What is the solution for y^{''}+2y^'+10y=10x+2 ?

The solution for y^{''}+2y^'+10y=10x+2 is y=e^{-x}(c_{1}cos(3x)+c_{2}sin(3x))+x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+2y^'+10y=10x+2

y^{^{''}} + 2 y^{^{'}} + 10 y = 10 x + 2

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 10 y = 10 x + 2

Solve linear ODE:

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

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

What is the solution for y^{''}+2y^'+10y=10x+2 ?
The solution for y^{''}+2y^'+10y=10x+2 is y=e^{-x}(c_{1}cos(3x)+c_{2}sin(3x))+x