Page:Calculus Made Easy.pdf/262

Example 4.

Suppose that $\quad \quad \quad M\,dx+N\,dy=0$.

We could integrate this expression directly, if $M$ were a function of $x$ only, and $N$ a function of $y$ only; but, if both $M$ and $N$ are functions that depend on both $x$ and $y$, how are we to integrate it? Is it itself an exact differential? That is: have $M$ and $N$ each been formed by partial differentiation from some common function $U$, or not? If they have, then

$\left\{{\begin{aligned}{\frac {\partial U}{\partial x}}=M,\\{\frac {\partial U}{\partial y}}=N.\end{aligned}}\right.$

And if such a common function exists, then

${\frac {\partial U}{\partial x}}\,dx+{\frac {\partial U}{\partial y}}\,dy$

is an exact differential (compare p. 175).

Now the test of the matter is this. If the expression is an exact differential, it must be true that

${\frac {dM}{dy}}={\frac {dN}{dx}}$;

for then

${\frac {d(dU)}{dx\,dy}}={\frac {d(dU)}{dy\,dx}}$,

which is necessarily true.

Take as an illustration the equation

$(1+3xy)\,dx+x^{2}\,dy=0$.