What is the solution for y^{''}+4y=2sec^3(2t) ?

The solution for y^{''}+4y=2sec^3(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tan(2t)sin(2t))/4

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y^{''}+4y=2sec^3(2t)

y^{^{''}} + 4 y = 2 sec^{3} (2 t)

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

Solution steps

y^{''} (t) + 4 y = 2 sec^{3} (2 t)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e^{x} - 12

y^{^{''}} - 3 y^{^{'}} + 2 y = \frac{e ^{3 x}}{1 + e ^{x}}

\frac{d ^{2} y}{d t ^{2}} - 4 y = t cos (2 t)

y^{^{''}} + 25 y = 6 sin (x)

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

What is the solution for y^{''}+4y=2sec^3(2t) ?
The solution for y^{''}+4y=2sec^3(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+(tan(2t)sin(2t))/4