What is the solution for linear y^{''}+3y^'+2y=0,y(0)=1,y^'(0)=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

linear y^{''}+3y^'+2y=0,y(0)=1,y^'(0)=0

linear

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = 2 e^{- t} - e^{- 2 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = 2 e^{- t} - e^{- 2 t}

y = 2 e^{- t} - e^{- 2 t}

\frac{d ^{2} y}{d x ^{2}} + 5^{2} y = 0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3

2 y^{^{''}} + 9 y^{^{'}} - 5 y = 0

y^{^{''}} + 1 y = 0, y (π^{2}) = 3, y^{^{'}} (π^{2}) = - 1

- 2 y^{^{''}} + 4 y^{^{'}} = 0

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