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

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

Mathematics
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take dx to the othr side
You can split the eq
and then split

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Other answers:

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