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I need to learn how to solve differential equations using either the Exact Equation Approach and or the Special Integrating Factor methods. Below is a differential Equation to solve.

$(2xy^2 + \cos x) \text{d}x + (2x^2 y + \sin y)\text{d}y = 0$

I would appreciate it if you would include comments to explain steps taken. Thanks in advance


Following your example I did the following

Given $$ (2x + y).dx + ( x - 2y).dy = 0$$

a) $$ M(x,y)=2x + y, N(x,y)= x - 2y $$

b) check if the d.e is exact.
$\frac{\partial M}{\partial y}=\frac{\partial}{\partial y}\left(2x + y\right)= 1 =\frac{\partial}{\partial x}\left(x - 2y\right)=\frac{dN}{\partial x}$.

c) $$ f\left(x,y\right)=\int M(x,y)\text{d}x =\int(2x + y)\text{d}x=x^{2} + xy + g(y).$$

d) To find $g\left(y\right)$
$f_{y}\left(x,y\right)=\frac{\partial}{dy}\left(x^{2}+ xy + g(y)\right)=0 + x + g'\left(y\right).$

e) Upon comparing with $N\left(x,y\right)$, I find $g'\left(y\right)= - 2y$ which implies that $g\left(y\right)=- y^{2}+K$

Therefore, the general solution is $ x^2 + xy - y^2 =C$.

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    You have been given nice answers but just in the case you wondered what the word *exact* **really** means: it comes from differential geometry. A differential form $\omega$ is *exact* if there exist a potential form $\alpha$ such that $\omega = {\rm d} \alpha$ where ${\rm d}$ is an exterior derivative. On the other hand, the form is *closed* if ${\rm d} \omega = 0$. From the commutation of second partial derivatives one has ${\rm d}^2 \omega = 0$, i.e. every exact form is closed. But when working on ${\mathbb R}^n$ the opposite also holds: a closed form is exact, i.e. you can find a potential.2011-06-21
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    @user1065: You missed out the term $xy$ when computing $f_y(x,y)$. Otherwise, everything else is perfect!!2011-06-21

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