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

The solution for 4y^{''}+y=2cos(x) is y=c_{1}cos(x/2)+c_{2}sin(x/2)-2/3 cos(x)

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

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

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2}) - \frac{2}{3} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2}) - \frac{2}{3} cos (x)

y = c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2}) - \frac{2}{3} cos (x)

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

x^{^{''}} + 2 x^{^{'}} + 41 x = 12 cos (6 t)

y^{^{''}} + 16 y = 4 csc^{2} (4 t)

y^{^{''}} + y = x e^{x}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + (- 4) y^{^{'}} + 3 y = - 12

What is the solution for 4y^{''}+y=2cos(x) ?
The solution for 4y^{''}+y=2cos(x) is y=c_{1}cos(x/2)+c_{2}sin(x/2)-2/3 cos(x)