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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 7 y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 2

y^{^{'''}} + y^{^{'}} = sec^{2} (x)

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 9 + 2 sin (x)

y^{^{''}} - y = e^{x} sin (x)

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{(\frac{x}{2})}

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