What is the solution for y^{''}-y^'-6y=61cos(3x) ?

The solution for y^{''}-y^'-6y=61cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-61/78 sin(3x)-305/78 cos(3x)

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y^{''}-y^'-6y=61cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 61 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{61}{78} sin (3 x) - \frac{305}{78} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 61 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{61}{78} sin (3 x) - \frac{305}{78} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{61}{78} sin (3 x) - \frac{305}{78} cos (3 x)

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

y^{^{''}} - 14 y^{^{'}} + 49 y = x^{2}

y^{^{''}} + 16 y^{^{'}} = x + e^{2 x}

y^{^{''}} + 4 y^{^{'}} + 4 y = x e^{x}

y^{^{''}} - 7 y^{^{'}} + 10 y = 6 x + e^{2 x}

What is the solution for y^{''}-y^'-6y=61cos(3x) ?
The solution for y^{''}-y^'-6y=61cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-61/78 sin(3x)-305/78 cos(3x)