What is the solution for y^{''}-2y^'+5y=-20e^xsin(2x) ?

The solution for y^{''}-2y^'+5y=-20e^xsin(2x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+5e^xxcos(2x)

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y^{''}-2y^'+5y=-20e^xsin(2x)

y^{^{''}} - 2 y^{^{'}} + 5 y = - 20 e^{x} sin (2 x)

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 5 e^{x} x cos (2 x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = - 20 e^{x} sin (2 x)

Solve linear ODE:

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 5 e^{x} x cos (2 x)

y = e^{x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 5 e^{x} x cos (2 x)

2 y^{^{''}} - 2 y^{^{'}} + y = 1 + x + x^{2}

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

y^{^{''''}} - y^{^{'}} = x^{3} + x^{2} + 1

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

y^{^{''}} + y = 5 x \cdot sin (x)

What is the solution for y^{''}-2y^'+5y=-20e^xsin(2x) ?
The solution for y^{''}-2y^'+5y=-20e^xsin(2x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+5e^xxcos(2x)