 y = e^{5 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{72} e^{5 t}

y = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{72} e^{5 t}

y^{^{''}} + 9 y^{^{'}} + 8 y = e^{2 t}

y^{^{'''}} + 7 y^{^{''}} + 8 y^{^{'}} - 16 y = - 3 e^{x} - 2

y^{^{''''}} + 2 y^{^{''}} + y = (x - 1)^{2}

y^{^{''}} + y^{^{'}} - 6 y = 2 t

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 8 + 2 sin (x)

What is the solution for y^{'''}-2y^{''}-y^'+2y=e^{5t} ?
The solution for y^{'''}-2y^{''}-y^'+2y=e^{5t} is y=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t+1/72 e^{5t}