What is the solution for y^{''}-2y^'+65y=e^xcos(8x) ?

The solution for y^{''}-2y^'+65y=e^xcos(8x) is y=e^x(c_{1}cos(8x)+c_{2}sin(8x))+(e^xxsin(8x))/(16)

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y^{''}-2y^'+65y=e^xcos(8x)

y^{^{''}} - 2 y^{^{'}} + 65 y = e^{x} cos (8 x)

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 65 y = e^{x} cos (8 x)

Solve linear ODE:

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

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

x^{^{''}} + 8 x = 16 cos (4 t)

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

y^{^{''}} - 3 y^{^{'}} = x e^{\frac{3 x}{4}}

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

y^{^{''}} - 8 y^{^{'}} + 2 y = x e^{x}

What is the solution for y^{''}-2y^'+65y=e^xcos(8x) ?
The solution for y^{''}-2y^'+65y=e^xcos(8x) is y=e^x(c_{1}cos(8x)+c_{2}sin(8x))+(e^xxsin(8x))/(16)