What is the solution for y^{''}+4y=sin(2x)cos(2x) ?

The solution for y^{''}+4y=sin(2x)cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/24 sin(4x)

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y^{''}+4y=sin(2x)cos(2x)

y^{^{''}} + 4 y = sin (2 x) cos (2 x)

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

Solution steps

y^{''} (x) + 4 y = sin (2 x) cos (2 x)

Solve linear ODE:

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

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

\frac{d}{d x} (21 \cdot π \cdot x^{2})

y = ln (3 x^{2} + 7 y^{2})

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\frac{d}{d x} (0.02 x^{3} - 0.07 x^{2} + 8 x)

\frac{d}{d x} (cos^{4} (2 x))

What is the solution for y^{''}+4y=sin(2x)cos(2x) ?
The solution for y^{''}+4y=sin(2x)cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)-1/24 sin(4x)