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:
$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
From your DE, we see that:
$\displaystyle M(x,y)=6xy^3+\cos y\implies \frac{\partial M}{\partial y}=18xy^2-\sin y$
$\displaystyle N(x,y)=2kx^2y^2-x\sin y\implies \frac{\partial N}{\partial x}=4kxy^2-\sin y$
Thus,
$\displaystyle 18xy^2-\sin y=4kxy^2-\sin y$
Try taking it from here.
Hope that this makes sense!
--Chris