What is the solution for y^{''}+6y^'+8y=4sin(2t) ?

The solution for y^{''}+6y^'+8y=4sin(2t) is y=c_{1}e^{-2t}+c_{2}e^{-4t}+1/10 sin(2t)-3/10 cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+6y^'+8y=4sin(2t)

y^{^{''}} + 6 y^{^{'}} + 8 y = 4 sin (2 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{1}{10} sin (2 t) - \frac{3}{10} cos (2 t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 8 y = 4 sin (2 t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{1}{10} sin (2 t) - \frac{3}{10} cos (2 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{1}{10} sin (2 t) - \frac{3}{10} cos (2 t)

\int \frac{x ^{2} + 1}{( x - 7 ) ( x - 6 ) ^{2}} d x

y^{^{'}} = - 0.5 y + x

\int \frac{y}{2} d y

d er i v a t i v e f (x) = 5 x 2

\int 17 (tan (x)) (ln (cos (x))) d x

What is the solution for y^{''}+6y^'+8y=4sin(2t) ?
The solution for y^{''}+6y^'+8y=4sin(2t) is y=c_{1}e^{-2t}+c_{2}e^{-4t}+1/10 sin(2t)-3/10 cos(2t)