What is the solution for y^{''}+4y=-12sec(2t) ?

The solution for y^{''}+4y=-12sec(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-3ln(cos(2t))cos(2t)-6tsin(2t)

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y^{''}+4y=-12sec(2t)

y^{^{''}} + 4 y = - 12 sec (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 3 ln (cos (2 t)) cos (2 t) - 6 t sin (2 t)

Solution steps

y^{''} (t) + 4 y = - 12 sec (2 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 3 ln (cos (2 t)) cos (2 t) - 6 t sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - 3 ln (cos (2 t)) cos (2 t) - 6 t sin (2 t)

y = 4 - 2 x - x^{- 3}

x \to - 2 lim (\frac{x - 2}{x + 2})

y^{^{'}} = \frac{x e ^{x^{2}}}{2}

\frac{\partial}{\partial x} (2 x^{0.3} y^{0.3})

\int r^{2} sec^{2} (r^{3}) d r

What is the solution for y^{''}+4y=-12sec(2t) ?
The solution for y^{''}+4y=-12sec(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-3ln(cos(2t))cos(2t)-6tsin(2t)