Cauchy problem with real parameter?

Question

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.

Answer 1

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?