Prove two solutions of differential equation are the same

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?

• Could you also provide the differential equation? – A. A. Jul 25 '14 at 15:00
• @Adolfo I edited the question. – PML Jul 25 '14 at 15:09
• 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? – Hans Engler Jul 25 '14 at 15:11
• @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? – PML Jul 25 '14 at 15:17
• Well, you could pick specific boundary conditions in order to select the integration constants, and then plot each solution. – 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 $$C_{{1}}=\dfrac14\,{{\rm e}^{2\,iC_{{4}}}},\ C_{{2}}=-\ln \left( 2 \right) +C_{{3}}-iC_{{4}}$$
• Note that we can assume $C_2 = 0$ or $C_3 = 0$, since both are additive constants. – Hans Engler Jul 25 '14 at 15:20