What is the solution for y^{''}-2y^'+5y=e^xcos(2x) ?

The solution for y^{''}-2y^'+5y=e^xcos(2x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+(e^xxsin(2x))/4

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y^{''}-2y^'+5y=e^xcos(2x)

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} cos (2 x)

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = e^{x} cos (2 x)

Solve linear ODE:

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

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

\frac{d}{d x} (cos (3 x + 1))

\frac{d}{d x} (x 3 - x)

\frac{\partial}{\partial x} (- 3 x e^{2 x y})

\frac{d y}{d x} = x (8 - y)

\frac{d ^{2}}{d x ^{2}} (4 e^{x} cos (x))

What is the solution for y^{''}-2y^'+5y=e^xcos(2x) ?
The solution for y^{''}-2y^'+5y=e^xcos(2x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+(e^xxsin(2x))/4