What is the solution for y^{''}-y^'-2y=e^{-t},y(0)=-1,y^'(0)=2 ?

The solution for y^{''}-y^'-2y=e^{-t},y(0)=-1,y^'(0)=2 is y=(-1+13/9)e^{2t}-13/9 e^{-t}-(e^{-t}t)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = (- 1 + \frac{13}{9}) e^{2 t} - \frac{13}{9} e^{- t} - \frac{e ^{- t} t}{3}

Solution steps

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

Solve linear ODE:

y = (- 1 + \frac{13}{9}) e^{2 t} - \frac{13}{9} e^{- t} - \frac{e ^{- t} t}{3}

y = (- 1 + \frac{13}{9}) e^{2 t} - \frac{13}{9} e^{- t} - \frac{e ^{- t} t}{3}

2 y^{^{''}} + 9 y^{^{'}} - y = 22

y^{^{''}} - 2 y^{^{'}} + y = (3 + 5 x) e^{2 x}

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 2, y^{^{''}} (0) = - 1

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

y^{^{''}} - 2 y^{^{'}} + y = x^{\frac{- 1}{2}} \cdot e^{x}

What is the solution for y^{''}-y^'-2y=e^{-t},y(0)=-1,y^'(0)=2 ?
The solution for y^{''}-y^'-2y=e^{-t},y(0)=-1,y^'(0)=2 is y=(-1+13/9)e^{2t}-13/9 e^{-t}-(e^{-t}t)/3