i can't see at all how come (d/dx)(y'/sin^2y)=cosy/siny i would think that = [-2(y'^2)cosy/siny]+[y''/sin^2y] it's frustrating me please give me some help
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Originally Posted by pepsi i can't see at all how come (d/dx)(y'/sin^2y)=cosy/siny i would think that = [-2(y'^2)cosy/siny]+[y''/sin^2y] it's frustrating me please give me some help is there more to this problem than what you have posted? ... where/how did that first equation come about?
Originally Posted by pepsi i can't see at all how come (d/dx)(y'/sin^2y)=cosy/siny i would think that = [-2(y'^2)cosy/siny]+[y''/sin^2y] it's frustrating me please give me some help The second formula is right and the first formula is wrong. It's that simple. Unless, as skeeter has suggested, there is more to the problem.
Last edited by HallsofIvy; Mar 29th 2010 at 12:05 PM.
Originally Posted by skeeter is there more to this problem than what you have posted? ... where/how did that first equation come about? i tried uploading the full question (and answer) the bit i have trouble with is b) also before this, when finding the Euler equation with theta as the dependent variable i don't see why not just use the first integral, anyway, thanks
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