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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} - y = e^{\frac{x}{2}} + 5

y^{^{''}} - y^{^{'}} = 2 x e^{x}

y^{^{'''}} - 6 y^{^{''}} = 2 - cos (x)

y^{^{''}} + 7 y^{^{'}} - 12 y = e^{3 x} - 2 e^{x} - 5 x

y^{^{''}} - 6 y^{^{'}} + 9 y = t

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