Domain of a Differential Equation

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SUMMARY

The discussion centers on finding the particular solution to the differential equation dy/dx=(1+y)/x with the initial condition f(-1)=1. The solution is identified as y(x)=2|x|-1 for x<0 and y(x)=3x-1 for x>0. The domain is restricted to x<0 due to the discontinuity at x=0, which affects the uniqueness of the solution. A unique solution across all real numbers except 0 would require two initial conditions, one for each disconnected part of the domain.

PREREQUISITES
  • Understanding of differential equations and their solutions
  • Knowledge of initial value problems and uniqueness of solutions
  • Familiarity with concepts of continuity and discontinuity in functions
  • Basic calculus, particularly differentiation and integration techniques
NEXT STEPS
  • Study the uniqueness theorem for ordinary differential equations
  • Explore the concept of discontinuities in differential equations
  • Learn about initial value problems and their implications on solution domains
  • Investigate the role of multiple initial conditions in defining solution domains
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Mathematics students, educators, and professionals working with differential equations, particularly those focusing on initial value problems and solution uniqueness.

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Find the particular solution y=f(x) to dy/dx=(1+y)/x given that f(-1)=1 and state its domain.

My answer was: 2|x|-1, x ∈ ℝ/{0}

Apparently, the domain should be x<0

Solving the differential equation was not an issue at all, but I have no idea why the domain is restricted to x<0. I understand that zero cannot be included because dy/dx fails to exist at x=0, but why are positive real numbers not included?

In general, as well, how is the domain of a particular differential equation found that is not obvious (e.g. found by simply looking at dy/dx)
 
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With the domain of a "particular" solution for a diff.eq, we mean those x's for which the given intial conditions yield a UNIQUE solution.

In your case, an equally good solution is:

y(x)=2|x|-1, x less than 0, and y(x)=3x-1 for x greater than 0.

The discontinuity at x=0 wreaks havoc on the uniqueness of the solution that normally is guaranteed by specifying a SINGLE initial condition.
 
Notice that if you were given exactly the same differential equation with the condition f(1)= 1, the domain would be the set of all positive numbers.
 
Ah, I see now. For an initial condition problem, the domain of the solution curve is the set of all x that satisfies the differential equation and contains the initial point. So, since there's a singularity at x=0, only the left half of the curve is the solution. Is my reasoning correct?
 
Indeed.

If you were to have a unique solution covering all of R except 0, you'd need TWO initial conditions, one in each of the two disconnected parts of the domain.
 

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