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

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

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

4 y^{^{''}} + 2 y = 4 x^{3} + 5 x^{2} - x + 7

y^{^{''}} - 3 y^{^{'}} + 2 y = cos (x), y^{^{'}} (0) = 2, y (0) = 0

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = 2 x^{2} + 4 sin (x)

y^{^{'''}} - 12 y^{^{''}} + 48 y^{^{'}} - 64 y = 4 e^{4 x}

y^{^{''}} - 3 y^{^{'}} + 2 y = t e^{- 3 t}

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