What is the solution for y^{''}-4y=6cos(3x+30) ?

The solution for y^{''}-4y=6cos(3x+30) is y=c_{1}e^{2x}+c_{2}e^{-2x}-6/13 cos(3x+30)

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y^{''}-4y=6cos(3x+30)

y^{^{''}} - 4 y = 6 cos (3 x + 3 0^{\circ})

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{6}{13} cos (3 x + 3 0^{\circ})

Solution steps

y^{''} (x) - 4 y = 6 cos (3 x + 3 0^{\circ})

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{6}{13} cos (3 x + 3 0^{\circ})

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{6}{13} cos (3 x + 3 0^{\circ})

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 2 e^{4 x} + 3 cos (4 x) - 2 x + 1

x^{^{''}} - 6 x^{^{'}} + 9 x = 5 t e^{3 t}

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

y^{^{''}} - 3 y^{^{'}} + 7 y = t^{2} + cos (t)

What is the solution for y^{''}-4y=6cos(3x+30) ?
The solution for y^{''}-4y=6cos(3x+30) is y=c_{1}e^{2x}+c_{2}e^{-2x}-6/13 cos(3x+30)