What is the solution for 0.1y^{''}+0.4y^'+2y=0 ?

The solution for 0.1y^{''}+0.4y^'+2y=0 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0.1y^{''}+0.4y^'+2y=0

0.1 y^{^{''}} + 0.4 y^{^{'}} + 2 y = 0

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

0.1 y^{''} (t) + 0.4 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

\frac{d ^{2} x}{d t ^{2}} - x \frac{d x}{d x} = 0

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

144 y^{^{''}} - 24 y^{^{'}} + y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = - \frac{8}{5}, y^{^{'}} (0) = - 5

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 2, y (0) = - 7

What is the solution for 0.1y^{''}+0.4y^'+2y=0 ?
The solution for 0.1y^{''}+0.4y^'+2y=0 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))