CALCULUS initial value problem HELP PLEASE :)?
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CALCULUS initial value problem HELP PLEASE :)?

[From: ] [author: ] [Date: 14-05-12] [Hit: ]
so, applying the above rule,dv/dx = (1/x)y - (1/x)v, or,......
http://i.snag.gy/hMVVv.jpg

Can someone explain how you perform that step, with the black arrow next to it?

THANKS

-
As you probably know, the product rule states that if
P(x) = u(x)*v(x)
dP/dx = udv/dx + vdu/dx = uv' + vu'

As part of your solution you had v = y/x = yx^-1, so, applying the above rule,
dv/dx = y*-x^-2 + x^-1*dy/dx which can be rearranged as
dv/dx = (1/x)y' - (1/x)(y/x)
Just recognise that y/x = v and you have it
dv/dx = (1/x)y' - (1/x)v, or,
v' = (1/x)(y' - v)

and so on to

v'x + v = y' = -2v^3 + v (from earlier)
Cancelling v on both sides
xdv/dx = -2v^3 and separating variables
dv/v^3 = -2dx/x etc etc

Regards - Ian H
1
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