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Exact Differential Equations and Integrating Factors cheat sheet - grade college

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Calculus Grade college

Exact Differential Equations and Integrating Factors Cheat Sheet

A printable reference covering exactness tests, potential functions, implicit solutions, and integrating factors for college.

TopicsDerivativesIntegralsFunctionsAlgebraCalculusStudy Guide
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Exact differential equations use partial derivatives to recognize when a first order differential equation comes from a single potential function. This cheat sheet helps students test exactness, build the potential function, and write the implicit solution clearly. It is especially useful for solving equations written in differential form, where algebra and partial integration can become confusing.

The goal is to make the method systematic and easy to check.

The main form is M(x,y) dx+N(x,y) dy=0M(x,y)\,dx + N(x,y)\,dy = 0, which is exact when My=NxM_y = N_x on a suitable region. If exact, there is a function ψ(x,y)\psi(x,y) such that ψx=M\psi_x = M and ψy=N\psi_y = N, and the solution is ψ(x,y)=C\psi(x,y)=C. If the equation is not exact, an integrating factor such as μ(x)\mu(x) or μ(y)\mu(y) may make it exact.

The most important shortcuts test whether My−NxN\frac{M_y-N_x}{N} depends only on xx or whether Nx−MyM\frac{N_x-M_y}{M} depends only on yy.

Key Facts

  • A first order equation in differential form is written as M(x,y) dx+N(x,y) dy=0M(x,y)\,dx + N(x,y)\,dy = 0.
  • The equation M(x,y) dx+N(x,y) dy=0M(x,y)\,dx + N(x,y)\,dy = 0 is exact when ∂M∂y=∂N∂x\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} on a region where the needed derivatives are continuous.
  • For an exact equation, find a potential function ψ(x,y)\psi(x,y) satisfying ψx=M\psi_x=M and ψy=N\psi_y=N.
  • The implicit general solution of an exact equation is ψ(x,y)=C\psi(x,y)=C, where CC is an arbitrary constant.
  • A common way to build ψ\psi is ψ(x,y)=∫M(x,y) dx+g(y)\psi(x,y)=\int M(x,y)\,dx+g(y), then use ψy=N\psi_y=N to find g′(y)g'(y).
  • If My−NxN=f(x)\frac{M_y-N_x}{N}=f(x) depends only on xx, then an integrating factor is μ(x)=e∫f(x) dx\mu(x)=e^{\int f(x)\,dx}.
  • If Nx−MyM=g(y)\frac{N_x-M_y}{M}=g(y) depends only on yy, then an integrating factor is μ(y)=e∫g(y) dy\mu(y)=e^{\int g(y)\,dy}.
  • For a linear equation dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x), the integrating factor is μ(x)=e∫P(x) dx\mu(x)=e^{\int P(x)\,dx}.

Vocabulary

Differential form
A first order differential equation written as M(x,y) dx+N(x,y) dy=0M(x,y)\,dx+N(x,y)\,dy=0.
Exact equation
An equation M dx+N dy=0M\,dx+N\,dy=0 is exact when it can be written as dψ=0d\psi=0 for some potential function ψ(x,y)\psi(x,y).
Potential function
A function ψ(x,y)\psi(x,y) whose differential is dψ=ψx dx+ψy dyd\psi=\psi_x\,dx+\psi_y\,dy.
Integrating factor
A nonzero function μ\mu that multiplies a differential equation to make it exact or easier to integrate.
Implicit solution
A solution written as a relation such as ψ(x,y)=C\psi(x,y)=C instead of solving explicitly for yy.
Exactness condition
The test My=NxM_y=N_x used to determine whether M(x,y) dx+N(x,y) dy=0M(x,y)\,dx+N(x,y)\,dy=0 is exact.

Common Mistakes to Avoid

  • Testing exactness with the wrong derivatives is incorrect because the condition is My=NxM_y=N_x, not Mx=NyM_x=N_y.
  • Forgetting the unknown function after partial integration is incorrect because ∫M(x,y) dx\int M(x,y)\,dx may still need an added term g(y)g(y).
  • Treating yy as a constant in every step is wrong because yy is constant only when integrating with respect to xx, while xx is constant when integrating with respect to yy.
  • Using μ(x)=e∫My−NxN dx\mu(x)=e^{\int \frac{M_y-N_x}{N}\,dx} when My−NxN\frac{M_y-N_x}{N} still contains yy is invalid because that formula requires dependence on xx only.
  • Stopping after finding an integrating factor is incomplete because the multiplied equation μM dx+μN dy=0\mu M\,dx+\mu N\,dy=0 must still be solved as an exact equation.

Practice Questions

  1. 1 Determine whether (2xy+3) dx+(x2+4y) dy=0(2xy+3)\,dx+(x^2+4y)\,dy=0 is exact, and if it is exact, find the implicit solution.
  2. 2 Solve (y+2x) dx+x dy=0(y+2x)\,dx+x\,dy=0 by first testing exactness and then finding ψ(x,y)=C\psi(x,y)=C.
  3. 3 For (2y) dx+(3x+4y) dy=0(2y)\,dx+(3x+4y)\,dy=0, test whether an integrating factor of the form μ(y)\mu(y) exists using Nx−MyM\frac{N_x-M_y}{M}.
  4. 4 Explain why an integrating factor can change a non-exact equation into an exact equation without changing the solution curves when μ(x,y)≠0\mu(x,y)\neq 0.

Understanding Exact Differential Equations and Integrating Factors

An exact equation has a useful geometric meaning. The two coefficient functions describe the tiny change in one quantity as x and y change. If the equation is exact, that quantity depends only on the current location in the x y plane.

It does not depend on the route taken to reach that location. This is similar to height on a hill. Walking from one point to another changes your height by the same total amount no matter which trail you use.

The solution curves are level curves of the potential function. Along one of these curves, the potential stays constant even though x and y may both change.

The region where an equation is studied matters. Matching partial derivatives is reliable on a region without holes, provided the functions behave well there. A missing point can create a problem.

For example, functions involving division by x or y are not defined on the corresponding axis. A result found in one region may not apply across that axis. Students should state or at least notice domain restrictions before integrating.

Another common mistake is treating the integration constant as an ordinary number too early. When integrating with respect to x, the missing part may still depend on y. That dependence carries the information needed to match the other coefficient.

An integrating factor is a multiplier chosen to repair an equation that nearly has the path independence property. Multiplying changes both coefficient functions, so the product rule creates extra derivative terms. Those extra terms can make the cross derivatives agree.

The multiplier must be nonzero on the region being used. Otherwise it could add or remove solution points. The familiar integrating factor for a linear equation is not a separate trick.

It comes from the same idea of turning the left side into the total derivative of a product. This connection helps students remember why an exponential appears instead of viewing it as a formula to memorize.

Exact differentials appear whenever a system has a state quantity. In physics, potential energy depends on position, while work may depend on the path unless the force is conservative. In thermodynamics, temperature, pressure, and volume describe states, but heat and work are path dependent quantities.

These examples explain why it matters to tell apart a total change in a state function from a path dependent change. After finding an implicit solution, differentiate it with respect to x to check that it returns the original differential equation.

Then apply any initial condition to select one constant level curve. Watch for solutions lost when dividing by an expression that might be zero, since those special curves may need to be tested separately.