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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 15 y = 7 + 4 x

y^{^{''}} + 4 y = - 5 sin (2 t) + 2 cos (2 t)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = 2 e^{2 x}

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 15 y = 3

y^{^{''}} - 6 y^{^{'}} - 27 y = - 57 t + 54 t^{2}

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