What is the solution for y^{''}-y^'-6y=58cos(3x) ?

The solution for y^{''}-y^'-6y=58cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-29/39 sin(3x)-145/39 cos(3x)

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y^{''}-y^'-6y=58cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 58 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{29}{39} sin (3 x) - \frac{145}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 58 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{29}{39} sin (3 x) - \frac{145}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{29}{39} sin (3 x) - \frac{145}{39} cos (3 x)

y^{^{''}} - 9 y = (x^{2} - 7) sin (3 x)

y^{^{''}} + 36 y = csc^{2} (6 x)

y^{^{''}} + 6 y^{^{'}} - 40 y = 1 - 3 x + x^{2}

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{2 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 45 y = 6, y (0) = 5, y^{^{'}} (0) = 9

What is the solution for y^{''}-y^'-6y=58cos(3x) ?
The solution for y^{''}-y^'-6y=58cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-29/39 sin(3x)-145/39 cos(3x)