What is the solution for y^{''}+4y= 5/2 cos(2x) ?

The solution for y^{''}+4y= 5/2 cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(5xsin(2x))/8

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y^{''}+4y= 5/2 cos(2x)

y^{^{''}} + 4 y = \frac{5}{2} cos (2 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{5 x sin ( 2 x )}{8}

Solution steps

y^{''} (x) + 4 y = \frac{5}{2} cos (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{5 x sin ( 2 x )}{8}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{5 x sin ( 2 x )}{8}

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = 1 - 4 x^{3}

y^{^{''''}} + 2 y^{^{''}} + y = 7 + cos (2 t)

x^{^{''}} + 6 x^{^{'}} + 34 x = 68, x (0) = 5, x^{^{'}} (0) = 7

y^{^{''}} + 9 y = 13 sec^{2} (3 t)

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

What is the solution for y^{''}+4y= 5/2 cos(2x) ?
The solution for y^{''}+4y= 5/2 cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(5xsin(2x))/8