# I have tried two different techniques to find the exact solution of BVP but cant run the codes in mathemtica Im trying to find the exact solution for the singular BVP $$y^{\prime\prime}(t)=-\frac{2}{t}y^{\prime}(t)+y^{5}(t)$$ with the Boundary Conditions $$y(1)=\sqrt{\frac{3}{4}}$$ and $$y^{\prime}(0)=0$$ and $$0. I know that the exact solution of this equation is $$y(t)=\sqrt{\frac{3}{3+t^2}}$$. Now I want to find the exact solution using mathemtica software. I tried the following first: (1)

DSolve[{y''[t] == -(2/t) (y'[t]) + (y[t])^5, y == Sqrt[3/4], y' == 0}, y[t], t]


but I cant find the exact solution like above. After this, I tried

SOL = DSolve[{y''[t] == -(2/t) (y'[t]) + (y[t])^5}, y[t],  t] (* General solution *)
F = FindRoot[{(y[t] /. SOL[] /. t -> 1 /. C -> c1 /.C -> c2) ==Sqrt[3/4], (y'[t] /. SOL[] /. t -> 0 /. C -> c1 /.C -> c2) == 0}, {c1, 1/2}, {c2, 1/2}] // Chop // Rationalize
(* Finding constans c1 and c2 *)
SOL /. C -> c1 /.C -> c2 /. F(* Paste constants c1 and c2 to general solution *)
(*{{y[t]\->Sqrt(3/(3+t)^2)}*)


• Second comment: you said that you know the analytic solution and you gave a function. Even when Mathematica cannot get the answer from DSolve it is able to verify it when we impose it. In this case, it does not. If you run y''[t] == -(2/t) (y'[t]) + (y[t])^5 /. y -> (Sqrt[3/(3 + #^2)] &) // FullSimplify you will see that it does not yield True. Could you please check the solution you provided for the y[t]?
– kcr
Feb 23 at 20:47
• The main problem is that "your exact solution" does not satisfy the original equation. I can say that you haven't checked if this is a solution Feb 24 at 5:07
• Given that exact solution, your equation is wrong. You need a negative sign before y[t]^5. Here is why: If you differentiate \sqrt{3} \sqrt{\frac{1}{t^2+3}} once, you get -\sqrt{3} t \left(\frac{1}{t^2+3}\right)^{3/2}. If you differentiate twice, you get \sqrt{3} \left(\frac{1}{t^2+3}\right)^{5/2} \left(2 t^2-3\right). Now the closest form you can get it in the main equation: after multiplying the first derivative to (2/t) and dividing by y[t]^5 you will get -1. And yes, Mathematica can't solve this by normal DSolve without other advanced techniques that I am not aware of. Feb 24 at 16:04
• First No way to find exact solution by Mathematica. Second give us exact solution of this equation, a sceenshot from article(I do not have 42.99 euros), because as has been mentioned many times, your exact solution does not satisfy the original equation. Feb 24 at 17:51
• The Singh and Kumar reference confirms Mariusz's claim above that the equation is $u''+2/x u'=u^5$: See reference (no euros needed): sciencedirect.com/science/article/abs/pii/…
– josh
Feb 24 at 19:51

Using perturbation method for the equation:

$$y''(x)+\frac{2 y'(x)}{x}+y(x)^5=0$$

If solution is y[x]=Sqrt[3/(3 + x^2)] for x=0 is: 1.

Then we have initial conditions:y=1,and y'=0

ClearAll["*"]; Remove["*"]

eq = y''[x] + 2/x*y'[x] + \[Delta]*y[x]^5;

(*Perturbation method*)

ord = 10;
eqs = CoefficientList[
Normal[Series[
eq /. y ->
Function[x,
Evaluate[Sum[y[i][x] \[Delta]^i, {i, 0, ord}]]], {\[Delta], 0,
ord}]], \[Delta]];yiList = {};
Do[AppendTo[yiList,
MapAt[Factor,
DSolve[{(eqs[[k + 1]] /. yiList /. Log[Exp[x]] -> x) == 0,
y[k] == If[k == 0, 1, 0], y[k]' == If[k == 0, 0, 0]},
y[k][x], x][[1, 1]], 2]], {k, 0, ord - 2}];

(*Series solution:*)

sol2 = Sum[y[i][x] \[Delta]^i, {i, 0, ord - 2}] /. yiList /. \[Delta] -> 1

(*Converting series solution to function:*)

FullSimplify[FindGeneratingFunction[
Table[SeriesCoefficient[sol2 + O[x]^18, n], {n, 0, 10}], x],
Assumptions -> x > 0]

(*Sqrt/Sqrt[3 + x^2]*)


With this metod we can solve:$$y''(x)+\frac{2 y'(x)}{x}-y(x)^5=0$$ with: y=1,and y'=0.

Exact solution is:$$y(x)=\frac{1}{\sqrt{1-\frac{x^2}{3}}}$$