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:

$$0=\frac{\mathrm{d}F(\gamma (x))}{\mathrm{d}x}=F_x+F_y\frac{\mathrm{d}y}{\mathrm{d}x}.$$

Multiplying through by $\mathrm{d}x$ we obtain $$0=F_x\mathrm{d}x+F_y\mathrm{d}y.$$ Therefore, if we were not given the original function F but only an equation of the form:

$$M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0,$$

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:

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

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: $$\int M(x,y)dx=\int F_x(x,y)dx=F(x,y)+a(y)$$ $$\int N(x,y)dy=\int F_y(x,y)dy=F(x,y)+b(x)$$

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 $x$):$a^{\prime }(y)=-\int M_y(x,y)\mathrm{d}x+N(x,y)$and then integrate in $y$:$a(y)=\int [-\int M_y(x,y)dx+N(x,y)]\mathrm{d}y+c.$

(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 $$\begin{array}{cc}\frac{\partial a}{\partial x}=& \int \frac{\partial }{\partial x}[-\int M_y(x,y)\mathrm{d}x+N(x,y)]\mathrm{d}y\\ =& \int [-M_y(x,y)+N_x(x,y)]\mathrm{d}x\\ =& \int 0\mathrm{d}x\\ =& 0\end{array}$$