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)

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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_{1}^{3} x - 1 d x

\frac{d y}{d x} = \frac{e ^{x - y}}{1 + e ^{x}}

\int (1 + 4 x)^{- \frac{1}{2}} d x

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