What is the solution for y^{''}-y^'-6y=22cos(3x) ?

The solution for y^{''}-y^'-6y=22cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-11/39 sin(3x)-55/39 cos(3x)

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y^{''}-y^'-6y=22cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 22 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{11}{39} sin (3 x) - \frac{55}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 22 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{11}{39} sin (3 x) - \frac{55}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{11}{39} sin (3 x) - \frac{55}{39} cos (3 x)

9 y^{^{''}} + 6 y^{^{'}} + y = 2, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = t

y^{^{''''}} + 4 y^{^{''}} = sin (2 t) + t e^{3 t} + 4

y^{^{''}} - y^{^{'}} - 12 = 0

y^{^{'''}} + y = sin (x)

What is the solution for y^{''}-y^'-6y=22cos(3x) ?
The solution for y^{''}-y^'-6y=22cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-11/39 sin(3x)-55/39 cos(3x)