What is the solution for y^{''}+2y^'+y=3x+8 ?

The solution for y^{''}+2y^'+y=3x+8 is y=c_{1}e^{-x}+c_{2}xe^{-x}+3x+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+2y^'+y=3x+8

y^{^{''}} + 2 y^{^{'}} + y = 3 x + 8

y = c_{1} e^{- x} + c_{2} x e^{- x} + 3 x + 2

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = 3 x + 8

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + 3 x + 2

y = c_{1} e^{- x} + c_{2} x e^{- x} + 3 x + 2

What is the solution for y^{''}+2y^'+y=3x+8 ?
The solution for y^{''}+2y^'+y=3x+8 is y=c_{1}e^{-x}+c_{2}xe^{-x}+3x+2