What is the solution for y^{''}-y^'-6y=53cos(3x) ?

The solution for y^{''}-y^'-6y=53cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-53/78 sin(3x)-265/78 cos(3x)

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y^{''}-y^'-6y=53cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 53 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{53}{78} sin (3 x) - \frac{265}{78} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 53 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{53}{78} sin (3 x) - \frac{265}{78} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{53}{78} sin (3 x) - \frac{265}{78} cos (3 x)

y^{^{''}} - y^{^{'}} - 2 y = x^{2} + cos (x)

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 t}}{t}

y^{^{''}} - 5 y^{^{'}} + 7 y = 3 e^{5 t}

y^{^{''}} + y = sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{2 x} tan^{2} (x)

What is the solution for y^{''}-y^'-6y=53cos(3x) ?
The solution for y^{''}-y^'-6y=53cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-53/78 sin(3x)-265/78 cos(3x)