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This might be an easy question in the theory of ordinary differential equation. But since I know very little about it, I posted it here and hope to get some answers or references.

Consider the equation $$x\phi'(x)-c\phi(x)=xf(x), x>0$$ where $c$ is some constant.

My question is that under what (growth) conditions, the solution $\phi(x)$ has the following property: the limit $$\lim_{x\to 0}x^{-c}\phi(x)$$ exists and nonzero?

Thank you very much for your help and sorry if this question is not appropriate here.

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Robert answered it, but in general, when faced with a problem of this type one can define, say, $a(x)=x^{-c}\phi(x)$, so $\phi(x)=x^c a(x)$, and then substitute into the differential equation for $\phi$, obtaining a differential equation for $a$, which you can use to understand the asymptotics of $a$. –  Will Sawin Jan 9 '12 at 23:00
    
Thank you very much for your comment. –  user1832 Jan 9 '12 at 23:13

1 Answer 1

The solution to your linear differential equation is $\phi(x) = x^c F(x)$ where $F(x)$ is any antiderivative of $f(x) x^{-c}$. In order for some solution to have $\lim_{x \to 0} x^{-c} \phi(x)$ exist, the necessary and sufficient condition is that the improper integral $\int_0^\varepsilon f(x)\ x^{-c}\ dx$ converges for some $\varepsilon > 0$. Then the limit will exist for all the solutions. However, the limit will be $0$ for one of those solutions, namely $\phi(x) = x^c \int_0^x f(t)\ t^{-c}\ dt$.

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Thank you very much for your answer. –  user1832 Jan 9 '12 at 23:13

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