What is the solution for y^{''}-y^'-6y=14sin(3) ?

The solution for y^{''}-y^'-6y=14sin(3) is y=c_{1}e^{3t}+c_{2}e^{-2t}-(7sin(3))/3

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y^{''}-y^'-6y=14sin(3)

y^{^{''}} - y^{^{'}} - 6 y = 14 sin (3)

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

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = 14 sin (3)

Solve linear ODE:

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

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

y^{^{''}} - 4 y + 3 y = 9 x^{2}

y^{^{''}} + 6 y^{^{'}} - 40 y = - 2 e^{3 x} + cos (2 x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 e^{(- t)} sin (2 t)

x^{^{''}} - 4 x^{^{'}} = 20 cos (5 t) - 4 sin (5 t) - 7

x^{^{''}} + 4 x^{^{'}} + 13 x = t e^{t}

What is the solution for y^{''}-y^'-6y=14sin(3) ?
The solution for y^{''}-y^'-6y=14sin(3) is y=c_{1}e^{3t}+c_{2}e^{-2t}-(7sin(3))/3