18 Jun 2025

Exact and Inexact Differential Equations and Integrating Factors

The exactness condition, construction of a potential function, and integrating factors depending on one variable.

bsc semester-ii mathematical-physics exact-equations integrating-factors

Write a first-order equation as

\[M(x,y)dx+N(x,y)dy=0.\]

It is exact if there is a scalar function $\Psi(x,y)$ such that

\[d\Psi=\Psi_xdx+\Psi_y dy=Mdx+Ndy.\]

Thus $M=\Psi_x$ and $N=\Psi_y$. If the second partial derivatives are continuous, then

\[\boxed{M_y=\Psi_{xy}=\Psi_{yx}=N_x}.\]

On a simply connected region this condition is also sufficient. The solution is the level curve

\[\boxed{\Psi(x,y)=C}.\]

Constructing the potential

Consider

\[(2xy+3)dx+(x^2+4y)dy=0.\]

Since

\[M_y=2x=N_x,\]

the equation is exact. Integrate $M$ with respect to $x$, treating $y$ as constant:

\[\Psi=\int(2xy+3)dx=x^2y+3x+g(y).\]

Differentiate this expression with respect to $y$ and compare it with $N$:

\[\Psi_y=x^2+g'(y)=x^2+4y.\]

Hence $g’(y)=4y$, $g=2y^2$, and

\[\boxed{x^2y+3x+2y^2=C}.\]

Turning an inexact equation into an exact one

If $M_y\ne N_x$, seek a nonzero integrating factor $\mu$ such that

\[\mu Mdx+\mu Ndy=0\]

is exact. If $\mu=\mu(x)$, exactness requires

\[\frac{\partial(\mu M)}{\partial y} =\frac{\partial(\mu N)}{\partial x},\]

so

\[\mu M_y=\mu' N+\mu N_x.\]

Therefore, when the right-hand side depends on $x$ alone,

\[\boxed{ \frac{\mu'}{\mu}=\frac{M_y-N_x}{N}, \qquad \mu(x)=\exp\!\int\frac{M_y-N_x}{N}dx}.\]

Similarly, if $\mu=\mu(y)$,

\[\boxed{ \frac{1}{\mu}\frac{d\mu}{dy}=\frac{N_x-M_y}{M}, \qquad \mu(y)=\exp\!\int\frac{N_x-M_y}{M}dy}.\]

Example

For

\[(2y-x)dx+xdy=0,\]

$M_y=2$ and $N_x=1$, so the equation is inexact. Moreover,

\[\frac{M_y-N_x}{N}=\frac1x.\]

Thus $\mu=e^{\int dx/x}=\lvert x\rvert$ on an interval not crossing $x=0$; an irrelevant constant sign lets us take $\mu=x$. Multiplication gives

\[(2xy-x^2)dx+x^2dy=0.\]

Now

\[\frac{\partial}{\partial y}(2xy-x^2)=2x =\frac{\partial}{\partial x}(x^2).\]

Integrating the new $M$ with respect to $x$ gives

\[\Psi=x^2y-\frac{x^3}{3}+g(y).\]

Since $\Psi_y=x^2+g’(y)=x^2$, $g$ is constant. Hence

\[\boxed{x^2y-\frac{x^3}{3}=C}.\]

The exactness tests and recovered potentials are verified in the Unit II Maxima worksheet.