What is the solution for y^{''}+64y=sin(8x) ?

The solution for y^{''}+64y=sin(8x) is y=c_{1}cos(8x)+c_{2}sin(8x)-(xcos(8x))/(16)

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y^{''}+64y=sin(8x)

y^{^{''}} + 64 y = sin (8 x)

y = c_{1} cos (8 x) + c_{2} sin (8 x) - \frac{x cos ( 8 x )}{16}

Solution steps

y^{''} (x) + 64 y = sin (8 x)

Solve linear ODE:

y = c_{1} cos (8 x) + c_{2} sin (8 x) - \frac{x cos ( 8 x )}{16}

y = c_{1} cos (8 x) + c_{2} sin (8 x) - \frac{x cos ( 8 x )}{16}

y^{^{''}} + 6 y = t^{4}, y (0) = 0, y^{^{'}} (0) = 0

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

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

y^{^{''}} - y = 5, e^{- 4 x} + 2 x y (0) = 0

y^{^{''}} + 6 y = 6 t^{2} - 3, y (0) = 0, y^{^{'}} (0) = - 3

What is the solution for y^{''}+64y=sin(8x) ?
The solution for y^{''}+64y=sin(8x) is y=c_{1}cos(8x)+c_{2}sin(8x)-(xcos(8x))/(16)