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Is it possible to get simbolical solution of this differential equation: $$\dfrac{\mathrm{d}}{\mathrm{d}z}(x_0x_1) = -\dfrac{8}{r_i-z(r_i-1)}\dfrac{\mathrm{d} x_0}{\mathrm{d}z} $$ where if $z=1: x_0=1, x_1=0$. I tried to solve it like this, but it does not work:

sol = DSolve[
  x1'[z] == (-8 * x0'[z]/(ri - z (ri - 1)) - x0'[z]*x1[z])/x0[z], x1, 
  z]

I tried to integrate it by hand, but I am making some mistakes because I can see that the numerical and my solution are different (probably something between $\mathrm{d}z$ and $r_i-z(r_i-1)$).

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  • $\begingroup$ Condition x1==0 and use inside DSolve as function x1[z] are contradictory! $\endgroup$ Apr 22, 2022 at 15:47
  • $\begingroup$ Why is contradictory? $\endgroup$
    – iviviviviv
    Apr 22, 2022 at 15:49
  • $\begingroup$ You use the same symbol as parameter and function! $\endgroup$ Apr 22, 2022 at 15:52
  • $\begingroup$ Do you mean z is the same symbol as parameter and function? Or some other symbol? $\endgroup$
    – iviviviviv
    Apr 22, 2022 at 15:56
  • $\begingroup$ Your initial conditions should be written as x0[1] == 1, x1[1] == 0; however, you have two functions with only one equation. Your system is underdetermined. $\endgroup$
    – Bob Hanlon
    Apr 22, 2022 at 15:56

1 Answer 1

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Assumig you want solve the TeXform of the equation:

Integrate both sides

x0[z]x1[z]-x0[1]x1[1]== -Integrate[-8 /(ri - zz (ri - 1)) x0'[zz],{zz,1,z}]

Knowing x1[1]==0 it follows

x1[z]== -Integrate[-8 /(ri - zz (ri - 1)) x0'[zz],{zz,1,z}]/x0[z]
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  • $\begingroup$ Yes, but Mathematica does not give solution, in solution it just write Integrate, or integral symbol. $\endgroup$
    – iviviviviv
    Apr 22, 2022 at 16:03
  • $\begingroup$ That's true, Mathematica provides a form for arbitrary function x0[z] $\endgroup$ Apr 22, 2022 at 16:05
  • $\begingroup$ Well, I have x0[z], it is x0[z] = (1 + 64 beta (1 - 1/(ri - z (ri - 1))^3)/(3 (ri - 1)))^0.5. Now your x0[z]x1[z]==c-Integrate[-8 /(ri - z (ri - 1)) x0'[z],z] gives me soluton with Hypergepmetric function, so it probably means it is not so easy to integrate this by hand, as I thought? $\endgroup$
    – iviviviviv
    Apr 22, 2022 at 16:08
  • $\begingroup$ What do you know about parameter ri and beta? Are there singularities of the integrand? $\endgroup$ Apr 22, 2022 at 16:24
  • $\begingroup$ ri and beta are not dependent on z, they are constants. $\endgroup$
    – iviviviviv
    Apr 22, 2022 at 17:03

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