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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+y=t^{(-1)}e^{(-t)}

y^{^{''}} + 2 y^{^{'}} + y = t^{(- 1)} e^{(- t)}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = t^{(- 1)} e^{(- t)}

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{3 t}, y (0) = 3, y^{^{'}} (0) = 1

y^{^{''}} + 6 y^{^{'}} = 5 x

(D^{2} + 2 D - 3) y = cos (2 x)

y^{^{''}} - 8 y^{^{'}} = 6 e^{2 x} - 6 x^{2} - 9 x - 14

y^{^{''}} + 10 y^{^{'}} + 41 = 0

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