What is the solution for y^{''}+5y^'+4y=3-2x ?

The solution for y^{''}+5y^'+4y=3-2x is y=c_{1}e^{-x}+c_{2}e^{-4x}-x/2+11/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+5y^'+4y=3-2x

y^{^{''}} + 5 y^{^{'}} + 4 y = 3 - 2 x

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{x}{2} + \frac{11}{8}

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 y = 3 - 2 x

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{x}{2} + \frac{11}{8}

y = c_{1} e^{- x} + c_{2} e^{- 4 x} - \frac{x}{2} + \frac{11}{8}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 54 x + 18

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 5 e^{2 x}

y^{^{''}} - y^{^{'}} - 2 y = 7 x + 4

y^{^{''}} + y^{^{'}} = cos (\frac{t}{2})

y^{^{''}} - 4 y = t^{2} + 1

What is the solution for y^{''}+5y^'+4y=3-2x ?
The solution for y^{''}+5y^'+4y=3-2x is y=c_{1}e^{-x}+c_{2}e^{-4x}-x/2+11/8