exact - 5

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle (-xy \sin x + 2y \cos x) \ dx + 2x \cos x \ dy = 0\)

 

[logistic_guy]

Let

\(\displaystyle M = -xy\sin x + 2y\cos x\)

And

\(\displaystyle N = 2x\cos x\)

Then,

\(\displaystyle \frac{\partial M}{\partial y} = (-x\sin x + 2\cos x) \neq (-2x\sin x + 2\cos x)= \frac{\partial N}{\partial x}\)

Since the partials are not equal, then the differential equation is \(\displaystyle \textcolor{red}{\bold{not}} \ \textcolor{blue}{\bold{exact}}\). And we have to do more work to solve it.

 

[logistic_guy]

Let us try this trick.

\(\displaystyle \frac{M_y - N_x}{N} = \frac{-x\sin x + 2\cos x + 2x\sin x - 2\cos x}{2x\cos x} = \frac{x\sin x}{2x\cos x} = \frac{1}{2}\tan x\)