Question

Find an appropriate substitution and solve (x-y)^2y'=1
see attachment

Answer 1

sorryyy I froze..... I'll attach what I got. I was going to use partial fractions until I got a cube

Answer 2

I am...it's subsitution like the formula du/f(1,u)-u = 1/x dx

Answer 3

I did that and got a cube...I tr4ied factor by grouping and it didn't budge

Answer 4

unless I distributed the negative wrong... but wait the negative sign has to be distributed especially if it's on the right hand side of the equation

Answer 5

it would've been x^2-2xy+y^2 if I hadn't done F(1,u)

Answer 6

unless let no multiply by 1/x but again I end up with (1-y/x)^2 and if u = y/x
(1-u)^2

Answer 7

How did you know what to substitute?

Answer 8

O_O

Answer 9

oh wait a sec... u = (x-y)

Answer 10

How did you pick \(y=ux\)?

Answer 11

-__- one sec let me redo this on paint

Answer 12

Maybe \(y=x-u\)

Answer 13

And \(dy = dx-\frac {du}{dx}dx\)

Answer 14

crud reached a dead end when I did partial fractions I had a =1
A+ B = 0
A-B = 0

Answer 15

Partial fractions? Where did you get to so far?

Answer 16

You have \[
1-u' = \frac{1}{u^2}
\]

Answer 17

It's been a while since I've done this sort of thing.

Answer 18

if only the partial fraction would work!!!!!!!! OOOOOO!!!!

Answer 19

:(

Answer 20

u = x-y.......................

Answer 21

So \(y=f(x,u)=x-u\) and \[
\frac{dy}{dx} = \frac{d}{dx}(x-u)\frac{dx}{dx} + \frac{d}{du}(x-u)\frac{du}{dx}
\]

Answer 22

but that just leaves y' = 1/(y)^2
d/dx(x-u)dx/dx+d/du(x-u)du/dx = 1/y^2 what a messss

Answer 23

Hmm, let me think about this for a bit.

Answer 24

same... I was doing partial fractions a bit and only got one part of it to show up

Answer 25

@UsukiDoll again, please answer me what is your course?

Answer 26

differential equationss

Answer 27

aha so, why don't you substitute u = x -y? so du = 1 - dy/dx\(rightarrow\) dy/dx = 1 - du/dx
yours become \(u^2 (1- \frac{du}{dx})=1\) and then \(u^2 -1= \frac{du}{dx}\) then \(\dfrac{du}{u^2-1}=dx\)

Answer 28

partial fraction for LHS . that's it

Answer 29

don't get what I mean?

Answer 30

omggggggggggggggg that saved my booty!!!!!!! x()()()()

Answer 31

How did you get \(u^2-1=\frac{du}{dx}\)?

Answer 32

\[
u^2-1 =u^2\frac{du}{dx}
\]