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

The solution for y^{''}+4y=sin(2t)cos(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-1/24 sin(4t)

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{24} sin (4 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{24} sin (4 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{24} sin (4 t)

\frac{( e ^{- s} )}{s ^{5}}

x^{2} + y^{2} = 2

\frac{\partial}{\partial x} (3 x (4 + y)^{- 1})

\int_{0}^{6} \frac{1}{36 + x ^{2}} d x

\frac{\partial}{\partial x} (e^{3 y})

What is the solution for y^{''}+4y=sin(2t)cos(2t) ?
The solution for y^{''}+4y=sin(2t)cos(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)-1/24 sin(4t)