What is the solution for (D^4-1)*y=sin(2x) ?

The solution for (D^4-1)*y=sin(2x) is y=c_{1}e^x+c_{2}e^{-x}+c_{3}cos(x)+c_{4}sin(x)+1/15 sin(2x)

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(D^4-1)*y=sin(2x)

(D^{4} - 1) \cdot y = sin (2 x)

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) + \frac{1}{15} sin (2 x)

Solution steps

(D^{4} - 1) y = sin (2 x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) + \frac{1}{15} sin (2 x)

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) + \frac{1}{15} sin (2 x)

y^{^{''}} + y^{^{'}} - 7 y = 42 t

y^{^{''}} + y = 2, y (0) = 2, y (0) = 0

y^{^{''}} - 6 y^{^{'}} - 9 y = x^{2} + e^{x}

x^{^{''}} + 5 x^{^{'}} + 4 x = t^{2} + 3 t + 5

y^{^{''}} - 2 y^{^{'}} + y = 2 sin (3 x)

What is the solution for (D^4-1)*y=sin(2x) ?
The solution for (D^4-1)*y=sin(2x) is y=c_{1}e^x+c_{2}e^{-x}+c_{3}cos(x)+c_{4}sin(x)+1/15 sin(2x)