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I want to solve below ODE:

$\partial_x y(x) = \frac{a \: g(x)}{5 x^2}+ \frac{c}{x^2}$

In which $a$ and $c$ are constants and $g(x)$ is as below:

g[x_] := 1/(11664 a^4)*E^(-(((5 + 3 Sqrt[30]) a)/(35 x))) (36 (-5 c E^(((5 + 3 Sqrt[30]) a)/(35 x)) (324 a^2 - 630 a x - 7595 x^2) + 324 a^4 (E^((6 Sqrt[6/5] a)/(7 x)) (18 a^2 - 21 Sqrt[30] a x + 245 x^2) C1 + (18 a^2 + 21 Sqrt[30] a x + 245 x^2) C2)) + 245 c (E^((6 Sqrt[6/5] a)/(7 x)) (18 Sqrt[30] a^2 - 630 a x + 245 Sqrt[30] x^2) ExpIntegralEi[((5 - 3 Sqrt[30]) a)/(35 x)] - (18 Sqrt[30] a^2 + 630 a x + 245 Sqrt[30] x^2) ExpIntegralEi[((5 + 3 Sqrt[30]) a)/(35 x)]));

C1 = -((2790 c + 5832 a^4 C2 - 245 Sqrt[30] c ArcCoth[3Sqrt[6/5]])/(5832 a^4));

C2 =-(5 c (44100 (434 + 67 a) E^(1/35 (5 + 3 Sqrt[30]) a) + 2 E^(6/7 Sqrt[6/5]a) (558 (-17150 + a (245 (-5 + 6 Sqrt[30]) + 3 a (35 (-18 + Sqrt[30]) + 6 (-5 + 3 Sqrt[30]) a))) + 245 (3430 Sqrt[30] +  a (245 (-36 + Sqrt[30]) + 18 a (-35 + 21 Sqrt[30] - 18 a + Sqrt[30] a))) ArcCoth[3 Sqrt[6/5]]) + 245 (3430 Sqrt[30] + a (245 (-36 + Sqrt[30]) + 18 a (7 (-5 + 3 Sqrt[30]) + (-18 + Sqrt[30]) a))) E^( 6/7 Sqrt[6/5] a) ExpIntegralEi[1/35 (5 - 3 Sqrt[30]) a] - 245 (3430 Sqrt[30] + a (245 (36 + Sqrt[30]) + 18 a (35 + 21 Sqrt[30] + 18 a +  Sqrt[30] a))) ExpIntegralEi[1/35 (5 + 3 Sqrt[30]) a]))/(11664 a^4 (17150 + 1225 a + 1470 Sqrt[30] a + 1890 a^2 + 105 Sqrt[30] a^2 + 90 a^3 + 54 Sqrt[30]a^3 + (-17150 + a (245 (-5 + 6 Sqrt[30]) + 3 a (35 (-18 + Sqrt[30]) + 6 (-5 + 3 Sqrt[30]) a))) E^(6/7 Sqrt[6/5] a)));

I used below code, but it can not be run:

Simplify[DSolve[yy'[x] == a a20/5 g[x]/x^2 + c/x^2, yy[x], x ]]

Could anyone help me to solve the equation? The boundary condition for $y$ are:

$\partial_xy(x=1)=-s$

and

$y(x=\infty)=w$

which $w$ and $s$ are known constants.

If there is no analytic solution: $c$ is a constant which should be find by a boundary condition. Also, values for $a$ should be chosen larger than 1 in magnitude. For example $a=2$ or $a=-2$.

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  • $\begingroup$ psitwo is not defined. $\endgroup$ – anderstood Feb 22 '18 at 18:56
  • $\begingroup$ Sorry, it was a typo. I edit it @anderstood $\endgroup$ – Wat Watson Feb 22 '18 at 18:57
  • $\begingroup$ DSolve returns {{yy[x] -> -(a20/(58320 a^3 x)) - c/x + C[1]}}. C1 and C2 are not used. $\endgroup$ – anderstood Feb 22 '18 at 19:02
  • $\begingroup$ C1 and C2 are ysed in definition of g(x). and this solution is not correct because $g$ is a function of $x$ @anderstood $\endgroup$ – Wat Watson Feb 22 '18 at 19:05
  • $\begingroup$ OK I see, I edited your code so that the copy and paste now gives the right expression for g. Would you be OK with a numerical solution instead of a closed-form (which might not exist)? That would require values for a,b,c. $\endgroup$ – anderstood Feb 22 '18 at 19:11
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Assuming s=w=a20=1 the following works:

c = c /. First@Solve[(a*g[x]/(5 x^2) + c/x^2 /. x -> 1 // N) == -s, c]
sol[x_] = 
NDSolveValue[yy'[x] == a/5*g[x]/x^2 + c/x^2 && yy[10^10] == w, yy[x], {x, 1, 10}]

Plot[sol[x], {x, 1, 10}]

You can update with your actual values that you forgot to indicate.

Regarding the closed-form solution: I think you ODE is too complicated for a closed-form to exist, or to be found by Mathematica. But that's just a belief, not a proof...

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  • $\begingroup$ I put s = 1; w = 100; a = 2; and I plot the solution in LogLogPlot but y-axis of the output is not logarithmic. Do you know where is the problem? LogLogPlot[sol[x], {x, 1, 100}] $\endgroup$ – Wat Watson Feb 22 '18 at 19:56
  • $\begingroup$ If you have a new question, feel free to start a new post. (Try LogLogPlot[sol[x], {x, 1, 100}, PlotRange -> Full]) $\endgroup$ – anderstood Feb 22 '18 at 20:00
  • $\begingroup$ I think there is a problem in your answer. When I use below code, boundary condition at infinity (here 500) is not satisfied in the output: s = -1; w = 1000; a = 2; c = c /. First@ Solve[(a*g[x]/(5 x^2) + c/x^2 /. x -> 1 // N) == -s, c] sol[x_] = NDSolveValue[yy'[x] == a/5*g[x]/x^2 + c/x^2 && yy[500] == w, yy[x], {x, 1, 10}] Plot[sol[x] , {x, 1, 500}, PlotRange -> Full] $\endgroup$ – Wat Watson Feb 22 '18 at 20:29
  • $\begingroup$ @WatWatson In NDSolveValue, you need to adjust {x, 1, 10} to the range of interest (here, {x, 1, 500}). $\endgroup$ – anderstood Feb 22 '18 at 20:33

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