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

The solution for y^{''}-y^'-2y=sin(2x) is y=c_{1}e^{2x}+c_{2}e^{-x}-3/20 sin(2x)+1/20 cos(2x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3}{20} sin (2 x) + \frac{1}{20} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3}{20} sin (2 x) + \frac{1}{20} cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3}{20} sin (2 x) + \frac{1}{20} cos (2 x)

y^{^{''}} - y^{^{'}} = e^{t} cos (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = e^{5 x} ln (4 + x^{2})

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 9

y^{^{''}} - 10 y^{^{'}} + 9 y = 5 t, y (0) = - 1, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + 2 y = \frac{e ^{x}}{sin ( x )}

What is the solution for y^{''}-y^'-2y=sin(2x) ?
The solution for y^{''}-y^'-2y=sin(2x) is y=c_{1}e^{2x}+c_{2}e^{-x}-3/20 sin(2x)+1/20 cos(2x)