What is the solution for y^{''}+16y=sec^3(4t) ?

The solution for y^{''}+16y=sec^3(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)+(tan(4t)sin(4t))/(32)

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y^{''}+16y=sec^3(4t)

y^{^{''}} + 16 y = sec^{3} (4 t)

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{tan ( 4 t ) sin ( 4 t )}{32}

Solution steps

y^{''} (t) + 16 y = sec^{3} (4 t)

Solve linear ODE:

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{tan ( 4 t ) sin ( 4 t )}{32}

y = c_{1} cos (4 t) + c_{2} sin (4 t) + \frac{tan ( 4 t ) sin ( 4 t )}{32}

y^{^{''}} - 36 y = 12, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} - y = k, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 25 y = sinh (5 x)

(D^{2} + 1) y = sin^{2} (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = - x e^{4} x

What is the solution for y^{''}+16y=sec^3(4t) ?
The solution for y^{''}+16y=sec^3(4t) is y=c_{1}cos(4t)+c_{2}sin(4t)+(tan(4t)sin(4t))/(32)