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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 + lim (\frac{x ^{2}}{8} - \frac{6}{x})

\frac{\partial}{\partial x} (sin (ln (x)))

\frac{\partial}{\partial y} (x^{- 7} y^{2} + x y^{2} + 3 x y)

\frac{d y}{d x} = - 10 y, y (0) = 5

\int_{0}^{\frac{1}{2}} 8 x (\frac{1}{2} - x) d x

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