What is the solution for y^{''}+6y^'+34y=68 ?

The solution for y^{''}+6y^'+34y=68 is y=e^{-3t}(c_{1}cos(5t)+c_{2}sin(5t))+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+6y^'+34y=68

y^{^{''}} + 6 y^{^{'}} + 34 y = 68

y = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t)) + 2

Solution steps

y^{''} (t) + 6 y^{'} (t) + 34 y = 68

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t)) + 2

y = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t)) + 2

x^{^{''}} + 16 x = cos (2 t)

(8 x^{^{''}} + 4 x^{^{'}} - 24) = 48 e^{2 t}

y^{^{''}} + 2 y^{^{'}} = 2 e^{- 2 t}

2 y^{^{''}} + y^{^{'}} - y = 3 x + 4

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2} - 4

What is the solution for y^{''}+6y^'+34y=68 ?
The solution for y^{''}+6y^'+34y=68 is y=e^{-3t}(c_{1}cos(5t)+c_{2}sin(5t))+2