What is the solution for y^{''}-y^'-6y=28cos(3x) ?

The solution for y^{''}-y^'-6y=28cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-14/39 sin(3x)-70/39 cos(3x)

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y^{''}-y^'-6y=28cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 28 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{14}{39} sin (3 x) - \frac{70}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 28 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{14}{39} sin (3 x) - \frac{70}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{14}{39} sin (3 x) - \frac{70}{39} cos (3 x)

u^{^{''}} + 3 u^{^{'}} + 5 H (t) = e^{- 5 t}

y^{^{''}} - 4 y^{^{'}} = 3 e^{2 x}, y (0) = 0, y^{^{'}} (0) = 0

2 y^{^{''}} + 50 y = 2 cos (5 x) cos (5 x)

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

y^{^{''}} + 2 y^{^{'}} + 2 y = sec (x) e^{- x}

What is the solution for y^{''}-y^'-6y=28cos(3x) ?
The solution for y^{''}-y^'-6y=28cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-14/39 sin(3x)-70/39 cos(3x)