Math Problem Statement

how can you identify if the equation is homogeneous and exact DE

Solution

To determine whether a first-order differential equation (DE) is homogeneous or exact, we follow specific criteria for each.

1. Homogeneous Differential Equation

A differential equation of the form: M(x,y)dx+N(x,y)dy=0M(x, y) \, dx + N(x, y) \, dy = 0 is homogeneous if both functions M(x,y)M(x, y) and N(x,y)N(x, y) are homogeneous of the same degree. That is, when each term in both M(x,y)M(x, y) and N(x,y)N(x, y) can be rewritten as a function of yx\frac{y}{x}.

Steps to Identify Homogeneity:

  • Replace xx by λx\lambda x and yy by λy\lambda y in M(x,y)M(x, y) and N(x,y)N(x, y).
  • If both M(λx,λy)M(\lambda x, \lambda y) and N(λx,λy)N(\lambda x, \lambda y) can be factored as λkM(x,y)\lambda^k M(x, y) and λkN(x,y)\lambda^k N(x, y), where kk is the same for both, then the DE is homogeneous of degree kk.

Example:

(3x2+2xy)dx+(2x2+y2)dy=0(3x^2 + 2xy) \, dx + (2x^2 + y^2) \, dy = 0

  • Replacing xx by λx\lambda x and yy by λy\lambda y: M(λx,λy)=3(λx)2+2(λx)(λy)=λ2(3x2+2xy)M(\lambda x, \lambda y) = 3(\lambda x)^2 + 2(\lambda x)(\lambda y) = \lambda^2 (3x^2 + 2xy) N(λx,λy)=2(λx)2+(λy)2=λ2(2x2+y2)N(\lambda x, \lambda y) = 2(\lambda x)^2 + (\lambda y)^2 = \lambda^2 (2x^2 + y^2) Both are homogeneous of degree 2, so the equation is homogeneous.

2. Exact Differential Equation

A differential equation of the form: M(x,y)dx+N(x,y)dy=0M(x, y) \, dx + N(x, y) \, dy = 0 is exact if there exists a function ψ(x,y)\psi(x, y) such that: ψx=M(x,y)andψy=N(x,y)\frac{\partial \psi}{\partial x} = M(x, y) \quad \text{and} \quad \frac{\partial \psi}{\partial y} = N(x, y)

Condition for Exactness:

To check if the DE is exact, verify that the following condition holds: My=Nx\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} If this equality is true, then the DE is exact, and there exists a potential function ψ(x,y)\psi(x, y) whose total differential leads to the given equation.

Example:

Consider the equation: (2xy+y2)dx+(x2+2xy)dy=0(2xy + y^2) \, dx + (x^2 + 2xy) \, dy = 0

  • Compute My\frac{\partial M}{\partial y} and Nx\frac{\partial N}{\partial x}: My=2x+2y,Nx=2x+2y\frac{\partial M}{\partial y} = 2x + 2y, \quad \frac{\partial N}{\partial x} = 2x + 2y Since My=Nx\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, the equation is exact.

Summary of Steps:

  1. Homogeneous DE:

    • Check if M(x,y)M(x, y) and N(x,y)N(x, y) are homogeneous of the same degree.
  2. Exact DE:

    • Verify that My=Nx\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.

Would you like more details on a specific example or method? Here are 5 related questions:

  1. How do you solve a homogeneous differential equation once identified?
  2. What is the integrating factor if a DE is not exact?
  3. How do you find the potential function for an exact DE?
  4. Can a DE be both homogeneous and exact? How would you approach such a case?
  5. What are some common methods to test for non-exact differential equations?

Tip: If a differential equation is not exact, you can often multiply by an integrating factor to make it exact.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Homogeneous Differential Equations
Exact Differential Equations
Multivariable Calculus

Formulas

M(x, y) dx + N(x, y) dy = 0
Homogeneous condition: M(λx, λy) = λ^k M(x, y) and N(λx, λy) = λ^k N(x, y)
Exact condition: ∂M/∂y = ∂N/∂x

Theorems

Homogeneous Degree Theorem
Exactness Condition

Suitable Grade Level

Undergraduate - First Year Calculus