What is the solution for y^{''}+4y^'+3=5sin(t) ?

The solution for y^{''}+4y^'+3=5sin(t) is y=c_{1}+c_{2}e^{-4t}-5/17 sin(t)-20/17 cos(t)-(3t)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'+3=5sin(t)

y^{^{''}} + 4 y^{^{'}} + 3 = 5 sin (t)

y = c_{1} + c_{2} e^{- 4 t} - \frac{5}{17} sin (t) - \frac{20}{17} cos (t) - \frac{3 t}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 = 5 sin (t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - \frac{5}{17} sin (t) - \frac{20}{17} cos (t) - \frac{3 t}{4}

y = c_{1} + c_{2} e^{- 4 t} - \frac{5}{17} sin (t) - \frac{20}{17} cos (t) - \frac{3 t}{4}

What is the solution for y^{''}+4y^'+3=5sin(t) ?
The solution for y^{''}+4y^'+3=5sin(t) is y=c_{1}+c_{2}e^{-4t}-5/17 sin(t)-20/17 cos(t)-(3t)/4