Interval of existence and uniqueness of a separable 1st ODE

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SUMMARY

The discussion centers on the existence and uniqueness of solutions for the first-order ordinary differential equation (ODE) defined by y' = ((x-1)/(x^2))*(y^2) with the initial condition y(1)=1. The function f(x,y) is continuous for x0≠0, ensuring the existence of solutions according to Picard's Theorem. The unique solution is found to be y(x) = x/(ln|x| + 1), valid in the interval 1/e < x < ∞. The interval of uniqueness is confirmed as x ∈ (-∞, 0) ∪ (0, ∞).

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  • Understanding of first-order ordinary differential equations (ODEs)
  • Familiarity with Picard's Theorem
  • Knowledge of separation of variables technique
  • Basic calculus, specifically integration techniques
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Apothem
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Problem:
y'=((x-1)/(x^2))*(y^2) , y(1)=1 . Find solutions satisfying the initial condition, and determine the intervals where they exist and where they are unique.

Attempt at solution:
Let f(x,y)=((x-1)/(x^2))*(y^2), which is continuous near any (x0,y0) provided x0≠0 so a solution with y(x0)=y0 is guaranteed to exist when x0≠0
The partial derivative with respect to y of f(x,y) is 2y*((x-1)/(x^2)) which is continuous near any (x0,y0) provided x0≠0 so a solution with y(x0)=y0 is unique when x0≠0 (By Picard's Theorem)

From this the interval in which the solutions are unique is x∈(-∞,0)∪(0,∞).

Solving the differential equation and using the initial condition y(1)=1 we see that y(x)=x/(ln|x|+1). The interval of existence is 1/e < x < ∞

Is this right, or...?

Thanks for any help!
 
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Apothem said:
(1)=1 we see that y(x)=x/(ln|x|+1)

are you sure?

because by separation of variables of ##y'=\frac{x-1}{x^2}y^2## we have ##\int \frac{1}{y^2}\,=\,\int \frac{1}{x}-\frac{1}{x^2}dx## that gives ##y=-\frac{1}{\ln{|x|}+\frac{1}{x}+c}## ...
 

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