Question

The solution of differential equation..
(x^2-y^2)dx +2xy dy=0

Answer 1

Is the ans
1/2 ( y/x)^2=ln(c/x)

Answer 2

that's and exact equation

Answer 3

No, it is not,kid

Answer 4

why nit?

Answer 5

\(\partial M/\partial y=-2y\neq \partial N/\partial x= 2y\)

Answer 6

Yes , that equation is not right

Answer 7

So, you have to find \(\mu (x)\) or \(\mu (y)\) or \(\mu(xy)\) to have is exact
I worked on it, \(\mu (x)=1/x^2\) works well

Answer 8

yes, i miss that minus

Answer 9

Sorry ! I didn't take that 2 into consideration..

Answer 10

@Pawanyadav use (1/x^2) to multiple into both sides and then solve as usual

Answer 11

I do the same but can't reach the answer

Answer 12

Show your work, please.

Answer 13

I got this
y^2+x^2=cx

Answer 14

yeah old man is right

Answer 15

I got logx= log (1+ (y/x)^2)+c
Whats next

Answer 16

so what is the problem here, you made it exact

Answer 17

Not log x ..
It has to be log (1/x)

Answer 18

and solved it?

Answer 19

kid, then solve for f(x,y)

Answer 20

yes i know old man, I was talking to the poster

Answer 21

kid, take over, please :)

Answer 22

@samigupta8 you got it right

Answer 23

\[(1-\frac{y^2}{x^2})dx+2\frac{y^2}{x}dy=0\]

Answer 24

which one did you integrate

Answer 25

@samigupta how its log (1/x)

Answer 26

that supposed to have only y not y^2

Answer 27

I got these equations
v/2 + xdv/dx +1/2v=0
So now u take all v terms on left side and x terms on right..
Solve it u will get the desired ans

Answer 28

v is y/x here

Answer 29

\[f(x,y)=\int 1-\frac{y^2}{x^2}dx=x+y^2/x+h(y)=c\]

Answer 30

Then, take derivative of the above result w.r.t y
it is = 2y/x +h'(y)

Answer 31

yeah i know, just looking for something

Answer 32

Ok, kid, please, give him step by step. (Whole thing, please).

Answer 33

On dividing by x^2 dx,we got
(1-(y/x)^2)=-2(y/x)y'
Now put y=tx
I got
(1-t^2)/-2t=t+x(dt/dx)

Answer 34

Then
(t^2-1)/2t -t=x(dt/dx)

Answer 35

one second i will write the procedure

Answer 36

Finally
dx/x=(-2t/(1+t^2))dt
On integrate
-logx =log (1+t^2)+c

Answer 37

Thanks ,, finally I got it

Answer 38

Thanks old man and kid ..your method is little difficult..

Answer 39

what method are you using ?

Answer 40

Its all above..

Answer 41

yeah don't get, just tell me the name, i will check on it

Answer 42

Homogenisation of equation

Answer 43

oh i see

Answer 44

First convert to homogeneous equation than solve by variable separable form

Answer 45

so you get the solution i wrote

Answer 46

yeah i have seen that before just don;t recall the whole procedure now

Answer 47

No ,I can't open this page

Answer 48

it a word document

Answer 49

OK ..I'll get i

Answer 50

i made it a PDF

Answer 51

Thanks , now I can opeopen

Answer 52

I never solve any questions like this before

Answer 53

bravo, kid