Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation
$$
x d y-y d x=0
$$
which is not exact becomes so on multiplication by $1 / y^{2}$, for then we
$$
\frac{x}{y^{2}} d y-\frac{1}{y} d x=0
$$
which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation
$$
M d x+N d y=0
$$
to make it exact.
I. If $M x+N y \neq 0$ and the equation is homogeneous, then $\frac{1}{M x+N y}$ is an I.F.
II. If the equation $M d x+N d y=0$ is not exact but is of the form $f_{1}(x y) y d x+f_{2}(x y) x d y=0$, then $\frac{1}{M x-N y}$ is an
I.F., provided $M x-N y \neq 0$
III. When $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}$ is a function of $x$ alone, say $f(x)$, then I.F. $=e^{\int f(x) d x}$
IV. When $\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}$ is a function of $y$ alone, say $f(y)$, then I.F. $=e^{\int f(y) d y}$
The integrating factor to make the differential equation $\left(x y^{2}-e^{\frac{1}{x^{x}}}\right) d x-x^{2} y d y=0$ exact is
(A) $\frac{1}{x}$
(B) $\frac{1}{x^{2}}$
(C) $\frac{1}{x^{3}}$
(D) $\frac{1}{x^{4}}$