Proving v^2 = n^2 (A^2-x^2) from v dv/dx = -n^2x

[rhyso88]

hey guys, just hpoping sum1 could point me in the rite direction...i can't wrk out how to proove v^2=n^2 (A^2-x^2) from

v dv/dx = -n^2x

obviously you have to integrate...but i don't know how they get the A into the equation??

what i have tried is intergating both sides with recpect to x first

v dv/dx x = (-n^2x^2)/2 and given we know dv/dx . dx/ dt = -n^2x and v = dx/dt the sub into have

...well it brings me no where actually hence this post

thanks for you help

 

[Eds]

Integrate both sides wrt x, giving you:

integral v dv = -n^2 integral x dx. This, with intial conditions( v=0 at x = a), gives you the required quadratic.

Next up,

learn to spell and write English.

 

[rhyso88]

well thanks for that, and I am sorry about the english...just typing quickly and didn't realize correct grammar is required in a Maths forum...but never the less thanks