What is the solution for y^{''}+2y^'+y=-y^{''}+5y+2y^'-y ?

The solution for y^{''}+2y^'+y=-y^{''}+5y+2y^'-y is y=c_{1}e^{sqrt(3/2)t}+c_{2}e^{-sqrt(3/2)t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+2y^'+y=-y^{''}+5y+2y^'-y

y^{^{''}} + 2 y^{^{'}} + y = - y^{^{''}} + 5 y + 2 y^{^{'}} - y

y = c_{1} e^{\frac{3}{2} t} + c_{2} e^{- \frac{3}{2} t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = - y^{''} (t) + 5 y + 2 y^{'} (t) - y

Solve linear ODE:

y = c_{1} e^{\frac{3}{2} t} + c_{2} e^{- \frac{3}{2} t}

y = c_{1} e^{\frac{3}{2} t} + c_{2} e^{- \frac{3}{2} t}

y^{^{''}} - 6 y^{^{'}} + 32 y = 0

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

y^{^{''}} - 12 y^{^{'}} + 45 y = 0, y (0) = 5, y^{^{'}} (0) = - 3

10 y^{^{''}} + 45 y^{^{'}} + 2500 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 6

What is the solution for y^{''}+2y^'+y=-y^{''}+5y+2y^'-y ?
The solution for y^{''}+2y^'+y=-y^{''}+5y+2y^'-y is y=c_{1}e^{sqrt(3/2)t}+c_{2}e^{-sqrt(3/2)t}