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

The solution for y^{''}+y^'-2y=4sin(2x) is y=c_{1}e^x+c_{2}e^{-2x}-3/5 sin(2x)-1/5 cos(2x)

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

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

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{3}{5} sin (2 x) - \frac{1}{5} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{3}{5} sin (2 x) - \frac{1}{5} cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{3}{5} sin (2 x) - \frac{1}{5} cos (2 x)

\int e^{- 0.89 x} d x

\frac{\partial}{\partial y} (- y cos (x))

\frac{s + 1}{( s + 1 ) ( s ^{2} + 2 )}

y = - x^{2} + 2 x + 2, y = x - 4

\int \frac{d ^{2} y}{1} d x^{2}

What is the solution for y^{''}+y^'-2y=4sin(2x) ?
The solution for y^{''}+y^'-2y=4sin(2x) is y=c_{1}e^x+c_{2}e^{-2x}-3/5 sin(2x)-1/5 cos(2x)