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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'}} + y (\frac{3}{x} + \frac{x}{2}) = 0

f (x) = \frac{1}{7 x - 4}

x^{3} + x + 12 x^{\frac{1}{2}}

\frac{d}{d x} ((5 x + 1)^{3} (6 x^{2} + 3 x + 2)^{5})

\frac{d}{d x} (\frac{sec ^{2} ( x ^{3} )}{2})^{4}

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