Chapter 1 · First-Order ODEs

Non-Exact Equations & Integrating Factors

Making Non-Exact Equations Exact

Dr. Mohamed Mabrok · Qatar University

When an Equation is NOT Exact

For the equation $M(x,y)\,dx + N(x,y)\,dy = 0$, we check: is $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$?

The Problem

      If $\dfrac{\partial M}{\partial y} \neq \dfrac{\partial N}{\partial x}$, the equation is NOT exact — and we can't use the exact equation method directly.

But there's a powerful trick: multiply the entire equation by a clever function $\mu(x,y)$ called an integrating factor.

\mu(x,y) \cdot M\,dx + \mu(x,y) \cdot N\,dy = 0

The multiplied equation IS exact, so it becomes solvable!

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

Case 1: Integrating Factor $\mu(x)$

If the integrating factor depends only on $x$ (not on $y$), we have a simple formula.

Case 1 Formula

      If $\dfrac{1}{N}\left(\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x}\right) = h(x)$ depends only on $x$, then:
    
      $$\mu(x) = e^{\int h(x)\,dx}$$

Key insight: The function $h(x)$ must have NO $y$ terms — if it contains $y$, this case doesn't apply.

Quick Check

      Compute $\dfrac{1}{N}\left(\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x}\right)$. If the result is a function of $x$ alone (all $y$ terms cancel), you're in Case 1!

Case 2: Integrating Factor $\mu(y)$

If the integrating factor depends only on $y$ (not on $x$), we have a second formula.

Case 2 Formula

      If $\dfrac{1}{M}\left(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\right) = k(y)$ depends only on $y$, then:
    
      $$\mu(y) = e^{\int k(y)\,dy}$$

Note the difference: The subtraction order is reversed compared to Case 1: it's $\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}$, not the other way around.

Quick Check

      Compute $\dfrac{1}{M}\left(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\right)$. If the result is a function of $y$ alone (all $x$ terms cancel), you're in Case 2!

The 6-Step Integrating Factor Algorithm

STEP 1
        Verify NOT exact

        Check that $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$
      

STEP 2
        Try Case 1: $\mu(x)$

        Compute $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$
      

STEP 3
        Try Case 2: $\mu(y)$

        If Case 1 fails, try $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$
      

STEP 4
        Multiply by $\mu$

        Multiply the original equation by the integrating factor
      

STEP 5
        Verify NOW exact

        Check that $\frac{\partial(\mu M)}{\partial y} = \frac{\partial(\mu N)}{\partial x}$ ✓
      

STEP 6
        Solve as exact equation

        Use the exact equation method to find $F(x,y) = C$
      

Worked Example 1: Case 1

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

STEP 1 Check if exact

M = 3xy + y^2, \quad N = x^2 + xy$$ $$\frac{\partial M}{\partial y} = 3x + 2y, \quad \frac{\partial N}{\partial x} = 2x + y

NOT exact! Since $3x+2y \neq 2x+y$.

STEP 2 Try Case 1: compute $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$

h(x) = \frac{1}{x^2+xy} \cdot [(3x+2y) - (2x+y)] = \frac{x+y}{x(x+y)} = \frac{1}{x}

✓ Success! $h(x) = \frac{1}{x}$ depends only on $x$.

Worked Example 1 (continued)

Finding and Applying $\mu(x)$

STEP 3 Compute $\mu(x) = e^{\int h(x)\,dx}$

\mu(x) = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x

STEP 4 Multiply the original equation by $\mu = x$

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

STEP 5–6 Verify exact & solve

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

Worked Example 2: Case 2

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

STEP 1–2 Check Case 1

M = y^2 + y, \quad N = -x$$ $$h(x) = \frac{1}{-x}[(2y+1) - (-1)] = \frac{2y+2}{-x}

✗ Case 1 fails (contains $y$ terms).

STEP 3 Try Case 2: $k(y) = \frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$

k(y) = \frac{1}{y^2+y}[(-1) - (2y+1)] = \frac{-2y-2}{y(y+1)} = \frac{-2(y+1)}{y(y+1)} = -\frac{2}{y}

✓ Success! $\mu(y) = e^{\int -\frac{2}{y}\,dy} = e^{-2\ln y} = \dfrac{1}{y^2}$

Tips & Deeper Connections

Tip: Look for Common Factors

      Before jumping to Case 1 or 2, check if $M$ and $N$ share a common factor. Dividing by it first might reveal hidden structure.

Linear Equations as Special Case

      Linear equations $\frac{dy}{dx} + P(x)y = Q(x)$ rewrite as:

      $[P(x)y - Q(x)]\,dx + dy = 0$

      The integrating factor $\mu(x) = e^{\int P(x)\,dx}$ is exactly Case 1!

Important Note

      Integrating factors are NOT unique. Any constant multiple of an integrating factor is also an integrating factor. Usually we pick the simplest form.

Key Takeaways

1️⃣

        Try Case 1 First

        Compute $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$. If it depends only on $x$, use $\mu(x) = e^{\int h(x)\,dx}$.

2️⃣

        If Case 1 Fails, Try Case 2

        Compute $\frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})$. If it depends only on $y$, use $\mu(y) = e^{\int k(y)\,dy}$.

3️⃣

        After Finding $\mu$, Solve as Exact

        Multiply the equation by $\mu$ and use the exact equation method to find $F(x,y) = C$.

4️⃣

        Linear Equations Are Special

        The standard linear equation integrating factor $e^{\int P(x)\,dx}$ is exactly Case 1!

\text{Case 1:} \quad \mu(x) = e^{\int \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})\,dx}$$ $$\text{Case 2:} \quad \mu(y) = e^{\int \frac{1}{M}(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})\,dy}