What is the solution for y^{''}+2y^'+y=19e^{-t} ?

The solution for y^{''}+2y^'+y=19e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(19e^{-t}t^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+y=19e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 19 e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{19 e ^{- t} t ^{2}}{2}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 19 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{19 e ^{- t} t ^{2}}{2}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{19 e ^{- t} t ^{2}}{2}

\frac{d}{d x} (- 3 e^{x})

\frac{d ^{2}}{d x ^{2}} (x^{\frac{6}{7}} + 4 x)

\frac{\partial}{\partial x} (k!)

\frac{d}{d x} (\frac{6}{1 + e ^{- x}})

\int_{1}^{6} \frac{x ^{2} + 4}{7 x - x ^{2}} d x

What is the solution for y^{''}+2y^'+y=19e^{-t} ?
The solution for y^{''}+2y^'+y=19e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(19e^{-t}t^2)/2