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📘 Key Concepts: Exact Equations

The equation $M(x,y)\,dx + N(x,y)\,dy = 0$ is exact if:

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

If exact, there exists $F(x,y)$ such that $dF = M\,dx + N\,dy$, and the solution is $F(x,y) = C$.

Finding F: Integrate $M$ with respect to $x$ (treating $y$ as constant), then differentiate with respect to $y$ and match to $N$ to find any $y$-only function $g(y)$.

If not exact: Find an integrating factor $\mu$ such that multiplying through makes it exact.
— If $\dfrac{\partial M/\partial y - \partial N/\partial x}{N}$ is a function of $x$ only: $\mu = e^{\int (\cdot)\,dx}$
— If $\dfrac{\partial N/\partial x - \partial M/\partial y}{M}$ is a function of $y$ only: $\mu = e^{\int (\cdot)\,dy}$

Problem: Basic Exact Equation

Test for exactness and solve: $(2xy + 3)\,dx + (x^2 - 4y)\,dy = 0$.

DE – Exact – Problem 1 – Setup DE – Exact – Problem 1 – Diagram DE – Exact – Problem 1 – Diagram
Solution Solution Solution Solution

Problem: Exact Equation with Initial Condition

Solve $(2xy^2 + y)\,dx + (2x^2y + x)\,dy = 0$, $y(1) = 2$.

DE – Exact – Problem 2 – Setup DE – Exact – Problem 2 – Diagram DE – Exact – Problem 2 – Diagram
Solution Solution Solution Solution

Problem: Exact Equation with Exponential Terms

Test for exactness and solve: $(e^x \sin y + 2x)\,dx + (e^x \cos y)\,dy = 0$.

DE – Exact – Problem 3 – Setup DE – Exact – Problem 3 – Diagram DE – Exact – Problem 3 – Diagram
Solution Solution Solution Solution

Problem: Non-Exact — Find Integrating Factor

Show that $(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0$ is not exact. Find an integrating factor $\mu(x)$ and solve the resulting exact equation.

DE – Exact – Problem 4 – Setup DE – Exact – Problem 4 – Diagram DE – Exact – Problem 4 – Diagram
Solution Solution Solution Solution

Problem: Exact Equation Check and Solution

Solve (2x + y)dx + (x + 2y)dy = 0.

Here, M = 2x + y and N = x + 2y. Since M_y = 1 and N_x = 1, the equation is exact.

$$F=\int(2x+y)\,dx=x^2+xy+g(y)$$
$$F_y=x+g'(y)=x+2y \Rightarrow g(y)=y^2$$

Answer: $x^2+xy+y^2=C$.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w294

MSTE - Differential Equations / Applications of Differential Equations / MSTE April 2025

What is the equation of the curve passing through the point $(3, -2)$ and having a slope at any point $(x, y)$ equal to $(x^{2} + y^{2})/(y^{3} - 2xy)$?

  1. $4x^{3} + 10xy^{2} - 3y^{4} = 180$
  2. $4x^{3} + 2xy^{2} - 13y^{4} = -76$
  3. $4x^{3} + 12xy^{2} - 3y^{4} = 204$
  4. $4x^{3} + 12xy^{2} - 13y^{4} = 44$
Given a slope, integrate to recover the curve.

$\dfrac{dy}{dx} = \dfrac{x^{2} + y^{2}}{y^{3} - 2xy}$
$(y^{3} - 2xy)\,dy = (x^{2} + y^{2})\,dx$
$y^{3}\,dy - x^{2}\,dx = y^{2}\,dx + 2xy\,dy$
$y^{3}\,dy - x^{2}\,dx = y^{2}\,dx + x(2y\,dy)$

Recognize the right side as an exact differential, $d(uv) = u\,dv + v\,du$:
$y^{3}\,dy - x^{2}\,dx = d(xy^{2})$

Integrating,
$\dfrac{1}{4}y^{4} - \dfrac{1}{3}x^{3} = xy^{2} + C$

At (3, −2):
$\dfrac{1}{4}(-2)^{4} - \dfrac{1}{3}(3)^{3} = 3(-2)^{2} + C$
$C = -17$

Thus,
$\dfrac{1}{4}y^{4} - \dfrac{1}{3}x^{3} = xy^{2} - 17$
$3y^{4} - 4x^{3} = 12xy^{2} - 204$
$\boxed{4x^{3} + 12xy^{2} - 3y^{4} = 204}$