Verify whether y^2 is an IF of dx/y+2x/y^2.dy=0?
1 Answer
Aug 11, 2018
It is not
Explanation:
If this is already exact , then:
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:
#{(f_(xy) = 1),(f_(yx) = 2),(f_(xy) ne f_(yx)):}#
Ditto: not exact.
Try instead an I.F. that is
#{(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#