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I am learning differential equation and the uniqueness theorem for 1st order ODE said that if $y' = f(x,y)$ where $f(x,y)$ is Lipschitz w.r.t. $y$ or $\frac{df}{dy}$ is continuous, then the ODE has at most one solution. But my question is not about this theorem.

Now, I find this homogeneous ODE: $$ y' = \frac{xy + y^2 + x^2}{x^2}$$ The usual technique applies and a solution is found. My question is that

Is $\frac{xy + y^2 + x^2}{x^2}$ Lipschitz?

Do we have another solution for this ODE so that the contrapositive of the uniqueness theorem is verified?

I tried this $$ \frac{f(x,y)-f(x,0)}{x-0}=\frac{xy+y^2}{x^3}$$ Can I argue that when $y$ tends to infinity, the above diverges?

Thanks in advance.

Community
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Nighty
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1 Answers1

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$$ \frac{\partial f}{\partial y}=\frac{x+2\,y}{x^2}. $$ It is continuous on $\{(x,y):x>0\}$, and hence $f(x,y)$ is (locally) Lipschitz with respect to $y$ on the same set. The existence and uniqueness theorem tells you that given $x_0>0$ and $y_0\in\mathbb{R}$ there is a unique solution defined on some interval around $x_0$ such that $y(x_0)=y_0$. To find it, you impose that condition on the general solution to find the value of the integration constant $C$.

Julián Aguirre
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  • Thanks for replying. Very often, the range of $x$ is not given in the question of an ODE. Then how should we address the question of being Lipschitz so as to give a 'complete' answer? – Nighty Feb 11 '15 at 13:23
  • When you are asked to find the general solution, usually no attention is paid to the region where the equation is defined. You may even find solutions to which the existence theorem does not apply. For instance, $y=-x$ is a solution of the equation defined at $x=0$, even if the function $f(x,y)$ is not continuous at $(0,0)$. In short: when finding the general solution, do not worry about the existence theorem. – Julián Aguirre Feb 11 '15 at 13:29