What is the solution for y^{''}-y^'-6y=5cos(3x) ?

The solution for y^{''}-y^'-6y=5cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-5/78 sin(3x)-25/78 cos(3x)

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y^{''}-y^'-6y=5cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 5 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{5}{78} sin (3 x) - \frac{25}{78} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 5 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{5}{78} sin (3 x) - \frac{25}{78} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{5}{78} sin (3 x) - \frac{25}{78} cos (3 x)

y^{^{''}} - 18 y^{^{'}} + 81 y = t^{9} e^{9 t}

y^{^{''}} - y^{^{'}} = 3 - 4 x, y (0) = - 3, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} \cdot ln (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 3 x - 2

y^{^{''}} + 25 y = x^{2}

What is the solution for y^{''}-y^'-6y=5cos(3x) ?
The solution for y^{''}-y^'-6y=5cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-5/78 sin(3x)-25/78 cos(3x)