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anonymous

  • 4 years ago

xy'-y=tan(y/x) Book says y=x[arcsin(cx+2npi)] or y=x[-arcsin(cx+(2n-1)pi)], & y=0 I get y=0 (easy) but when i solve i get y=x[arcsin(ce^-(1/x)+2npi)]

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  1. anonymous
    • 4 years ago
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    i also understand the shift of equation as a result of the integral of \[\int\limits_{}^{}\cot (v)= \ln \left| \sin(v) \right|\] the positive value y=x[arcsin(cx+2npi)] negative value y=x[-arcsin(cx+(2n-1)pi)]

  2. TuringTest
    • 4 years ago
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    you are in good company, wolfram has your answer http://www.wolframalpha.com/input/?i=xy%27-y%3Dtan%28y%2Fx%29 not sure if that's somehow the same as the books answer, but I don't see how...

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