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

The solution for 4y^{''}+y=sin(2x) is y=c_{1}cos(x/2)+c_{2}sin(x/2)-1/15 sin(2x)

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

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

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2}) - \frac{1}{15} sin (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2}) - \frac{1}{15} sin (2 x)

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2}) - \frac{1}{15} sin (2 x)

(D^{2} - 4 D + 4) y = 8 (x^{2} + e^{2 x} + sin (2 x))

y^{^{''}} + y = cos (x - \frac{π}{2}), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3} x cos (e^{x})

y^{^{''}} + 5 y^{^{'}} + y = 4 t^{2} + 8 t + 8

y^{^{''}} + 3 y^{^{'}} = x^{2} sin (x) cos (x)

What is the solution for 4y^{''}+y=sin(2x) ?
The solution for 4y^{''}+y=sin(2x) is y=c_{1}cos(x/2)+c_{2}sin(x/2)-1/15 sin(2x)