What is the solution for y^{''}+4y=6cos(2x) ?

The solution for y^{''}+4y=6cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(3xsin(2x))/2

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y^{''}+4y=6cos(2x)

y^{^{''}} + 4 y = 6 cos (2 x)

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

Solution steps

y^{''} (x) + 4 y = 6 cos (2 x)

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} - 10 y = e^{2 x} + x e^{- 5 x}

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 6 - e^{x} + e^{5 x}

y^{^{''''}} + 2 y^{^{''}} + y = 7 + cos (3 t)

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

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

What is the solution for y^{''}+4y=6cos(2x) ?
The solution for y^{''}+4y=6cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+(3xsin(2x))/2