Finding a value for a y(0) = a in a IVP

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Discussion Overview

The discussion revolves around finding a value for \( a \in \mathbb{R} \) in the context of an initial value problem (IVP) defined by a specific differential equation and initial condition. Participants explore methods for solving the equation and determining conditions under which a solution exists.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant seeks to find an implicit solution to the differential equation and a suitable value for \( a \) that does not lead to an undefined expression.
  • Another participant points out that the initial expression lacks an equals sign, questioning whether it is indeed a differential equation.
  • A later reply suggests a substitution \( u=xy \) and provides a derivative relationship \( u'=y+xy' \), prompting further exploration of the problem.

Areas of Agreement / Disagreement

Participants have not reached a consensus, as there are differing interpretations of the original expression and its classification as a differential equation. The discussion remains unresolved regarding the approach to finding the value of \( a \) and the nature of the solution.

Contextual Notes

There are limitations related to the clarity of the original expression and the assumptions about the existence of a solution based on the chosen value of \( a \). The discussion does not resolve these uncertainties.

nathancurtis111
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Find the value of a ∈ R for which the initial value problem:

(S) {(xy' +y)/(1+x2y2)=1
{y(0)= a

has a solution and, for this value of a, find explicitly the maximal solution of (S).

I'm assuming I first find the implicit solution of the original diff eq (which I'm not too sure how to do given the equation) and then find a number a which will not make the eq undefined? I'm not too sure how to go about this one.
 
Last edited:
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You have some derivatives in an expression there, but it's not a differential equation, because there's no equals sign. Is there more to the expression?
 
My apologies, it is equal to 1, will edit original post to include that.
 
Try the substitution $u=xy$. Then $u'=y+xy'$. Can you continue?
 

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