What is the solution for y^{''}-6y^'+9y=sin(3t) ?

The solution for y^{''}-6y^'+9y=sin(3t) is y=c_{1}e^{3t}+c_{2}te^{3t}+1/18 cos(3t)

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y^{''}-6y^'+9y=sin(3t)

y^{^{''}} - 6 y^{^{'}} + 9 y = sin (3 t)

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{1}{18} cos (3 t)

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = sin (3 t)

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{1}{18} cos (3 t)

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{1}{18} cos (3 t)

y^{^{''}} - 2 y^{^{'}} - 8 y = 6 + 3 x

2 y^{^{''}} + 4 y = 6 t

y^{^{''}} + 6 y^{^{'}} + 8 y = 8 t + 22, y (0) = - 7, y^{^{'}} (0) = 20

y^{^{''}} - 5 y^{^{'}} + 4 y = 8 x \cdot e^{2 x}

y^{^{''}} + 2 y^{^{'}} - 3 y = 2 e^{t}

What is the solution for y^{''}-6y^'+9y=sin(3t) ?
The solution for y^{''}-6y^'+9y=sin(3t) is y=c_{1}e^{3t}+c_{2}te^{3t}+1/18 cos(3t)