What is the solution for y^{''}+y^'-2y=8sin(2x) ?

The solution for y^{''}+y^'-2y=8sin(2x) is y=c_{1}e^x+c_{2}e^{-2x}-6/5 sin(2x)-2/5 cos(2x)

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y^{''}+y^'-2y=8sin(2x)

y^{^{''}} + y^{^{'}} - 2 y = 8 sin (2 x)

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{6}{5} sin (2 x) - \frac{2}{5} cos (2 x)

Solution steps

y^{''} (x) + y^{'} (x) - 2 y = 8 sin (2 x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{6}{5} sin (2 x) - \frac{2}{5} cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 2 x} - \frac{6}{5} sin (2 x) - \frac{2}{5} cos (2 x)

(d^{2} + 2 d + 1) y = cos^{2} (x)

(d^{2} + 3 d + 2) y = e - x + cos (x)

(d^{2} + 5 d + 6) y = e^{- 2} x sin^{2} (x)

y, (d^{2} + 4 d + 4) y = sin (t)

(d^{2} - 6 d + 8) y = (e^{2} x + 1)^{2}

What is the solution for y^{''}+y^'-2y=8sin(2x) ?
The solution for y^{''}+y^'-2y=8sin(2x) is y=c_{1}e^x+c_{2}e^{-2x}-6/5 sin(2x)-2/5 cos(2x)