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

The solution for y^{''}+4y^'+8y=0,y(0)=1,y^'(0)=1 is y=e^{-2t}(cos(2t)+3/2 sin(2t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{- 2 t} (cos (2 t) + \frac{3}{2} sin (2 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{- 2 t} (cos (2 t) + \frac{3}{2} sin (2 t))

y = e^{- 2 t} (cos (2 t) + \frac{3}{2} sin (2 t))

2 y^{^{''}} + 16 y = 0, y (0) = 0.125, y^{^{'}} (0) = 0

y^{^{''}} + 23 y^{^{'}} + 3 y = 0

y^{^{'''}} - 6 y^{^{''}} - 27 y^{^{'}} = 0

y^{^{''}} + 3 y^{^{'}} - 4 y = 0, y (0) = 7, y^{^{'}} (0) = 2

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} + 6 y = 0, y (0) = 2

What is the solution for y^{''}+4y^'+8y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+4y^'+8y=0,y(0)=1,y^'(0)=1 is y=e^{-2t}(cos(2t)+3/2 sin(2t))