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

The solution for y^{''}+2y^'+y=t^2+1-e^t is y=c_{1}e^{-t}+c_{2}te^{-t}+t^2-4t+7-1/4 e^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^2+1-e^t

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

y = c_{1} e^{- t} + c_{2} t e^{- t} + t^{2} - 4 t + 7 - \frac{1}{4} e^{t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + t^{2} - 4 t + 7 - \frac{1}{4} e^{t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + t^{2} - 4 t + 7 - \frac{1}{4} e^{t}

\frac{d}{d x} (3 tan (x) + co tan (3 x))

\frac{d}{d x} (cos (2 x) y)

y = x^{2} - 6, y = 3

y = e^{3 x}, y = e^{x}, [0, 2]

\frac{d}{d x} (\frac{24}{x ^{5}})

What is the solution for y^{''}+2y^'+y=t^2+1-e^t ?
The solution for y^{''}+2y^'+y=t^2+1-e^t is y=c_{1}e^{-t}+c_{2}te^{-t}+t^2-4t+7-1/4 e^t