What is the solution for x^{''}+3x^'+2x=sin(3t) ?

The solution for x^{''}+3x^'+2x=sin(3t) is x=c_{1}e^{-t}+c_{2}e^{-2t}-7/130 sin(3t)-9/130 cos(3t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+3x^'+2x=sin(3t)

x^{^{''}} + 3 x^{^{'}} + 2 x = sin (3 t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} - \frac{7}{130} sin (3 t) - \frac{9}{130} cos (3 t)

Solution steps

x^{''} (t) + 3 x^{'} (t) + 2 x = sin (3 t)

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 2 t} - \frac{7}{130} sin (3 t) - \frac{9}{130} cos (3 t)

x = c_{1} e^{- t} + c_{2} e^{- 2 t} - \frac{7}{130} sin (3 t) - \frac{9}{130} cos (3 t)

x^{^{''}} + 8 x^{^{'}} + 16 = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- t}

x^{^{''''}} - x^{^{''}} = 1

y^{^{''}} - 2 y^{^{'}} + y = 3 e^{t}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 4 y^{^{'}} - 5 y = 125

What is the solution for x^{''}+3x^'+2x=sin(3t) ?
The solution for x^{''}+3x^'+2x=sin(3t) is x=c_{1}e^{-t}+c_{2}e^{-2t}-7/130 sin(3t)-9/130 cos(3t)