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}

3 x^{^{''}} - 30 x^{^{'}} + 75 x = 2 t + 1

y^{^{''}} - 6 y^{^{'}} + 5 y = 5 x^{2} + 3 x - 16 - 9 e^{2 x}

y^{^{''}} - 9 y = (x^{2} - 6) sin (3 x)

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

y^{^{''}} + y^{^{'}} - 6 y = 12 e^{3 t} + 12 e^{- 2 t}

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