What is the solution for y^{''}+4y=-12sin(2x) ?

The solution for y^{''}+4y=-12sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+3xcos(2x)

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y^{''}+4y=-12sin(2x)

y^{^{''}} + 4 y = - 12 sin (2 x)

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

Solution steps

y^{''} (x) + 4 y = - 12 sin (2 x)

Solve linear ODE:

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

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

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

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = x e^{x} - e^{- x} + 7

y^{^{''}} + 36 y = csc (6 x)

y^{^{''}} + y = 2 + x^{2}

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 x + 1

What is the solution for y^{''}+4y=-12sin(2x) ?
The solution for y^{''}+4y=-12sin(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+3xcos(2x)