Question

\[{dy \over dx} = {\sqrt{x^2+y^2} - x \over y}\]
Solve the DE by using a substitution

Answer 1

perhaps, let y=vx

Answer 2

This equation is homogenous, correct?

Answer 3

Yes :)

Answer 4

Divide by x in numerator and denominator and do as what @experimentX told :)

Answer 5

divide by x?

Answer 6

Yes
\[{dy \over dx} = {\sqrt{1+(y/x)^2} - 1 \over (y/x)}\]
Now proceed :)

Answer 7

okay, I did that and then took y=vx so v=y/x and dy/dx = v + x dv/dx

Answer 8

\[v + x{dv \over dx} = {\sqrt{1 + v^2} - 1 \over v}\] and got this

Answer 9

Take v to RHS and then take LCM

Answer 10

now use variable separation technique.

Answer 11

I am getting log x = log (1- (y/x)) +log c

Answer 12

\[{v \over \sqrt{1 + v^2} - 1 - v^2} dv = {dx \over x}\] is what I came up with

Answer 13

@gmer , yes. take - sign out to write it as -(1+v^2)

Then take t^2 = 1+v^2 and integrate

Answer 14

I rhink 1 +v^2 = t would be better supposition, eliminates a step.

Answer 15

@experimentX , no it rather simplifies the problem :D

Answer 16

it --> t^2 = 1+ v^2

Answer 17

but then how do you find dt if you use t^2 ?

Answer 18

Simple
2 t dt = 2 v dv
tdt = vdv

Answer 19

Finally you get

\[\int\limits_{}^{}{ tdt \over t -t^2 } = \int\limits_{}^{}{ dt \over 1 -t }\]

Answer 20

@gmer , I hope you can complete the problem from here :D

Answer 21

looks like your supposition was better than mine ... LOL

Answer 22

gotta sleep good night!!

Answer 23

Yes. It is 7 am. I better catch up with some sleep too. :D

Answer 24

same here LOL:D

Answer 25

Looks like many Indians have got addicted to OS :P

Answer 26

Thanks guys, think that I got it

Answer 27

Hey @shivam_bhalla , mind checking over my work? It got kinda weird at the end and I'm unsure of it