Thread: finding a value to make a DE exact

Please help me out!!!

this is the question: Find the value for k so that the given DE is exact

(6x (y^3) + cos y )dx + (2k (x^2) (y^2) - x sin y)dy =0


thank you so much

Originally Posted by kithy

Please help me out!!!

this is the question: Find the value for k so that the given DE is exact

(6x (y^3) + cos y )dx + (2k (x^2) (y^2) - x sin y)dy =0


thank you so much

For a DE to be exact, the following must be true:



From your DE, we see that:





Thus,



Try taking it from here.

Hope that this makes sense!

--Chris

Ok, i equated the two equations and i came out with K= (18xy^2)/(4xy^2) and so i am assuming the xy^2 cancel out and we are left with k=9/2 is that right?

Originally Posted by kithy

Ok, i equated the two equations and i came out with K= (18xy^2)/(4xy^2) and so i am assuming the xy^2 cancel out and we are left with k=9/2 is that right?

Sorry for the late response. I've been in school all day...

Yes, that is correct. Does is make more sense now?

--Chris

Yes it really does, thank you so much, i understand it better now