Math 2.0

Ordinary differential equations Theorem

Test for exactness

Builds on Exact equation

Statement

Let R=(α,β)×(γ,δ)R = (\alpha, \beta) \times (\gamma, \delta) be an open rectangle in ℝ2\mathbb{R}^2. Let M,N:R→ℝM, N : R \to \mathbb{R} be continuous functions such that the partial derivatives ∂M/∂y\partial M / \partial y and ∂N/∂x\partial N / \partial x exist and are continuous on RR. Then the equation M dx+N dy=0M\, dx + N\, dy = 0 is exact on RR if and only if ∂M∂y=∂N∂xon R.\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \quad \text{on } R.

Remarks

The condition is necessary on each open set: if ψx=M\psi_x = M and ψy=N\psi_y = N, the mixed partial derivatives of ψ\psi are equal, so My=NxM_y = N_x. For the converse, fix (x0,y0)∈R(x_0, y_0) \in R and set ψ(x,y)=∫x0xM(s,y0) ds+∫y0yN(x,t) dt\psi(x, y) = \int_{x_0}^{x} M(s, y_0)\, ds + \int_{y_0}^{y} N(x, t)\, dt. The converse holds on each simply connected open set. It can fail on other sets: (−y dx+x dy)/(x2+y2)=0(-y\, dx + x\, dy)/(x^2 + y^2) = 0 satisfies the condition on ℝ2∖{0}\mathbb{R}^2 \setminus \{0\}, but it is not exact there.

Reading path

To understand this page from the ground up, read these 7 pages in order. The foundations come first.

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  1. Differential equation concept
  2. Function concept
  3. Interval concept
  4. Ordinary differential equation definition
  5. Solution of an ordinary differential equation definition
  6. Exact equation definition
  7. Test for exactness statement

Where this leads

1 page builds on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Exact_differential_equation, Identifying first-order exact differential equations
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Tag0MSCQNFTM8
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Kindstatement, theorem
WritingW0 stub
MSC34A05
StatusThis copy of the site has no build data. Run m2 status 0MSCQNFTM8.
UsesExact equation

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