Question

Initial value (x^2 -1)y' + 2xy^2 =0, y(0) = 1

Answer 1

\[(x^2 - 1)y' + 2 xy^2 =0 , y(0)=1\]

Answer 2

\[y'=\frac{ -2xy^2 }{ x^2-1 }\] I tried doing this but is no help --

Answer 3

no u are done with that

Answer 4

you are almost there, think "separation of variables"

Answer 5

I'm stuck with the y^2 on the other side, not sure how to move it(i think)

Answer 6

divide by y^2

Answer 7

\[\frac{ y' }{ y^2 } = \frac{ -2x }{ x^2-1 }\] oh wow -.-

Answer 8

lol :)

Answer 9

alright, thanks you.

Answer 10

you're welcome :P

Answer 11

\[\int\limits_{}^{}\frac{ dy }{ y^2 } = -\int\limits_{}^{}\frac{ 2x }{ x^2-1 }dx\]

Answer 12

\[-\frac{ 1 }{ y } = -\ln|x^2-1|+C\]

Answer 13

\[y=-\frac{ 1 }{ \ln|x^2-1| }\ +C\]

Answer 14

\[\frac{ 1 }{ \ln|0^2-1 |}-C = 1\]

Answer 15

\[\frac{ 1 }{ \ln|-1| }-C=1\]\[0-C=1\]\[C=-1\]

Answer 16

\[\frac{ 1 }{ \ln|x^2-1|}+1 = y\] Answer

Answer 17

@dan815 could you check my work please?

Answer 18

@dan815 is the ln|-1| alright? I feel that something is wrong but just might be that im sleepy

Answer 19

http://prntscr.com/6q6pi6

Answer 20

@dan815 ,Thanks you very much.