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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{x ^{5} - x ^{2}} d x

\frac{d}{d x} (\frac{- 2 x ^{2} + x + 4}{x ^{2} + 4 x - 1})

x \to 0 lim (\frac{- x ^{3}}{x ^{3}})

\int \frac{x}{9 - 2 x ^{4}} d x

\int \frac{2 x + 1}{x ^{3} - 3 x ^{2} + 3 x - 1} d x

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