Question

Verify whether y^2 is an IF of dx/y+2x/y^2.dy=0?

Answer 1

It is not

Explanation:

If this is already exact , then:

`underbrace(1/y)_(= f_x) dx + underbrace((2x)/y^2)_(= f_y) dy=underbrace(0)_(because df = f_x dx + f_y dy = 0)`

The mixed partials are not equal, so it is not exact.

• `{(f_(xy) = - 1/y^2),(f_(yx) = 2/y^2),(f_(xy) ne f_(yx)):}`

With your suggested I.F., it becomes:

`underbrace(y)_(= f_x) dx + underbrace(2x )_(= f_y) dy=0`

• `{(f_(xy) = 1),(f_(yx) = 2),(f_(xy) ne f_(yx)):}`

Ditto: not exact.

Try instead an I.F. that is `y^3`:

`underbrace(y^2)_(= f_x) dx + underbrace(2xy)_(= f_y) dy=0`

• `{(f_(xy) =2y),(f_(yx) = 2y),(bb(f_(xy) = f_(yx))):}`

This is exact.

Integrating:

• `int \ y^2 \ partial x = xy^2 + alpha(y)`

• `int \ 2xy \ partial y = xy^2 + beta(x)`

So there is solution:

• `qquad f(x,y) = xy^2 = C`