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anonymous

  • 5 years ago

solve diff eq sinxcosx*dy\dx=y+sinx

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  1. anonymous
    • 5 years ago
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    take dx to the othr side

  2. anonymous
    • 5 years ago
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    You can split the eq

  3. anonymous
    • 5 years ago
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    and then split

  4. anonymous
    • 5 years ago
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    dy\dx = y+sinx \ sin x cos x .... sinxcosx = it would look like this.. dy\dx = y + sin x \ sinxcosx =0 .. rewrite it to.. dy\dy - y\1\2 sin2x = sinx \sinxcosx dy\dy - y\1\2 sin2x = 1\cosx .. hope i did nt to any wrongs..

  5. anonymous
    • 5 years ago
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    hmmm...yeah its fine till here...

  6. anonymous
    • 5 years ago
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    i got the same :)

  7. anonymous
    • 5 years ago
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    go on ...type

  8. anonymous
    • 5 years ago
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    Then, you need to find the Integration factor.. and U\[U \times Y \int\limits_{}^{} U \times 1\div cosx... Their .. U = e ^{\int\limits_{}^{} 1\div (1\div 2 \times \sin (2x))}\]

  9. anonymous
    • 5 years ago
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    sry i forgott i Minus before U=e∫-1÷(1÷2×sin(2x)

  10. anonymous
    • 5 years ago
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    hmm....then..

  11. anonymous
    • 5 years ago
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    hey i think the person who asked the question is not here

  12. anonymous
    • 5 years ago
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    aha..

  13. anonymous
    • 5 years ago
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    Have u done the this?

  14. anonymous
    • 5 years ago
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    that makes it \[e ^{\int\limits_{}^{} 2/sin2x}\]

  15. anonymous
    • 5 years ago
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    yepp

  16. anonymous
    • 5 years ago
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    And that Integral, i am not sure of..

  17. anonymous
    • 5 years ago
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    you can take, u =sin 2x and solve

  18. anonymous
    • 5 years ago
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    aha, so u mean that e∫2/sin2x = sin2x

  19. anonymous
    • 5 years ago
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    no, not that 'U', lets take.. say p=sin2x, so that , int(1/p) = log p = log (sin 2x) ...got it?

  20. anonymous
    • 5 years ago
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    noe it gives e^ (log (sin 2x))

  21. anonymous
    • 5 years ago
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    *now

  22. anonymous
    • 5 years ago
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    :P

  23. anonymous
    • 5 years ago
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    ln (1\p)

  24. anonymous
    • 5 years ago
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    why ln (1/p) ?

  25. anonymous
    • 5 years ago
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    Okey, i see..

  26. anonymous
    • 5 years ago
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    \[\int\limits_{}^{} 1\div p =\ln p\]

  27. anonymous
    • 5 years ago
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    yupp

  28. anonymous
    • 5 years ago
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    what about the inner derivata? 2x

  29. anonymous
    • 5 years ago
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    1\2 must be 1\2lnp

  30. anonymous
    • 5 years ago
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    that 2 in the denominator o f '1/2' goes to the numerator as '2'

  31. anonymous
    • 5 years ago
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    Found it. ∫ 1/sin(2x) dx = ∫ csc(2x) dx. Let u = 2x <==> du = 2 dx. Then, the integral becomes: 1/2 ∫ csc(u) du = -ln|cot(u) + csc(u)|/2+ C = -ln|cot(2x) + csc(2x)|/2 + C. <== ANSWER

  32. anonymous
    • 5 years ago
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    good going :)

  33. anonymous
    • 5 years ago
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    finally, got the answer huh:)

  34. anonymous
    • 5 years ago
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    haha.. yeah.. long time ago, with math..

  35. anonymous
    • 5 years ago
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    Need to refresh my memory!

  36. anonymous
    • 5 years ago
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    ha ha ..anyways you found it

  37. anonymous
    • 5 years ago
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    @nabaz&thinker thnxs for the help buddies i am getting the same ! but the ans is given to b \[y cotx=c+lntan \left( x/2 \right)\] is it the same?

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