Chapter 1 · First-Order ODEs

Exact Differential Equations

Finding Potential Functions

Dr. Mohamed Mabrok · Qatar University

What is an Exact Equation?

An exact equation is one where we can recognize it as the differential of a potential function.

Standard Form

      Consider the differential form:
    
      $$M(x,y)\,dx + N(x,y)\,dy = 0$$

This equation is exact if there exists a function $F(x,y)$ such that:

\frac{\partial F}{\partial x} = M \quad \text{and} \quad \frac{\partial F}{\partial y} = N

Then the total differential is: $dF = M\,dx + N\,dy = 0$

Test for Exactness

The Exactness Criterion

      $M(x,y)\,dx + N(x,y)\,dy = 0$ is exact if and only if:
    
      $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

Why? By Clairaut's Theorem (equality of mixed partials):

\frac{\partial^2 F}{\partial y \, \partial x} = \frac{\partial^2 F}{\partial x \, \partial y}

If $\frac{\partial F}{\partial x} = M$, then $\frac{\partial^2 F}{\partial y \, \partial x} = \frac{\partial M}{\partial y}$
If $\frac{\partial F}{\partial y} = N$, then $\frac{\partial^2 F}{\partial x \, \partial y} = \frac{\partial N}{\partial x}$
By Clairaut: these must be equal!

7-Step Solution Method

STEP 1
        Identify $M$ and $N$

        From $M\,dx + N\,dy = 0$
      

STEP 2
        Check exactness

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

STEP 3
        Integrate $M$ w.r.t. $x$

        $F = \int M\,dx + g(y)$
      

STEP 4
        Differentiate $F$ w.r.t. $y$

        Compare with $N$
      

STEP 5
        Find $g(y)$

        Solve for $g'(y)$ from Step 4
      

STEP 6
        Write solution

        $F(x,y) = C$ (implicit)
      

STEP 7
        Apply initial conditions

        Use $y(x_0) = y_0$ to find $C$
      

Worked Example

Solve: $(2xy+3)\,dx + (x^2+4y)\,dy = 0$

STEP 1 Identify $M$ and $N$

M(x,y) = 2xy + 3 \quad , \quad N(x,y) = x^2 + 4y

STEP 2 Check exactness

\frac{\partial M}{\partial y} = 2x \quad , \quad \frac{\partial N}{\partial x} = 2x \quad \checkmark

STEP 3 Integrate $M$ w.r.t. $x$

F = \int (2xy + 3)\,dx = x^2 y + 3x + g(y)

Worked Example (continued)

Finding the Potential Function

STEP 4 Differentiate $F$ w.r.t. $y$ and match with $N$

\frac{\partial F}{\partial y} = x^2 + g'(y) = N = x^2 + 4y

STEP 5 Solve for $g'(y)$ and integrate

g'(y) = 4y \quad \Longrightarrow \quad g(y) = 2y^2

STEP 6 Write the implicit solution

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

IVP Example

Solve: $(2x+y)\,dx + (x-3y^2)\,dy = 0$, $y(1)=2$

Quick check: $\frac{\partial M}{\partial y} = 1$, $\frac{\partial N}{\partial x} = 1$ ✓

STEP 3 Integrate $M$ w.r.t. $x$

F = \int (2x+y)\,dx = x^2 + xy + g(y)

STEP 4-5 Match with $N$ and find $g(y)$

\frac{\partial F}{\partial y} = x + g'(y) = x - 3y^2 \quad \Rightarrow \quad g(y) = -y^3

STEP 6 General solution

$$x^2 + xy - y^3 = C$$

STEP 7 Apply $y(1)=2$

1 + 1(2) - 2^3 = C \quad \Rightarrow \quad C = -5 \quad \boxed{x^2 + xy - y^3 = -5}

Common Pitfalls to Avoid

MISTAKE 1

          Forgetting $g(y)$ is a function

          When integrating $M$ w.r.t. $x$, the "constant" depends on $y$: $g(y)$ not $C$.

MISTAKE 3

          Swapping the partial derivatives

          It's always: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

          Not the other way around!

MISTAKE 2

          Skipping the exactness check

          Always verify $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ before proceeding. Not exact equations require different methods!

Geometric Interpretation

Exact equations connect to conservative vector fields and level curves.

Level Curves

      The solutions $F(x,y) = C$ are level curves of the potential function $F$. Each curve is a complete solution family member.

Conservative Vector Fields

      The vector field $(M, N) = \nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}\right)$ is conservative — it's the gradient of a scalar potential function.

Key Observation

      The gradient vector $\nabla F = (M, N)$ is always perpendicular to the level curves $F(x,y) = C$. This is why solutions are implicit relations, not explicit functions.

Key Takeaways

TEST

        Exactness criterion

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

        By Clairaut's theorem on mixed partials

METHOD

        Solution steps

        Integrate $M$ w.r.t. $x$, find $g(y)$ via $N$

        Write $F(x,y) = C$ implicitly

SOLUTIONS

        Always implicit

        $F(x,y) = C$ — the potential function itself

        Don't try to solve for $y$ explicitly

FOUNDATION

        Clairaut's Theorem

        Equality of mixed partials underpins everything

        It's the bridge between $(M,N)$ and $F$

\text{Check exactness} \;\Longrightarrow\; \text{Find } F \;\Longrightarrow\; \text{Level curves } F(x,y)=C \;\Longrightarrow\; \text{Apply IC}