What is the solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 ?

The solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 is y=c_{1}e^{-3x}+c_{2}cos(x)+c_{3}sin(x)+1/8 e^x+4/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{'''}+3y^{''}+y^'+3y=e^x+4

y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} + 3 y = e^{x} + 4

y = c_{1} e^{- 3 x} + c_{2} cos (x) + c_{3} sin (x) + \frac{1}{8} e^{x} + \frac{4}{3}

Solution steps

y^{'''} (x) + 3 y^{''} (x) + y^{'} (x) + 3 y = e^{x} + 4

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} cos (x) + c_{3} sin (x) + \frac{1}{8} e^{x} + \frac{4}{3}

y = c_{1} e^{- 3 x} + c_{2} cos (x) + c_{3} sin (x) + \frac{1}{8} e^{x} + \frac{4}{3}

What is the solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 ?
The solution for y^{'''}+3y^{''}+y^'+3y=e^x+4 is y=c_{1}e^{-3x}+c_{2}cos(x)+c_{3}sin(x)+1/8 e^x+4/3