What is the solution for 0.2y^{''}+14y^'+1/(0.004)y=0 ?

The solution for 0.2y^{''}+14y^'+1/(0.004)y=0 is y=e^{-35t}(c_{1}cos(5t)+c_{2}sin(5t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0.2y^{''}+14y^'+1/(0.004)y=0

0.2 y^{^{''}} + 14 y^{^{'}} + \frac{1}{0.004} y = 0

y = e^{- 35 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

0.2 y^{''} (t) + 14 y^{'} (t) + \frac{1}{0.004} y = 0

Solve linear ODE:

y = e^{- 35 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{- 35 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

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

66 y^{^{'''}} + 17 y^{^{''}} + y^{^{'}} = 0

y^{^{'''}} - 15 y^{^{''}} + 27 y^{^{'}} + 243 y = 0

\frac{d ^{2} y}{d x ^{2}} (1) - \frac{d y}{d x} (5) + 4 y = 0

2 x^{^{''}} + 2 x^{^{'}} + 3 x = 0

What is the solution for 0.2y^{''}+14y^'+1/(0.004)y=0 ?
The solution for 0.2y^{''}+14y^'+1/(0.004)y=0 is y=e^{-35t}(c_{1}cos(5t)+c_{2}sin(5t))