What is the solution for y^{''}-2y^'+17y=e^xcos(4x) ?

The solution for y^{''}-2y^'+17y=e^xcos(4x) is y=e^x(c_{1}cos(4x)+c_{2}sin(4x))+(e^xxsin(4x))/8

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y^{''}-2y^'+17y=e^xcos(4x)

y^{^{''}} - 2 y^{^{'}} + 17 y = e^{x} cos (4 x)

y = e^{x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{e ^{x} x sin ( 4 x )}{8}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 17 y = e^{x} cos (4 x)

Solve linear ODE:

y = e^{x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{e ^{x} x sin ( 4 x )}{8}

y = e^{x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{e ^{x} x sin ( 4 x )}{8}

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

y^{^{''}} - 4 y^{^{'}} - 12 y = 12 t, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''''}} + 2 y^{^{''}} + y = 5 + cos (6 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = 3 e^{x}

y^{^{''}} + 36 y = 5 sec (6 x)

What is the solution for y^{''}-2y^'+17y=e^xcos(4x) ?
The solution for y^{''}-2y^'+17y=e^xcos(4x) is y=e^x(c_{1}cos(4x)+c_{2}sin(4x))+(e^xxsin(4x))/8