Question

Solve the diff eqn

Answer 1

\[\LARGE \sqrt{1+x^2+y^2+x^2 y^2} +xy \frac{dy}{dx}=0\]

Answer 2

\[\large => \sqrt{(1+x^2)(1+y^2)}+xy \frac{dy}{dx}=0\]
not sure how to proceed further

Answer 3

@hartnn

Answer 4

@hartnn

Answer 5

@hartnn

Answer 6

@hartnn

Answer 7

@hartnn

Answer 8

@hartnn

Answer 9

@hartnn

Answer 10

@hartnn

Answer 11

@hartnn

Answer 12

@hartnn

Answer 13

@hartnn

Answer 14

@hartnn

Answer 15

@hartnn

Answer 16

@hartnn

Answer 17

@shubhamsrg

Answer 18

@shubhamsrg

Answer 19

@shubhamsrg

Answer 20

1+x^2 + y^2 + x^2 y^2 = (1+x^2) + y^2 (1+x^2)

take it from here.
best of luck for your future.
gaand marao bc. :)

Answer 21

take what from here o.O

Answer 22

(1+x^2)(1+y^2) ho gaya
ab variable separable form ho gaya.
dy/dx wala term LHS le jake , x wala func. 1 sath, y wala 1 sath

Answer 23

maine bhi to yahi kia tha 1st step me ..-..

Answer 24

RHS* le jake

Answer 25

to fir integration me problem hai? variable separate ho gaye ?

Answer 26

\[\large \sqrt{1+x^2} \sqrt{1+y^2} + xy y'=0\]
\[\frac{ \sqrt{1+x^2}}{x}+\frac{\sqrt{1+y^2}}{y} +y'=0\]

Answer 27

+ kaha se aya ? :3

Answer 28

sorry *

Answer 29

y' = ... likh ke
y wale terms dy ke sath
x wale dx ke sath
and just phuck it

Answer 30

\[\large \frac{dy}{dx}=-( \frac{ \sqrt{1+x^2}}{x} \times \frac{ \sqrt{1+y^2}}{y} )\]
abe variable sup?

Answer 31

yo maen

Answer 32

and x= tan A and y= tan B o.O ? to integrate

Answer 33

x= tanA kar sakte ho
y= tanB ka zarurat nahi
simply 1+y^2 = B le lena

Answer 34

=B ? kya matlab?