What is the solution for y^{''}-y^'-6y=19sin(3x) ?

The solution for y^{''}-y^'-6y=19sin(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-95/78 sin(3x)+19/78 cos(3x)

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y^{''}-y^'-6y=19sin(3x)

y^{^{''}} - y^{^{'}} - 6 y = 19 sin (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{95}{78} sin (3 x) + \frac{19}{78} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 19 sin (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{95}{78} sin (3 x) + \frac{19}{78} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{95}{78} sin (3 x) + \frac{19}{78} cos (3 x)

x^{^{''}} - 3 x^{^{'}} + 2 x = t^{2} e^{t}

(D^{3} - D) y = 4 e^{(- x)} + 3 e^{2} x

y^{^{''}} + 36 y = 60 sin (6 t)

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 2 x = 2

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - y = x sinh (x)

What is the solution for y^{''}-y^'-6y=19sin(3x) ?
The solution for y^{''}-y^'-6y=19sin(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-95/78 sin(3x)+19/78 cos(3x)