What is the solution for y^{''}-y^'-2y=10sin(2x) ?

The solution for y^{''}-y^'-2y=10sin(2x) is y=c_{1}e^{2x}+c_{2}e^{-x}-3/2 sin(2x)+1/2 cos(2x)

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y^{''}-y^'-2y=10sin(2x)

y^{^{''}} - y^{^{'}} - 2 y = 10 sin (2 x)

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

Solution steps

y^{''} (x) - y^{'} (x) - 2 y = 10 sin (2 x)

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} + 12 y = 4 e^{2 x}

y^{^{''}} - 49 y = cosh (7 x)

y^{^{''}} + y = 2 cos^{2} (x)

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 9 + e^{\frac{x}{2}}

4 y^{^{''}} + y + 2 = 0, y (0) = 0, y^{^{'}} (0) = \frac{1}{2}

What is the solution for y^{''}-y^'-2y=10sin(2x) ?
The solution for y^{''}-y^'-2y=10sin(2x) is y=c_{1}e^{2x}+c_{2}e^{-x}-3/2 sin(2x)+1/2 cos(2x)