What is the solution for y^{''}+4y^'+3y=x-1 ?

The solution for y^{''}+4y^'+3y=x-1 is y=c_{1}e^{-x}+c_{2}e^{-3x}+x/3-7/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'+3y=x-1

y^{^{''}} + 4 y^{^{'}} + 3 y = x - 1

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{x}{3} - \frac{7}{9}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = x - 1

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{x}{3} - \frac{7}{9}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{x}{3} - \frac{7}{9}

What is the solution for y^{''}+4y^'+3y=x-1 ?
The solution for y^{''}+4y^'+3y=x-1 is y=c_{1}e^{-x}+c_{2}e^{-3x}+x/3-7/9