What is the solution for y^{''}+22y^'+121y=16.3u ?

The solution for y^{''}+22y^'+121y=16.3u is y=c_{1}e^{-11t}+c_{2}te^{-11t}+(163u)/(1210)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+22y^'+121y=16.3u

y^{^{''}} + 22 y^{^{'}} + 121 y = 16.3 u

y = c_{1} e^{- 11 t} + c_{2} t e^{- 11 t} + \frac{163 u}{1210}

Solution steps

y^{''} (t) + 22 y^{'} (t) + 121 y = 16.3 u

Solve linear ODE:

y = c_{1} e^{- 11 t} + c_{2} t e^{- 11 t} + \frac{163 u}{1210}

y = c_{1} e^{- 11 t} + c_{2} t e^{- 11 t} + \frac{163 u}{1210}

y^{^{''}} + y = 2 e^{- 8 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'''}} - 6 y^{^{'}} - 7 = 0

\frac{d ^{2} y}{d x ^{2}} + 9 y = 15 x^{2} - 9

y^{^{''}} + 6 y^{^{'}} + 8 y = 2 t + e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x}, y (1) = 0, y^{^{'}} (1) = 1

What is the solution for y^{''}+22y^'+121y=16.3u ?
The solution for y^{''}+22y^'+121y=16.3u is y=c_{1}e^{-11t}+c_{2}te^{-11t}+(163u)/(1210)