What is the solution for x^{''}+49x=99cos(4t) ?

The solution for x^{''}+49x=99cos(4t) is x=c_{1}cos(7t)+c_{2}sin(7t)+3cos(4t)

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x^{''}+49x=99cos(4t)

x^{^{''}} + 49 x = 99 cos (4 t)

x = c_{1} cos (7 t) + c_{2} sin (7 t) + 3 cos (4 t)

Solution steps

x^{''} (t) + 49 x = 99 cos (4 t)

Solve linear ODE:

x = c_{1} cos (7 t) + c_{2} sin (7 t) + 3 cos (4 t)

x = c_{1} cos (7 t) + c_{2} sin (7 t) + 3 cos (4 t)

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2} - 65

y^{^{''}} + 2 y^{^{'}} - 8 y = 4 e^{- 2 x} - e^{- x}

y^{^{''}} - 4 y^{^{'}} + 4 y = (x + 1) \cdot e^{2 x}

y^{^{''}} + 6 y^{^{'}} + 13 y = 10 e^{- 2 t}, y (0) = 3

4 y^{^{''}} + y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

What is the solution for x^{''}+49x=99cos(4t) ?
The solution for x^{''}+49x=99cos(4t) is x=c_{1}cos(7t)+c_{2}sin(7t)+3cos(4t)