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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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