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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = (2 x - 1) e^{4 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{2 e ^{x}}{x ^{2} - 1}

y^{^{''}} + 5 y^{^{'}} + 6 y = - 19 t e^{(2 t)}

y^{^{''}} + 9 y = g (t), y (0) = 4, y^{^{'}} (0) = 0

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