What is the solution for y^{''}+4y=2sec^3(2x) ?

The solution for y^{''}+4y=2sec^3(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(tan(2x)sin(2x))/4

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y^{''}+4y=2sec^3(2x)

y^{^{''}} + 4 y = 2 sec^{3} (2 x)

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

Solution steps

y^{''} (x) + 4 y = 2 sec^{3} (2 x)

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 6 = 0

y^{^{''}} - 2 y^{^{'}} + 65 y = e^{x} cos (8 x)

x^{^{''}} + 8 x = 16 cos (4 t)

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

y^{^{''}} - 3 y^{^{'}} = x e^{\frac{3 x}{4}}

What is the solution for y^{''}+4y=2sec^3(2x) ?
The solution for y^{''}+4y=2sec^3(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(tan(2x)sin(2x))/4