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Suppose we have the ODE ($N \ge 2$): $$ -\frac{1}{(\psi(r))^{N-1}}\left[(\psi(r))^{N-1} u^{\prime}(r)\right]^{\prime}=f(u(r)) \tag{0} $$ Then we will get
$$ (\psi(r))^{N-1} u^{\prime}(r)=-\int_{r_0}^r(\psi(s))^{N-1} f(u(s)) \mathrm{d} s \tag{1} $$

Then if $\psi(r) \rightarrow+\infty$, and the RHS of (1) is divergent, can we use L'Hôpital's Rule ?

If I use it then,
$$ \lim_{r \rightarrow +\infty} u^{\prime}(r)= \lim_{r \rightarrow +\infty} \frac{ - (\psi(r))^{N-1} f(u(r))} {(N-1)(\psi(r))^{N-2}\psi^{\prime}} \tag{2} $$

Did I misunderstand something ? Thank you very much !

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1 Answer 1

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Commenting via Answer Post , because it is too long ....

It is nice to try something new , though accuracy/consistency/precision is important ....

(A) You have not converted ODE (0) to (1) properly. You are missing the Constant of Integration somewhere.

(B) You have implicitly assumed $N-1$ is Positive to get $\infty/\infty$ , though when $N$ is Negative , we will have $\infty/0$ where we can not use LH rule.

(C) When we have $A(x) = B(x)$ , then $\lim A(x) = \lim B(x)$ , hence we can apply LH rule ( provided all the assumptions are valid !! ) to get the limit (2) , not the whole function $A(x)$.

(D) When we have $\lim A(x) = \lim B(x)$ , then we can not claim that $A(x) = B(x)$ , hence applying LH rule will not help us much to the whole function $A(x)$.

(E) There is no limit in the ODE which was valid over the whole Domain , hence when we take the limit , we are considering just that limit , not the whole Domain.

When I consider all these things , I can not make out much progress towards the ODE Solution out of this novel attempt ....

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