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I guys. I'm doing this problem: $$\begin{cases} y'(x)=\frac{2y(x)}{x}+3x^{\alpha} \\ y(1)=2\\ \end{cases} $$ with $\alpha\in \mathbb R^{+}$ constant.

I have started studying the existence and uniqueness of solutions. So $$f(x,y)=\frac{2y(x)}{x}+3x^{\alpha}$$ Has domain $D=\{(x,y)\in\mathbb R^{2} :x\neq0\}$.

I notice that $f(x,y$) is a continuous function on it's domain, so it's a derivable function and for this reason I can say that is locally Lipschitz continuous. So I can apply the local exinstance and uniqueness theorem.

Moreover $f(x,y)$ it's not Lipschitz in $y$ respect to $x$ on $D$, so it's not globally Lipschitz continuous and I can't apply the global existance and uniquess theorem for the solutions.

Now I want to find explicitly the solutions, depending on $\alpha$, and finally I want to find the value of $\alpha$ for which

$$\lim_{x\to \infty}y(x)=+\infty$$

Assuming all is correct from the beginning to this point, the equation could be solved by separation of variables:

$$\frac{y'(x)}{y(x)}=\frac{2}{x}+\frac{3x^{\alpha}}{y(x)}$$

Is this the most convenient possibility to solve this equation? Because I'm a little bit stucked at this point. Any suggestions to get the goal? Thanl you very much.

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The Cauchy problem has a unique global solution in $(0,+\infty)$ because the differential equation is linear.

Moreover, after multiplying by the integrating factor $1/x^2$ we have: $$\left(\frac{y(x)}{x^2}\right)'=\frac{y'(x)}{x^2}-\frac{2y(x)}{x^3}=3x^{\alpha-2}.$$ Then after integrating both side we obtain:

i) if $\alpha\not=1$, $$\frac{y(x)}{x^2}=\int 3x^{\alpha-2}\,dx=\frac{3x^{\alpha-1}}{\alpha-1}+C\implies y(x)=\frac{3x^{\alpha+1}}{\alpha-1}+Cx^2;$$

ii) if $\alpha=1$, $$\frac{y(x)}{x^2}=\int 3x^{-1}\,dx=3\ln(x)+C \implies y(x)=3x^2\ln(x)+Cx^2.$$ Can you take it from here?

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  • $\begingroup$ Thank you for your reply. I am thinking about the domain. For sure I have $\alpha>0$. But why $x$ couldn't be negative? If i pick (just for an example) $x=-1$ and $\alpha=2$ I would have $$3(-1)^{2}=9$$ That is a real positive number. What do you think about it? Am I strongly in error? Thank you. Edit: Maybe it gives some trouble when $0\leq\alpha\leq 1$ because we're going into $\mathbb C$. $\endgroup$ – muserock92 Feb 20 '18 at 15:07
  • $\begingroup$ Usually the domain of the solution of a Cauchy problem is an interval (a connected set) which contain the initial point ($1$ in this case). Moreover if $\alpha$ is irrational $x^{\alpha}$ is not defined for $x<0$ (what is $(-2)^{\pi}$?) $\endgroup$ – Robert Z Feb 20 '18 at 15:13
  • $\begingroup$ Yes, It's definitely the point of my edit. We don't want to work with compex numbers. $\endgroup$ – muserock92 Feb 20 '18 at 15:14
  • $\begingroup$ And what about existance and uniqueness of solutions? I just don't care about theorems, I just see that it's linear so for sure I have a unique global solution. Is it ok now? Thank you. $\endgroup$ – muserock92 Feb 20 '18 at 15:15
  • $\begingroup$ Yes! Moreover here we also found explicit solutions which are defined for $x>0$. $\endgroup$ – Robert Z Feb 20 '18 at 15:16

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