In a recent work I had to solve the following differential equation: $$ r x''(r)+r x'(r)^2+x'(r)-\frac{4}{r}=0~~. $$

To do so I used two methods and I got, using each, two solutions with different expressions. The solution of the differential equation is unique so those expressions must be the same.

Here are the two expressions: $$ \log \left(4 C_1\,r^4+1\right)+C_2-2 \log (r) $$ and $$ C_3+\log \left(\cos \left(C_4-2 i \log (r)\right)\right)~~, $$ where $C_1$,$C_2$,$C_3$ and $C_4$ are integration constants. I've been trying to prove that those solutions are the same but with no luck. Can somebody help me?

  • $\begingroup$ Could you also provide the differential equation? $\endgroup$ – A. A. Jul 25 '14 at 15:00
  • $\begingroup$ @Adolfo I edited the question. $\endgroup$ – PML Jul 25 '14 at 15:09
  • $\begingroup$ Have you tried to plot the two functions for different choices of $C1, \dots C_4$ - just to check that these are really the same? $\endgroup$ – Hans Engler Jul 25 '14 at 15:11
  • $\begingroup$ @HansEngler Since I don't know the relation between the constants of integration of the two solutions I can't just arbitrarily choose values for them and hope that the plot coincides, right? $\endgroup$ – PML Jul 25 '14 at 15:17
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    $\begingroup$ Well, you could pick specific boundary conditions in order to select the integration constants, and then plot each solution. $\endgroup$ – Semiclassical Jul 25 '14 at 15:21

Apply $\exp$ to both, express trig functions in terms of exponentials, and take the difference. I get $$ \left( \dfrac12\,{{\rm e}^{C_{{3}}+iC_{{4}}}}-4\,{{\rm e}^{C_{{2}}}}C_{{1} } \right) {r}^{2}+{\frac {\dfrac12\,{{\rm e}^{C_{{3}}-iC_{{4}}}}-{{\rm e}^{ C_{{2}}}}}{{r}^{2}}} $$ For this to be $0$ for all $r$, you need both coefficients to be $0$. This is true if $$ C_{{1}}=\dfrac14\,{{\rm e}^{2\,iC_{{4}}}},\ C_{{2}}=-\ln \left( 2 \right) +C_{{3}}-iC_{{4}} $$

  • $\begingroup$ Note that we can assume $C_2 = 0$ or $C_3 = 0$, since both are additive constants. $\endgroup$ – Hans Engler Jul 25 '14 at 15:20

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