Question

how to get a solution for \(\sqrt{1-y^2} dx + \sqrt{1-x^2}dy = 0\)?

Answer 1

seperable variables

Answer 2

Agreed! :)

Answer 3

is it possible to use linear DE?

Answer 4

subtract one to one side then divide by the GCF

\[-\sqrt{1-y^2}\sqrt{1-x^2}\]

Answer 5

why would you want to?

Answer 6

because it is a problem in the linear DE section of the book lol

Answer 7

i thnk that would make it harder than it should. Your book is just tryng to make you be able to distinguish between certain ways of solving DE's

Answer 8

literally there is numerous of ways you could solve this however seperation by variables would be the most logical way to do this

Answer 9

hmm ok..so how to do with variable separable?

Answer 10

hmm okay..

Answer 11

my bad it should be

\sqrt{1-x^2}dy=-\sqrt{1-y^2}dx\]

Answer 12

\[\sqrt{1-x^2}dy=-\sqrt{1-y^2}dx\]

Answer 13

then divide?

Answer 14

then find a greest commonfactor to cancel out each term that is next to the differentials

Answer 15

GCF??

Answer 16

it'd be basically multiply both sides by 1/both functions

Answer 17

oh so it is divide lol

Answer 18

i'll show you... it's a way of doing the dividing in one step

Answer 19

okay :S

Answer 20

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

Answer 21

see how the comon factor cancels it out on both sides

Answer 22

you're left with

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

Answer 23

hmm yep

Answer 24

now you can integrate both sides

Answer 25

so it's trig sub?

Answer 26

yep i believe that is arcsin

Answer 27

thanks :D