What is the solution for y^{''}+4y=-12sin(2x),y(0)=1.8,y^'(0)=5 ?

The solution for y^{''}+4y=-12sin(2x),y(0)=1.8,y^'(0)=5 is y=1.8cos(2x)+sin(2x)+3xcos(2x)

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y^{''}+4y=-12sin(2x),y(0)=1.8,y^'(0)=5

y^{^{''}} + 4 y = - 12 sin (2 x), y (0) = 1.8, y^{^{'}} (0) = 5

y = 1.8 cos (2 x) + sin (2 x) + 3 x cos (2 x)

Solution steps

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

Solve linear ODE:

y = 1.8 cos (2 x) + sin (2 x) + 3 x cos (2 x)

y = 1.8 cos (2 x) + sin (2 x) + 3 x cos (2 x)

y^{^{''}} - 11 y^{^{'}} + 24 y = 13 + 24 x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} - 2 y = - 260 sin (3 t)

y^{^{''}} - 9 y = 81 x^{3}

y^{^{''}} - 2 y^{^{'}} + 2 y = 3 e^{x} sin (x)

What is the solution for y^{''}+4y=-12sin(2x),y(0)=1.8,y^'(0)=5 ?
The solution for y^{''}+4y=-12sin(2x),y(0)=1.8,y^'(0)=5 is y=1.8cos(2x)+sin(2x)+3xcos(2x)