Solve: #d^2x//dt^2 + g sin theta t =0# if # theta=g//l# , and g and l are constants?

2 Answers
Ultrilliam
Apr 27, 2018

#x(t) = - g sin(g/1) t^2/2 + At + B#

Explanation:

You are asking for the solution to:

#(d^2x)/(dt^2) = - g sin(g/1) #

#implies (d x)/(dt) = - g sin(g/1) t + A#

#implies x(t) = - g sin(g/1) t^2/2 + At + B#

Steve M
Apr 28, 2018

# x = l^2/g \ sin theta t + At + B #

Explanation:

We have:

# (d^2x)/(dt^2) + g sintheta t = 0# where #theta=g/l#, a constant

Which we can write as:

# (d^2x)/(dt^2) = - g sintheta t #

We can "separate the variables" , to get:

# (dx)/(dt) = int \ - g sintheta t \ dt #

And integrating give us:

# (dx)/(dt) = g/theta cos theta t + A #

And repeating we get:

# x = int \ g/theta cos theta t + A \ dt #

So that:

# x = g/theta^2 sin theta t + At + B #

# \ \ = g/(g/l)^2 sin theta t + At + B #

# \ \ = g l^2/g^2 \ sin theta t + At + B #

# \ \ = l^2/g \ sin theta t + At + B #

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