What is the solution for y^{''}+3y^'+2y=6e^x ?

The solution for y^{''}+3y^'+2y=6e^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+3y^'+2y=6e^x

y^{^{''}} + 3 y^{^{'}} + 2 y = 6 e^{x}

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

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 6 e^{x}

Solve linear ODE:

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

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

q^{^{''}} + 100 q^{^{'}} + 2500 q = 30

y^{^{''}} - 7 y^{^{'}} + 4 y = x e^{x}

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - y - e^{x} = - x e^{x}

y^{^{''}} + 4 y = 20 (e^{x} - cos (2 x))

y^{^{''}} - 10 y^{^{'}} + 25 y = (x + 1) e^{x}

What is the solution for y^{''}+3y^'+2y=6e^x ?
The solution for y^{''}+3y^'+2y=6e^x is y=c_{1}e^{-x}+c_{2}e^{-2x}+e^x