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

Popular >

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}

\frac{d ^{2} y}{d x ^{2}} + k \frac{d y}{d x} - π y = 0

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

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = sin (3 x)

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

y^{^{''''}} - y = 0

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}