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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

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

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

100 y^{^{''}} + 180 y^{^{'}} + 145 y = 0

y^{^{''}} + 4 y = 0, y (0) = 5, y^{^{'}} (0) = - 3

(D^{2} + 64)^{2} y = 0

y^{^{''}} = a^{2} y

4 \cdot \frac{d ^{2} y}{d x ^{2}} + 12 \cdot \frac{d y}{d x} + 9 y = 0

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