What is the solution for y^{''}+6y^'+8y=130cos(3t) ?

The solution for y^{''}+6y^'+8y=130cos(3t) is y=c_{1}e^{-2t}+c_{2}e^{-4t}+36/5 sin(3t)-2/5 cos(3t)

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y^{''}+6y^'+8y=130cos(3t)

y^{^{''}} + 6 y^{^{'}} + 8 y = 130 cos (3 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{36}{5} sin (3 t) - \frac{2}{5} cos (3 t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 8 y = 130 cos (3 t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{36}{5} sin (3 t) - \frac{2}{5} cos (3 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 4 t} + \frac{36}{5} sin (3 t) - \frac{2}{5} cos (3 t)

x \to - \frac{1}{e} lim (x ln (e + \frac{1}{x}))

\int x 1 - 2 x d x

\frac{\partial}{\partial x} (\frac{x}{e ^{\frac{x}{y}}})

\frac{d}{d x} (- \frac{2}{( 6 x - 7 x ^{4} ) ^{3}})

\frac{d y}{d x} = lo g_{10} (x)

What is the solution for y^{''}+6y^'+8y=130cos(3t) ?
The solution for y^{''}+6y^'+8y=130cos(3t) is y=c_{1}e^{-2t}+c_{2}e^{-4t}+36/5 sin(3t)-2/5 cos(3t)