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

The solution for y^{''}+4y=sec^3(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(tan(2x)sin(2x))/8

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

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

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

Solution steps

y^{''} (x) + 4 y = 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 )}{8}

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

y = arctan (- x), at x = 3

\int 2 - x^{2} d x

\frac{d}{d x} (\frac{2}{tan ( x )})

x^{3} + 5 x^{2} - 63 x - 140

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

What is the solution for y^{''}+4y=sec^3(2x) ?
The solution for y^{''}+4y=sec^3(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(tan(2x)sin(2x))/8