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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 4 y = x sin (2 x)

2 y^{^{''}} - 7 y^{^{'}} + 5 y = - 29

y^{^{''}} + y = t^{2} + 2, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 5 y^{^{'}} + 4 y = 3 x^{2}

y^{^{''}} + 4 y^{^{'}} + 3 y = 7 e^{- x}

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