Question

General solution

Answer 1

huh? question? @LilySwan

Answer 2

She has given solution, we need to find what the question is.. :P

Answer 3

sorry network issue

Answer 4

It is okay..

Answer 5

There is a method to solve these types of Diff Equations..

Answer 6

If \(M.dx + N.dy = 0\) is non-exact, but it is in the form of \(y f(x,y)dx + x g(x,y)dy = 0\), and if \(Mx - Ny \ne 0\), then :
\[Integrating \; \; Factor = \frac{1}{Mx-Ny}\]

Answer 7

And General Solution is given by:
\[\int\limits (y \cdot f(x,y) \cdot IF) dx + \int\limits ( x \cdot g(x,y). IF) [\text{terms \not containing x}] dy = c\]

Answer 8

Your equation is :
\[y \sqrt{1+x^2}dx + x \sqrt{1+y^2} dy = 0\]

Answer 9

\[f(x,y) = \sqrt{1+x^2} \\ g(x,y) = \sqrt{1+y^2}\]

Answer 10

cant we take a more easier approach

Answer 11

I am also thinking the same.. :P

Answer 12

Okay, on your marks..

Answer 13

Your equation is :
\[y \sqrt{1+x^2}dx + x \sqrt{1+y^2} dy = 0\]
\[y \sqrt{1+x^2}dx = - x \sqrt{1 + y^2} dy \\ \frac{\sqrt{1+x^2}}{x} \cdot dx = \frac{-\sqrt{1+y^2}}{y} \cdot dy\]

Answer 14

Getting it? I just took second term on Right side in first step, then swap brought \(x\) and \(y\) on sides opposite to both, Good?

Answer 15

@unklerhaukus , save me from here..!!!

Answer 16

ok we used the separable method

Answer 17

Separable Method only if we can integrate those two big integrals, I gotta go now, but @ganeshie8 please help this girl with Differential Equations.. - water.

Answer 18

Or @iambatman