Question

\[xy'+y=y ^{2},y(1)=-1\]

Answer 1

i reached \[2\tanh ^{-1}(1-2y)=\ln x+c\]

Answer 2

any kind soul can guide me?

Answer 3

Let me call on people that may be able to answer this:
@across @lalaly @JamesJ @ash2326
it's just beyond me, sorry
look forward to seeing the solution :)

Answer 4

owh kk ty. actually i dont know what is tanh also. i reached the tanh based on the integration table ._. lol

Answer 5

looks like Bernoulli's eqn

Answer 6

http://en.wikipedia.org/wiki/Bernoulli_differential_equation

Answer 7

http://www.wolframalpha.com/input/?i=xy%27%2By+%3D+y%5E2%2C+y%281%29+%3D+-1

Answer 8

ou im using separable method. anyway, can explain how \[2\tanh ^{-1}(1-2y)=\ln x+c\]becomes\[\ln(1-y) -\ln y=\ln x+c\]
basically its the LHS tat i dont understand

Answer 9

http://upload.wikimedia.org/wikipedia/en/math/8/e/f/8ef979b2cbfb464fd813a557fdc8a4ec.png

Answer 10

ou.. ._. never knew there's such thing. TQ! =)

Answer 11

er, but if z=1-2y, why after becoming ln its 1-y

Answer 12

should be
1/2*2(ln(1+1-2y) - ln(2y)) = ln2(1-y) - ln2y = ln 2 + ln(1-y) - ln 2 - ln y

Answer 13

\[\tanh ^{-1}z=\frac12\ln(1+z)-\ln(1-(1-2y))\]\[\tanh ^{-1}z=\frac12[\ln(1+1-2y)-\ln(1-z)]=\ln2+\frac12\ln(1-y)-\frac12\ln y\]yep, experimentX beat me to it again
ln2 is a constant which can be absorbed into C is the point...

Answer 14

you missed ln 2 at last .. lol

Answer 15

owh. now i get it... ._. TQVM =)

Answer 16

d'oh, I was racing you too much :/
so they just cancel anyway...