dy/dx= (x+1)y/2x^2 where y(1)=1, for all x>=1 i ended up with y^2/2= 1/2 ln (2x^2)+1/4 tan-1 (x/2) + K anythoughts?
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I think this DE is separable Did you use this method?
Last edited by pickslides; Dec 15th 2009 at 02:20 PM. Reason: bad latex
i did it that way before and came up with: ln [y] = ln [2x] + 1/4 tan-1 (x/2) + K still doesnt seem right
I think that solution is better, from here raise each side of the equation to the power of . This will eliminate the on the left hand side. Before this, may I ask how you arrived at the term? I can't see this being correct.
table of integrals du / a^2 + u^2 = 1/a tan-1 u/a + C
I see, I'm not sure if you have applied that the correct way here. consider
i definitely overcomplicated that one obviously cheers
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