What is the solution for y^{''}+8y^'+16y=0,y(-1)=4,y^'(-1)=2 ?

The solution for y^{''}+8y^'+16y=0,y(-1)=4,y^'(-1)=2 is y=22e^{-4t-4}+18e^{-4t-4}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+16y=0,y(-1)=4,y^'(-1)=2

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 4, y^{^{'}} (- 1) = 2

y = 22 e^{- 4 t - 4} + 18 e^{- 4 t - 4} t

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = 22 e^{- 4 t - 4} + 18 e^{- 4 t - 4} t

y = 22 e^{- 4 t - 4} + 18 e^{- 4 t - 4} t

(D^{3} - 5 D^{2} + 8 D - 4) y = 0

y^{^{''}} + b \cdot y = 0, y (0) - 9 y^{^{'}} (0) = 0, y^{^{'}} (9) = 0

y^{^{'''}} + 6 y^{^{''}} + 9 y^{^{'}} + 4 y = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 6, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} = 0

What is the solution for y^{''}+8y^'+16y=0,y(-1)=4,y^'(-1)=2 ?
The solution for y^{''}+8y^'+16y=0,y(-1)=4,y^'(-1)=2 is y=22e^{-4t-4}+18e^{-4t-4}t