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Let $x_0,y_0 \in \mathbb R$ be given and consider the initial value problem:

$$y^{\prime}(x)=\cfrac{xy(x)}{x-1}$$$y(x_0)=y_0$.

For which values of $x_0,y_0$ is the solution unique?

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The Existence and Uniqueness Theorem says there is (locally) a unique solution as long as the right side of the differential equation is continuous and locally Lipschitz with respect to $y$ in a neighbourhood of $(x_0, y_0)$. That is the case here as long as $x_0 \ne 1$. On the other hand, for $x_0 = 1$ the right side of the equation is not even defined.

You don't need to solve the initial value problem to answer the question, but if you do solve it you should get $$ y(x) = y_0 e^{x-x_0} \frac{x-1}{x_0-1}$$

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  • $\begingroup$ Just realised that I was subbing in $x_0$ to y(x), which was why I was getting y=$y_0$. $\endgroup$
    – David
    Oct 31 '19 at 21:15

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