What is the solution for 10x^{''}+200x=0 ?

The solution for 10x^{''}+200x=0 is x=c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t)

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10x^{''}+200x=0

10 x^{^{''}} + 200 x = 0

x = c_{1} cos (25 t) + c_{2} sin (25 t)

Solution steps

10 x^{''} (t) + 200 x = 0

Solve linear ODE:

x = c_{1} cos (25 t) + c_{2} sin (25 t)

x = c_{1} cos (25 t) + c_{2} sin (25 t)

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 5, y^{^{'}} (0) = - 3

5 y^{^{''}} + 2 y^{^{'}} + 7 y = 0

3 y^{^{''}} + 8 y^{^{'}} + 3 y = 0

y^{^{''''}} - y^{^{''}} + 4 y = 0

x^{^{''}} - 4 x^{^{'}} + 4 x = 0

What is the solution for 10x^{''}+200x=0 ?
The solution for 10x^{''}+200x=0 is x=c_{1}cos(2sqrt(5)t)+c_{2}sin(2sqrt(5)t)