Ordinary Differential Equations/Exact equations

Introduction
Suppose the function $$F(x,y)$$ represents some physical quantity, such as temperature, in a region of the $$xy$$-plane. Then the level curves of F, where $$F(x,y)=\text{constant}$$, could be interpreted as isotherms on a weather map (i.e curves on a weather map representing constant temperatures). Along one of these curves, $$\gamma(x)=(x,y(x))$$, of constant temperature we have, by Chain rule and the fact that the temperature, F, is constant on these curves:

Multiplying through by $$\mathrm{d}x$$ we obtain Therefore, if we were not given the original function F but only an equation of the form:

we could set $$F_{x}:=M(x,y),F_{y}:=N(x,y)$$ and then by integrating figure out the original $$F$$.

Method formal steps
(1) First ensure that there is such an $$F$$, by checking the exactness-condition:

This is because if there was such an F, then $$\frac{\partial{}M}{\partial{}y}=\frac{\partial^{2}F}{\partial{}y\partial{}x}=\frac{\partial^{2}F}{\partial{}x\partial{}y}=\frac{\partial{}N}{\partial{}x},$$

where $$\frac{\partial}{\partial{}x}$$ and $$\frac{\partial}{\partial{}y}$$ simply denote the partial derivatives with respect to the variables $$x$$ and $$y$$ respectively (where we hold the other variable constant while taking the derivative).

(2)Second, integrate $$M,N$$ with respect to $$x,y$$ respectively: $$\int \!{}M(x,y)dx=\int \!{}F_{x}(x,y)dx= F(x,y)+a(y)$$

for some unknown functions $$a,b$$ (these play the role of constant of integration when you integrate with respect to a single variable). So to obtain $$F$$ it remains to determine either $$a$$ or $$b$$.

(3)Equate the above two formulas for $$F(x,y)$$:$$\int \!{}M(x,y)\mathrm{d}x+a(y)=F(x,y)=\int \!{}N(x,y)\mathrm{d}y+b(x).$$

(4) Since to find $$F$$ it suffices to determine $$a$$ or $$b$$, pick the integral that is easier to evaluate. Suppose that $$\int M(x,y)\mathrm{d}x$$ is easier to evaluate. To obtain $$a(y)$$ we differentiate both expression for $$F$$ in $$y$$ (for fixed $$x$$):$$a'(y)=-\int \!{}M_{y}(x,y)\mathrm{d}x+N(x,y) $$and then integrate in $$y$$:$$a(y)=\int\left[ -\int M_{y}(x,y)dx+N(x,y)\right]\mathrm{d}y+ c.  $$

(5)Observe that $$a$$ is only a function of $$y$$ since if we differentiate the expression we found for $$a$$ and use step $$1$$ we find that $$ \begin{align} \frac{\partial{}a}{\partial{}x}=&\int\!{}\frac{\partial}{\partial{}x}\left[-\int\!{}M_{y}(x,y)\mathrm{d}x+N(x,y)\right]\mathrm{d}y\\ =&\int\!{}\left[-M_{y}(x,y)+N_{x}(x,y)\right]\mathrm{d}x\\ =&\int\!{}0\mathrm{d}x\\ =&0 \end{align} $$