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Math101learn.math101.caExact Differential Equations
A rigorous method for exact first-order equations, potential functions, integrating checks, and solution domains.
Precise definition
An equation $M(x,y)\,dx+N(x,y)\,dy=0$ is exact on a region if there exists a potential $F$ with $F_x=M$ and $F_y=N$. Then solutions lie on level curves $F(x,y)=C$. On a simply connected region with continuous first partial derivatives, $M_y=N_x$ is sufficient as well as necessary for exactness.
Notation and mathematical language
Subscripts denote partial derivatives. Integrating $M$ with respect to $x$ gives $F(x,y)=\int M(x,y)\,dx+g(y)$; the 'constant' may depend on $y$. Comparing $F_y$ with $N$ determines $g'(y)$.
Conceptual picture
The differential $dF=F_xdx+F_ydy$ measures change in the potential. Setting $dF=0$ means solution curves remain on a constant-potential contour. Exactness is therefore a conservative-field condition in the plane.
Conditions and key results
Equality $M_y=N_x$ must hold throughout a suitable region, not only along one curve. Holes in the domain can defeat global potential conclusions despite matching cross-partials. An implicit solution may fail to define a single function $y(x)$ where $F_y=0$.
A reliable strategy
- Write the equation in differential form and identify $M$ and $N$ on a stated region.
- Compute $M_y$ and $N_x$; if they differ, this exact-equation method does not apply without an integrating factor.
- Integrate one component, retain an unknown function of the other variable, and determine it by comparison.
- Set $F=C$, apply the initial condition, and differentiate implicitly to verify the original equation.
Fully worked example
Interpretation and application
Exact equations model conserved energy and potentials. The constant $C$ labels a trajectory set by initial data. Conservation in the mathematical model does not by itself establish that a physical system has no dissipation; that is a modelling assumption requiring evidence.
Common mistakes
Verification and reasonableness
- Differentiate the recovered $F$ in both variables and recover exactly $M$ and $N$.
- Implicitly differentiate $F=C$ to confirm $M+Ny'=0$ where $N\ne0$.
- Substitute the initial point to verify the correct level curve.
Practice
- Is $(y+2x)dx+(x+3y^2)dy=0$ exact?
- After integrating $M$ with respect to $x$, what kind of constant is added?
- What is the implicit solution once $F$ is found?
Answers and brief solutions
- Yes; both cross-partials equal 1.
- An arbitrary function of $y$.
- $F(x,y)=C$.
Further deduction
If $M_y\ne N_x$, an integrating factor may restore exactness. For example, if $(M_y-N_x)/N$ depends only on $x$, then $\mu(x)=\exp\!\int[(M_y-N_x)/N]dx$ is a candidate. This is a conditional shortcut, not a universal guarantee. After multiplying, recompute both cross-partials; never assume that a guessed factor succeeded.
Level curves can develop vertical tangents where $F_y=N=0$ and $F_x\ne0$, because implicit differentiation gives $y'=-M/N$. The potential relation remains valid there even when a single graph $y(x)$ temporarily fails. Distinguishing the implicit curve from a graph prevents calling the conserved solution invalid at a vertical tangent.
Related topics
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
For $M=2xy+1$ and $N=x^2+5y$, what is the common cross-partial?
- $M_y=2x$.
- $N_x=2x$, so the equation is exact on a suitable region.
End of lesson
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