What is the solution for 4y^{''}+4y^'+y=e^{(-x)/2} ?

The solution for 4y^{''}+4y^'+y=e^{(-x)/2} is y=c_{1}e^{-x/2}+c_{2}e^{-x/2}x+(e^{-x/2}x^2)/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4y^{''}+4y^'+y=e^{(-x)/2}

4 y^{^{''}} + 4 y^{^{'}} + y = e^{\frac{- x}{2}}

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{- \frac{x}{2}} x ^{2}}{8}

Solution steps

4 y^{''} (x) + 4 y^{'} (x) + y = e^{\frac{- x}{2}}

Solve linear ODE:

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{- \frac{x}{2}} x ^{2}}{8}

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{- \frac{x}{2}} x ^{2}}{8}

What is the solution for 4y^{''}+4y^'+y=e^{(-x)/2} ?
The solution for 4y^{''}+4y^'+y=e^{(-x)/2} is y=c_{1}e^{-x/2}+c_{2}e^{-x/2}x+(e^{-x/2}x^2)/8