What is the solution for y^{''}+5y^'+6y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+5y^'+6y=0,y(0)=1,y^'(0)=2 is y=5e^{-2t}-4e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+5y^'+6y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = 5 e^{- 2 t} - 4 e^{- 3 t}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = 5 e^{- 2 t} - 4 e^{- 3 t}

y = 5 e^{- 2 t} - 4 e^{- 3 t}

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 25 y = 0

y^{^{''}} + k^{2} y = 0

y^{^{''}} - 17 y^{^{'}} + 72 y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{'''}} = y^{^{'}}

What is the solution for y^{''}+5y^'+6y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+5y^'+6y=0,y(0)=1,y^'(0)=2 is y=5e^{-2t}-4e^{-3t}