Question

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

Answer 1

take dx to the othr side

Answer 2

You can split the eq

Answer 3

and then split

Answer 4

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..

Answer 5

hmmm...yeah its fine till here...

Answer 6

i got the same :)

Answer 7

go on ...type

Answer 8

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

Answer 9

sry i forgott i Minus before U=e∫-1÷(1÷2×sin(2x)

Answer 10

hmm....then..

Answer 11

hey i think the person who asked the question is not here

Answer 12

aha..

Answer 13

Have u done the this?

Answer 14

that makes it \[e ^{\int\limits_{}^{} 2/sin2x}\]

Answer 15

yepp

Answer 16

And that Integral, i am not sure of..

Answer 17

you can take, u =sin 2x and solve

Answer 18

aha, so u mean that e∫2/sin2x = sin2x

Answer 19

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

Answer 20

noe it gives e^ (log (sin 2x))

Answer 21

*now

Answer 22

:P

Answer 23

ln (1\p)

Answer 24

why ln (1/p) ?

Answer 25

Okey, i see..

Answer 26

\[\int\limits_{}^{} 1\div p =\ln p\]

Answer 27

yupp

Answer 28

what about the inner derivata? 2x

Answer 29

1\2 must be 1\2lnp

Answer 30

that 2 in the denominator o f '1/2' goes to the numerator as '2'

Answer 31

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

Answer 32

good going :)

Answer 33

finally, got the answer huh:)

Answer 34

haha.. yeah.. long time ago, with math..

Answer 35

Need to refresh my memory!

Answer 36

ha ha ..anyways you found it

Answer 37

@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?