Mathematics.

first order odes

Exact Differential Equations

Differential Equations30 minDifficulty6 out of 10

You should know: first order differential equation

Overview

A first-order equation written as M(x,y)dx + N(x,y)dy = 0 is called exact if the left side is the total differential of some function F(x,y), i.e. dF = M dx + N dy. When this holds, the solution is simply the implicit curve F(x,y) = C, so no further integration technique is needed once F is found. The test for exactness is the mixed-partials condition ∂M/∂y = ∂N/∂x, which follows from equality of mixed partial derivatives of F. Exact equations generalize separable equations and are the starting point for the method of integrating factors when a given equation is not already exact.

Intuition

Think of M dx + N dy as measuring the change in some underlying quantity F — like elevation on a hillside, where M and N are the slopes in the x- and y-directions. If F really is a well-defined 'elevation' function, then the slope you measure moving east then north must match the slope from moving north then east first — that consistency check is exactly ∂M/∂y = ∂N/∂x. When it holds, solving the ODE is just recovering the elevation function F from its two partial derivatives, and every level curve F(x,y)=C is a solution.

Formal Definition

Definition

An equation M dx + N dy = 0 is exact exactly when the cross-partial condition holds, and then a potential function F exists with:

M(x,y) dx+N(x,y) dy=0M(x,y)\,dx + N(x,y)\,dy = 0
Standard form
∂M∂y=∂N∂x\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
Exactness test
Fx=M,Fy=NF_x = M, \quad F_y = N
Potential function conditions
F(x,y)=CF(x,y) = C
Implicit general solution

Worked Examples

  1. Identify M = 2xy+3, N = x²-1 and check exactness.

    ∂M∂y=2x=∂N∂x\frac{\partial M}{\partial y} = 2x = \frac{\partial N}{\partial x}

  2. Integrate M with respect to x to build F, holding y fixed; add an unknown function of y.

    F=∫(2xy+3) dx=x2y+3x+g(y)F = \int (2xy+3)\,dx = x^2y + 3x + g(y)

  3. Differentiate F with respect to y and match it to N to find g(y).

    Fy=x2+g′(y)=x2−1  ⟹  g′(y)=−1  ⟹  g(y)=−yF_y = x^2 + g'(y) = x^2 - 1 \implies g'(y) = -1 \implies g(y) = -y

  4. Assemble F and write the implicit solution.

    F(x,y)=x2y+3x−y=CF(x,y) = x^2y + 3x - y = C

Answer: x²y + 3x − y = C (verify: F_x = 2xy+3 = M, F_y = x²−1 = N, so dF = M dx + N dy = 0 exactly as required).

Practice Problems

Difficulty 6/10

Verify that (2x + y)dx + (x + 2y)dy = 0 is exact, then solve it.

Difficulty 6/10

Solve (y cos x + 2xe^y)dx + (sin x + x²e^y + 2)dy = 0.

Difficulty 6/10

In a certain conservative force field, the equation (2xy)dx + (x² - 1)dy = 0 describes level curves of the potential energy. Find the potential F(x,y) and confirm exactness.

Quiz

M dx + N dy = 0 is exact precisely when:
Once an equation is confirmed exact, the general solution is:
In (2xy+3)dx + (x²-1)dy = 0, after integrating M in x you get x²y+3x+g(y). How is g(y) found?

Summary

  • M dx + N dy = 0 is exact when ∂M/∂y = ∂N/∂x, meaning the left side is the total differential of some F(x,y).
  • Solve by integrating M in x (or N in y), then matching the leftover function of y (or x) against the other coefficient.
  • The general solution is the implicit level-curve family F(x,y) = C.

References

  1. WebsiteWikipedia — Exact differential equation