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

1 Answer
Aug 11, 2018

Answer:

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#