anonymous
  • anonymous
solve diff eq sinxcosx*dy\dx=y+sinx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
take dx to the othr side
anonymous
  • anonymous
You can split the eq
anonymous
  • anonymous
and then split

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anonymous
  • anonymous
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..
anonymous
  • anonymous
hmmm...yeah its fine till here...
anonymous
  • anonymous
i got the same :)
anonymous
  • anonymous
go on ...type
anonymous
  • anonymous
Then, you need to find the Integration factor.. and U\[U \times Y \int\limits_{}^{} U \times 1\div cosx... Their .. U = e ^{\int\limits_{}^{} 1\div (1\div 2 \times \sin (2x))}\]
anonymous
  • anonymous
sry i forgott i Minus before U=e∫-1÷(1÷2×sin(2x)
anonymous
  • anonymous
hmm....then..
anonymous
  • anonymous
hey i think the person who asked the question is not here
anonymous
  • anonymous
aha..
anonymous
  • anonymous
Have u done the this?
anonymous
  • anonymous
that makes it \[e ^{\int\limits_{}^{} 2/sin2x}\]
anonymous
  • anonymous
yepp
anonymous
  • anonymous
And that Integral, i am not sure of..
anonymous
  • anonymous
you can take, u =sin 2x and solve
anonymous
  • anonymous
aha, so u mean that e∫2/sin2x = sin2x
anonymous
  • anonymous
no, not that 'U', lets take.. say p=sin2x, so that , int(1/p) = log p = log (sin 2x) ...got it?
anonymous
  • anonymous
noe it gives e^ (log (sin 2x))
anonymous
  • anonymous
*now
anonymous
  • anonymous
:P
anonymous
  • anonymous
ln (1\p)
anonymous
  • anonymous
why ln (1/p) ?
anonymous
  • anonymous
Okey, i see..
anonymous
  • anonymous
\[\int\limits_{}^{} 1\div p =\ln p\]
anonymous
  • anonymous
yupp
anonymous
  • anonymous
what about the inner derivata? 2x
anonymous
  • anonymous
1\2 must be 1\2lnp
anonymous
  • anonymous
that 2 in the denominator o f '1/2' goes to the numerator as '2'
anonymous
  • anonymous
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
anonymous
  • anonymous
good going :)
anonymous
  • anonymous
finally, got the answer huh:)
anonymous
  • anonymous
haha.. yeah.. long time ago, with math..
anonymous
  • anonymous
Need to refresh my memory!
anonymous
  • anonymous
ha ha ..anyways you found it
anonymous
  • anonymous
@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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