What is the solution for y^{''}-y^'-6y=15sin(3x) ?

The solution for y^{''}-y^'-6y=15sin(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-25/26 sin(3x)+5/26 cos(3x)

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y^{''}-y^'-6y=15sin(3x)

y^{^{''}} - y^{^{'}} - 6 y = 15 sin (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{25}{26} sin (3 x) + \frac{5}{26} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 15 sin (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{25}{26} sin (3 x) + \frac{5}{26} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{25}{26} sin (3 x) + \frac{5}{26} cos (3 x)

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

y^{^{''}} - y^{^{'}} = 1 + 2 x + e^{x} + sin (x) + cos (x)

3 y^{^{''}} - 2 y^{^{'}} - 8 y = (2 x + 3) e^{5 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x ^{2}}

y^{^{''}} - 7 y^{^{'}} = 10

What is the solution for y^{''}-y^'-6y=15sin(3x) ?
The solution for y^{''}-y^'-6y=15sin(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-25/26 sin(3x)+5/26 cos(3x)