What is the solution for 3y^{''}+27y^'+42y=25sin(t) ?

The solution for 3y^{''}+27y^'+42y=25sin(t) is y=c_{1}e^{-2t}+c_{2}e^{-7t}+13/30 sin(t)-3/10 cos(t)

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3y^{''}+27y^'+42y=25sin(t)

3 y^{^{''}} + 27 y^{^{'}} + 42 y = 25 sin (t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 7 t} + \frac{13}{30} sin (t) - \frac{3}{10} cos (t)

Solution steps

3 y^{''} (t) + 27 y^{'} (t) + 42 y = 25 sin (t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 7 t} + \frac{13}{30} sin (t) - \frac{3}{10} cos (t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 7 t} + \frac{13}{30} sin (t) - \frac{3}{10} cos (t)

y^{^{'''}} - 4 y^{^{'}} = x + 3 cos (x) + e^{- 2 x}

y^{^{''}} - 9 y^{^{'}} + 8 y = x

y^{^{''}} + 14 y^{^{'}} + 50 y = 371 cos (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = sin (3 x) e^{2 x}

\frac{d ^{3} x}{d t ^{3}} + x = \frac{1}{2} t^{2} e^{t}

What is the solution for 3y^{''}+27y^'+42y=25sin(t) ?
The solution for 3y^{''}+27y^'+42y=25sin(t) is y=c_{1}e^{-2t}+c_{2}e^{-7t}+13/30 sin(t)-3/10 cos(t)