What is the solution for y^{''}+y^'+y=1+x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+y^'+y=1+x

y^{^{''}} + y^{^{'}} + y = 1 + x

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

Solution steps

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

Solve linear ODE:

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

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

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