What is the solution for 60x^{''}+240x=0 ?

The solution for 60x^{''}+240x=0 is x=c_{1}cos(2t)+c_{2}sin(2t)

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60x^{''}+240x=0

60 x^{^{''}} + 240 x = 0

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

Solution steps

60 x^{''} (t) + 240 x = 0

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y^{^{'}} = 0

y^{^{''}} + (\frac{1}{2}) y = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = e^{5 x}

y^{^{'''}} + y^{^{''}} = 0, y^{^{'}} (0) = 0, y^{^{''}} (0) = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 8, y^{^{'}} (0) = 7

What is the solution for 60x^{''}+240x=0 ?
The solution for 60x^{''}+240x=0 is x=c_{1}cos(2t)+c_{2}sin(2t)