What is the solution for y^{''}-4y^'-12y=sin(2t) ?

The solution for y^{''}-4y^'-12y=sin(2t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-1/20 sin(2t)+1/40 cos(2t)

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y^{''}-4y^'-12y=sin(2t)

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{1}{20} sin (2 t) + \frac{1}{40} cos (2 t)

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = sin (2 t)

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{1}{20} sin (2 t) + \frac{1}{40} cos (2 t)

y = c_{1} e^{6 t} + c_{2} e^{- 2 t} - \frac{1}{20} sin (2 t) + \frac{1}{40} cos (2 t)

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\int \frac{ln ( x ^{18} )}{x} d x

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\frac{\partial}{\partial y} (\frac{x ^{2} y}{x ^{2} + y ^{2}})

\frac{d y}{d x} = \frac{e ^{- 2 x}}{1 + e ^{- 2 x}}

What is the solution for y^{''}-4y^'-12y=sin(2t) ?
The solution for y^{''}-4y^'-12y=sin(2t) is y=c_{1}e^{6t}+c_{2}e^{-2t}-1/20 sin(2t)+1/40 cos(2t)