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 $N$:

\[\Psi_y=x^2+g^{\prime}(y)=x^2+4y.\]

Hence $g^{\prime}(y)=4y$, $g=2y^2$, 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

\[\mu M_y=\mu^{\prime} N+\mu N_x.\]

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

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

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 $\Psi_y=x^2+g^{\prime}(y)=x^2$, $g$ 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.

Solved Problems

1. An exact initial-value problem

Solve

\[(2x+y)dx+(x+2y)dy=0, \qquad y(0)=1.\]

Here $M=2x+y$ and $N=x+2y$, so

\[M_y=1=N_x.\]

Integrating $M$ with respect to $x$,

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

Then

\[\Psi_y=x+g^{\prime}(y)=x+2y,\]

which gives $g=y^2$. The implicit solution is

\[x^2+xy+y^2=C.\]

The initial condition gives $C=1$, hence

\[\boxed{x^2+xy+y^2=1}.\]

At $x=1/2$ this becomes $4y^2+2y-3=0$. The branch connected continuously to $y(0)=1$ is

\[\boxed{y\!\left(\frac12\right)=\frac{-1+\sqrt{13}}4=0.65139}.\]

Both partial derivatives of $\Psi=x^2+xy+y^2$ reproduce $M$ and $N$, and the selected point gives $\Psi=1$ exactly.

2. An integrating factor depending on y

Solve, on $x>0$ and $y>0$,

\[y\,dx+3x\,dy=0, \qquad y(1)=2.\]

Now $M=y$, $N=3x$, $M_y=1$, and $N_x=3$. Therefore

\[\frac{N_x-M_y}{M}=\frac2y,\]

which depends only on $y$. The integrating factor is

\[\mu(y)=\exp\!\int\frac{2}{y}dy=y^2.\]

After multiplication,

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

Thus $xy^3=C$. The initial condition gives $C=8$, and positivity selects

\[\boxed{y=2x^{-1/3}}.\]

Indeed, $y+3xy^{\prime}=0$. The solution remains real and positive on the stated interval and diverges as $x\to0^+$, consistent with the conserved product $xy^3=8$.

Descriptive Questions

1. Derive the exactness condition $M_y=N_x$ from the existence of a potential function.
2. Explain why the condition $M_y=N_x$ is sufficient only after suitable regularity and domain assumptions are imposed.
3. Derive the tests for integrating factors depending only on $x$ and only on $y$.
4. Explain why multiplying an integrating factor by a nonzero constant does not change the solution curves.

Numerical Problems

1. For $(3x^2+2y)dx+(2x+4y^3)dy=0$, find the potential constant on the curve through $(1,1)$. Final answer: $\Psi=x^3+2xy+y^4$ and $C=4$.
2. Evaluate the line integral of $(2x+y)dx+(x+2y)dy$ from $(0,0)$ to $(1,2)$ along any smooth path. Final answer: $\Delta(x^2+xy+y^2)=7$.
3. Find an integrating factor depending on $x$ for $4y\,dx+x\,dy=0$. If $y(1)=16$, find $y(2)$. Final answer: $\mu=x^3$, $x^4y=16$, and $y(2)=1$.
4. On the unit circle traversed counterclockwise, evaluate $\oint Mdx+Ndy$ for $M=-y/(x^2+y^2)$ and $N=x/(x^2+y^2)$. Final answer: $2\pi$; the nonzero closed integral shows why the punctured plane does not admit a single-valued global potential for this form.

The problem solutions and exact evaluations are verified in the Unit II problem-check worksheet.

References

1. Exact differential equation — Wikipedia
2. William E. Boyce, Richard C. DiPrima, and Douglas B. Meade, Elementary Differential Equations and Boundary Value Problems, 11th ed., chapter “First-Order Differential Equations,” Wiley.
3. Erwin Kreyszig, Advanced Engineering Mathematics, 10th ed., Chapter 1, “First-Order ODEs,” Wiley.
4. Mary L. Boas, Mathematical Methods in the Physical Sciences, 3rd ed., Chapter 8, “Ordinary Differential Equations,” Wiley.