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

The solution for y^{''}-y^'-6y=6cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-1/13 sin(3x)-5/13 cos(3x)

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

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

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{13} sin (3 x) - \frac{5}{13} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{13} sin (3 x) - \frac{5}{13} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{1}{13} sin (3 x) - \frac{5}{13} cos (3 x)

y^{^{''}} - 6 y + y^{^{'}} = e^{x}

y^{^{''}} - 4 y = - 12 sin (2 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 4 y = tan (x)

y^{^{''}} - 4 y = 24

y^{^{''}} + y^{^{'}} + y = 2 cos (2 t) - 7 sin (2 t)

What is the solution for y^{''}-y^'-6y=6cos(3x) ?
The solution for y^{''}-y^'-6y=6cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-1/13 sin(3x)-5/13 cos(3x)