Question

help with separable eqns

Answer 1

\[x^{2}y' =1-x^{2}+y^{2}-x^{2}y ^{2}\]

Answer 2

so you are saying this can be done by just seperation of variables?
if so I'm going to look a little harder at that method

Answer 3

im not sure. its a prob from a section called separable eqns

Answer 4

oh

Answer 5

I see

Answer 6

I think

Answer 7

we can factor by grouping on the right hand side

Answer 8

\[1(1-x^2)+y^2(1-x^2) \\ (1-x^2)(1+y^2)\]

Answer 9

\[x^2y'=(1-x^2)(1+y^2) \\ \frac{1}{1+y^2} dy=\frac{1-x^2}{x^2} dx\]

Answer 10

now you are ready for integration

Answer 11

arctan(y)+C= -x-(1/x)+C ?

Answer 12

\[\arctan(y)+C_2=\int\limits x^{-2} dx+\int\limits (-1 ) dx \\ \arctan(y)+C_2=\frac{-1}{x}-x+C_1\]
looks great but you only need one constant really

Answer 13

\[\arctan(y)=\frac{-1}{x}-x+C\]

Answer 14

so then tan of both sides?

Answer 15

if you want to solve for y

Answer 16

I need to find the general solution (implicit if necessary, explicit if convenient)

Answer 17

well it is convenient here

Answer 18

by doing exactly what you said

Answer 19

so then \[y= \tan^{-1} (\frac{ -1 }{ x}-x+C)\] ?

Answer 20

oh no

Answer 21

I thought you were going to do tan( ) of both sides

Answer 22

typo

Answer 23

\[\arctan(y)=\frac{-1}{x}-x+C \\ \tan(\arctan(y))=\tan(\frac{-1}{x}-x+C) \\ y=\tan(\frac{-1}{x}-x+C)\]

Answer 24

it is supposed to be tan bc tan would cancel the arctan

Answer 25

my error

Answer 26

ok I just wanted to make sure :p

Answer 27

Thank you.

Answer 28

np :)

Answer 29

Care to help with another?