What is the solution for y^{''}+4y=3cos(2.1x) ?

The solution for y^{''}+4y=3cos(2.1x) is y=c_{1}cos(2x)+c_{2}sin(2x)-7.31707…cos(2.1x)

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y^{''}+4y=3cos(2.1x)

y^{^{''}} + 4 y = 3 cos (2.1 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 7.31707 \dots cos (2.1 x)

Solution steps

y^{''} (x) + 4 y = 3 cos (2.1 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 7.31707 \dots cos (2.1 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) - 7.31707 \dots cos (2.1 x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x}, y (\frac{π}{2}) = 0, y (0) = 0

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

y^{^{''}} + 16 y = 6 sin (4 t), y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 6 x + 2 - 3 e^{2}

y^{^{''}} + 2 y^{^{'}} - 3 y = - 20 sin (x)

What is the solution for y^{''}+4y=3cos(2.1x) ?
The solution for y^{''}+4y=3cos(2.1x) is y=c_{1}cos(2x)+c_{2}sin(2x)-7.31707…cos(2.1x)