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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4y^{''}+4y^'+y=3xe^x

4 y^{^{''}} + 4 y^{^{'}} + y = 3 x e^{x}

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{x} x}{3} - \frac{4}{9} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{x} x}{3} - \frac{4}{9} e^{x}

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{x} x}{3} - \frac{4}{9} e^{x}

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