What is the solution for y^{''}+2y^'+0.75x=0 ?

The solution for y^{''}+2y^'+0.75x=0 is y=c_{1}+c_{2}e^{-2x}-0.1875x^2+0.1875x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+2y^'+0.75x=0

y^{^{''}} + 2 y^{^{'}} + 0.75 x = 0

y = c_{1} + c_{2} e^{- 2 x} - 0.1875 x^{2} + 0.1875 x

Solution steps

y^{''} (x) + 2 y^{'} (x) + 0.75 x = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} - 0.1875 x^{2} + 0.1875 x

y = c_{1} + c_{2} e^{- 2 x} - 0.1875 x^{2} + 0.1875 x

What is the solution for y^{''}+2y^'+0.75x=0 ?
The solution for y^{''}+2y^'+0.75x=0 is y=c_{1}+c_{2}e^{-2x}-0.1875x^2+0.1875x