What is the solution for y^{''}-y^'-6y=12cos(3x) ?

The solution for y^{''}-y^'-6y=12cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-2/13 sin(3x)-10/13 cos(3x)

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y^{''}-y^'-6y=12cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 12 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{2}{13} sin (3 x) - \frac{10}{13} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 12 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{2}{13} sin (3 x) - \frac{10}{13} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{2}{13} sin (3 x) - \frac{10}{13} cos (3 x)

x^{^{''}} + 5 x^{^{'}} + 6 x = 10

y^{^{''}} - 7 y^{^{'}} + 9 y = x, y^{1} (0) = e x

y^{^{''}} - 8 y + 20 y = 100 x^{2} - 26 x e^{x}

y^{^{''}} - 4 y^{^{'}} + y = (- 3 x + 4) sin (x)

(D^{2} + D + 1) y = x e^{x}

What is the solution for y^{''}-y^'-6y=12cos(3x) ?
The solution for y^{''}-y^'-6y=12cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-2/13 sin(3x)-10/13 cos(3x)