What is the solution for y^{''}+4y=8cos(2x)+12sin(2x) ?

The solution for y^{''}+4y=8cos(2x)+12sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2xsin(2x)-3xcos(2x)

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y^{''}+4y=8cos(2x)+12sin(2x)

y^{^{''}} + 4 y = 8 cos (2 x) + 12 sin (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x sin (2 x) - 3 x cos (2 x)

Solution steps

y^{''} (x) + 4 y = 8 cos (2 x) + 12 sin (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x sin (2 x) - 3 x cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + 2 x sin (2 x) - 3 x cos (2 x)

x^{^{''}} + 4 x^{^{'}} + 3 x = 1

(D - 1)^{2} y = (e^{2 x} + sin (2 x))

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} + 36 y = 10 cos (6 x) + 15 sin (6 x)

25 y^{^{''}} + 20 y^{^{'}} + 4 y = 4 x^{2} - 3 x + 2

What is the solution for y^{''}+4y=8cos(2x)+12sin(2x) ?
The solution for y^{''}+4y=8cos(2x)+12sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2xsin(2x)-3xcos(2x)