What is the solution for y^{''}+2y^'-3y=5sin(3x) ?

The solution for y^{''}+2y^'-3y=5sin(3x) is y=c_{1}e^x+c_{2}e^{-3x}-1/3 sin(3x)-1/6 cos(3x)

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y^{''}+2y^'-3y=5sin(3x)

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 sin (3 x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = 5 sin (3 x)

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 1 = 0

y^{^{'''}} + 3 y^{^{''}} - 4 y^{^{'}} - 12 y = 10 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} + 9 y = tan (3 x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 96 e^{- 5 t} - 24 e^{- 3 t} + 18

y^{^{''}} + 6 y^{^{'}} + 9 y = x e^{2 x}

What is the solution for y^{''}+2y^'-3y=5sin(3x) ?
The solution for y^{''}+2y^'-3y=5sin(3x) is y=c_{1}e^x+c_{2}e^{-3x}-1/3 sin(3x)-1/6 cos(3x)