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

The solution for y^{''}-2y^'+5y=e^x*cos(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^x*cos(2x)

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} \cdot 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} \cdot 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}

y^{^{''}} - 2 y^{^{'}} + y = 6 e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 13 + e x

y^{^{''}} + 10 y^{^{'}} + 25 y = 30 x + 3

y^{^{''}} - 3 y^{^{'}} + 2 y = sin (2 x)

y^{^{''}} + 9 y = \frac{3}{2} sin (2 t)

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