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Given the differential equation $$xdy/dx=2y$$ my textbook says that applying the existence and uniqueness theorem obtains us $$y(x)=Cx^2$$ since $f(x,y)=2y/x$ and the partial derivative of f with respect to y is $2/x$.

I don’t understand at all what kind of substitution happened to get a general solution for $y$

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  • $\begingroup$ Actually, the general solution is $y(x)=C x^2$ for $x \ge 0$, $y(x)=D x^2$ for $x < 0$, where $C$ and $D$ can be different constants. The existence and uniqueness theorem for $dy/dx = f(x,y) = 2y/x$ doesn't hold at $x=0$, since $f$ is undefined there. $\endgroup$ Sep 3 at 6:04
  • $\begingroup$ Also, the phrase “my textbook” is rather useless information. If you're going to mention a book, please tell us which book it is (and which page). $\endgroup$ Sep 3 at 6:05
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Assuming that $y\not\equiv 0$, it results that:

\begin{align*} xy' = 2y & \Longleftrightarrow \frac{y'}{y} = \frac{2}{x}\\\\ & \Longleftrightarrow \ln|y| = 2\ln|x| + c\\\\ & \Longleftrightarrow \ln|y| = \ln|x|^{2} + c\\\\ & \Longleftrightarrow y(x) = \pm \exp(c)x^{2}\\\\ & \Longleftrightarrow y(x) = Cx^{2} \end{align*} where $C\in\mathbb{R}\backslash\{0\}$.

However $y\equiv 0$ is also a solution. Hence $y(x) = Cx^{2}$, for every $C\in\mathbb{R}$.

Hopefully this helps!

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