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

find_the_fun

Active member
Feb 1, 2012
166
Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point \(\displaystyle (x_0, y_0)\) in the region.

\(\displaystyle x \frac{dy}{dx} = y\)

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 \(\displaystyle (x_0, y_0)\) in its interior. If f(x,y) and \(\displaystyle \frac{\partial{d} f}{\partial{d} y}\) are continuous on R then there exists some interval \(\displaystyle I_0: (x_0-h, x_0+h), h>0\) contained in [a/b] and a unique function y(x) defined on \(\displaystyle I_0\) that is a solution of the initial value problem.

That's a bit difficult to digest. How do I proceed?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.