What is the solution for y^{''}-4y^'-12y=3cos(x) ?

The solution for y^{''}-4y^'-12y=3cos(x) is y=c_{1}e^{6x}+c_{2}e^{-2x}-12/185 sin(x)-39/185 cos(x)

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y^{''}-4y^'-12y=3cos(x)

y^{^{''}} - 4 y^{^{'}} - 12 y = 3 cos (x)

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{12}{185} sin (x) - \frac{39}{185} cos (x)

Solution steps

y^{''} (x) - 4 y^{'} (x) - 12 y = 3 cos (x)

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{12}{185} sin (x) - \frac{39}{185} cos (x)

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{12}{185} sin (x) - \frac{39}{185} cos (x)

y^{^{''}} + y^{^{'}} + 3 y = 2 x - x^{2}

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{x ^{2} + 1}

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

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

y^{^{''}} - 2 y^{^{'}} + y = 14 x^{\frac{3}{2}} e^{x}

What is the solution for y^{''}-4y^'-12y=3cos(x) ?
The solution for y^{''}-4y^'-12y=3cos(x) is y=c_{1}e^{6x}+c_{2}e^{-2x}-12/185 sin(x)-39/185 cos(x)