# I Interval of existence and uniqueness of a separable 1st ODE

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1. Sep 28, 2016

### Apothem

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!

2. Sep 28, 2016

### Ssnow

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 $y=-\frac{1}{\ln{|x|}+\frac{1}{x}+c}$ ...