What is the solution for y^{''}-4y^'-12y=sin(2x) ?

The solution for y^{''}-4y^'-12y=sin(2x) is y=c_{1}e^{6x}+c_{2}e^{-2x}-1/20 sin(2x)+1/40 cos(2x)

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y^{''}-4y^'-12y=sin(2x)

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (2 x)

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) + \frac{1}{40} cos (2 x)

Solution steps

y^{''} (x) - 4 y^{'} (x) - 12 y = sin (2 x)

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) + \frac{1}{40} cos (2 x)

y = c_{1} e^{6 x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) + \frac{1}{40} cos (2 x)

f (x) = \frac{2}{x - 2}, at x = - 4

x \to 0 lim (\frac{cos ( x ) - x}{x})

y = x^{3}, y = x^{\frac{1}{2}}

\frac{\partial}{\partial x} (x^{3} + 4 x^{2} + x^{4} + 3 x^{5})

\int_{0}^{6 π} x^{2} sin (2 x) d x

What is the solution for y^{''}-4y^'-12y=sin(2x) ?
The solution for y^{''}-4y^'-12y=sin(2x) is y=c_{1}e^{6x}+c_{2}e^{-2x}-1/20 sin(2x)+1/40 cos(2x)