What is the solution for y^{''}-y^'-6y=17cos(3x) ?

The solution for y^{''}-y^'-6y=17cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-17/78 sin(3x)-85/78 cos(3x)

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y^{''}-y^'-6y=17cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 17 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{17}{78} sin (3 x) - \frac{85}{78} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 17 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{17}{78} sin (3 x) - \frac{85}{78} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{17}{78} sin (3 x) - \frac{85}{78} cos (3 x)

y^{^{''}} + 3 y = x^{3} + 1

y^{^{''}} - 4 y^{^{'}} - 4 t + 4 = 0, y (0) = 2, y^{^{'}} (0) = 9

y^{^{''}} - 5 y^{^{'}} + 6 y = 5 \cdot e^{2 x}

y^{^{''}} - y^{^{'}} - 6 y = 2 cos (3 x)

2 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 6 y = 14 e^{x}

What is the solution for y^{''}-y^'-6y=17cos(3x) ?
The solution for y^{''}-y^'-6y=17cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-17/78 sin(3x)-85/78 cos(3x)