Finding region of xy plane for which differential equation has a unique solution

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SUMMARY

The discussion focuses on determining a region of the xy-plane where the differential equation x (dy/dx) = y has a unique solution passing through a specific point (x_0, y_0). It establishes that for a rectangular region R defined by a <= x <= b and c <= y <= d, if the functions f(x,y) and ∂f/∂y are continuous, a unique solution exists. The analysis confirms that in the regions D (x > 0) and D' (x < 0), the conditions for continuity are satisfied, thus confirming the existence of unique solutions in both regions.

PREREQUISITES
  • Understanding of differential equations, specifically first-order equations.
  • Familiarity with the concept of unique solutions in the context of initial value problems.
  • Knowledge of continuity conditions for functions of two variables.
  • Basic understanding of the xy-plane and its geometric representation.
NEXT STEPS
  • Study the existence and uniqueness theorem for first-order differential equations.
  • Explore the implications of continuity in differential equations using examples.
  • Learn about the graphical interpretation of solutions in the xy-plane.
  • Investigate the behavior of solutions in different regions of the xy-plane.
USEFUL FOR

Mathematics students, educators, and professionals working with differential equations, particularly those focusing on initial value problems and their geometric interpretations.

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Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point [math](x_0, y_0)[/math] in the region.

[math]x \frac{dy}{dx} = y[/math]

What does an xy-plane have to do with anything? I looked up the definition of unique solutions and here it is

Let R be a rectangular region in the xy-planed defined by a <=x<=b, c<=y<=d that contains the point [math](x_0, y_0)[/math] in its interior. If f(x,y) and [math]\frac{\partial{d} f}{\partial{d} y}[/math] are continuous on R then there exists some interval [math]I_0: (x_0-h, x_0+h), h>0[/math] contained in [a/b] and a unique function y(x) defined on [math]I_0[/math] that is a solution of the initial value problem.

That's a bit difficult to digest. How do I proceed?
 
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In each of the regions $D\equiv x>0$ and $D'\equiv x<0$ the differential equation is equivalent to $y'=f(x,y)=\dfrac{y}{x},$ and in both regions, $f$ and $\dfrac{\partial f}{\partial y}=\dfrac{1}{x}$ are continuous, so and according to a well known theorem, $D$ and $D'$ are solutions to your question.
 

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