What is the solution for y^{''}-y^'-2y=sin(3t) ?

The solution for y^{''}-y^'-2y=sin(3t) is y=c_{1}e^{2t}+c_{2}e^{-t}-11/130 sin(3t)+3/130 cos(3t)

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y^{''}-y^'-2y=sin(3t)

y^{^{''}} - y^{^{'}} - 2 y = sin (3 t)

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{11}{130} sin (3 t) + \frac{3}{130} cos (3 t)

Solution steps

y^{''} (t) - y^{'} (t) - 2 y = sin (3 t)

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{11}{130} sin (3 t) + \frac{3}{130} cos (3 t)

y = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{11}{130} sin (3 t) + \frac{3}{130} cos (3 t)

y^{^{''}} + 6 y = 4 t^{2} - 3, y (0) = 0, y^{^{'}} (0) = - 7

y^{^{''}} + 10 y^{^{'}} + 16 y = 2581 cos (5 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{x}}{1 + e ^{x}}

y^{^{''}} + y^{^{'}} + \frac{1}{4} y = e^{x} (sin (3 x) - cos (3 x))

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{( 1 + x ^{2} )}

What is the solution for y^{''}-y^'-2y=sin(3t) ?
The solution for y^{''}-y^'-2y=sin(3t) is y=c_{1}e^{2t}+c_{2}e^{-t}-11/130 sin(3t)+3/130 cos(3t)