What is the solution for y^{''}-y^'-6y=14cos(3x) ?

The solution for y^{''}-y^'-6y=14cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-7/39 sin(3x)-35/39 cos(3x)

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y^{''}-y^'-6y=14cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 14 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{7}{39} sin (3 x) - \frac{35}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 14 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{7}{39} sin (3 x) - \frac{35}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{7}{39} sin (3 x) - \frac{35}{39} cos (3 x)

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

y^{^{'''}} + 7 y^{^{''}} - 8 y = x e^{x} + 12

y^{^{''}} + 4000 y^{^{'}} + 4000000 y = 36

\frac{d ^{2} y}{d x ^{2}} + 14 \frac{d y}{d x} + 49 y = 4 e^{5 x}

y^{^{''}} + 5 y^{^{'}} - y = 11

What is the solution for y^{''}-y^'-6y=14cos(3x) ?
The solution for y^{''}-y^'-6y=14cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-7/39 sin(3x)-35/39 cos(3x)