What is the solution for y^{''}-y^'-6y=70cos(3x) ?

The solution for y^{''}-y^'-6y=70cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-35/39 sin(3x)-175/39 cos(3x)

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y^{''}-y^'-6y=70cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 70 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{35}{39} sin (3 x) - \frac{175}{39} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 70 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{35}{39} sin (3 x) - \frac{175}{39} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{35}{39} sin (3 x) - \frac{175}{39} cos (3 x)

y^{^{''}} + 4 y^{^{'}} + 13 y = - 2 sin (3 t)

2 y^{^{''}} + 3 y - 2 y = 6 x^{2} + 5 x - 8

\frac{3}{2} \frac{d ^{2} y}{d x ^{2}} + \frac{1}{80} \frac{d y}{d x} = g

x^{^{''}} + 2 x^{^{'}} + 5 x = e^{t}

y^{^{''}} + 40 y^{^{'}} + 4000 y = 800 sin (40 x)

What is the solution for y^{''}-y^'-6y=70cos(3x) ?
The solution for y^{''}-y^'-6y=70cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-35/39 sin(3x)-175/39 cos(3x)