# finding a value to make a DE exact

• June 11th 2008, 08:09 AM
kithy
finding a value to make a DE exact

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
• June 11th 2008, 08:17 AM
Chris L T521
Quote:

Originally Posted by kithy

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:

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

From your DE, we see that:

$M(x,y)=6xy^3+\cos y\implies \frac{\partial M}{\partial y}=18xy^2-\sin y$

$N(x,y)=2kx^2y^2-x\sin y\implies \frac{\partial N}{\partial x}=4kxy^2-\sin y$

Thus,

$18xy^2-\sin y=4kxy^2-\sin y$

Try taking it from here.

Hope that this makes sense! :D

--Chris
• June 11th 2008, 08:56 AM
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?
• June 11th 2008, 06:03 PM
Chris L T521
Quote:

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? :D

--Chris
• June 11th 2008, 07:55 PM
kithy
Yes it really does, thank you so much, i understand it better now (Rofl)