What is the solution for y^{''}+y=3cos(2x) ?

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

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y^{''}+y=3cos(2x)

y^{^{''}} + y = 3 cos (2 x)

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

Solution steps

y^{''} (x) + y = 3 cos (2 x)

Solve linear ODE:

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

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

x^{^{''}} + 3 x^{^{'}} + 2 x = sin (t)

y^{^{''}} + 4 y = 2 sec (2 t)

\frac{d ^{2} y}{d x ^{2}} + 4 y = x sin (2 x)

2 y^{^{''}} - 7 y^{^{'}} + 5 y = - 29

y^{^{''}} + y = t^{2} + 2, y (0) = 1, y^{^{'}} (0) = - 1

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