Solving an ODE with parameters and taking the limit of the solution

Question

I am very new to Mathematica and already spent a lot of time trying to do this but failed.

I am trying to solve an ODE:

solution = DSolve[{-((m (1 + m) + 4/(9 (-2/3 + t) t)) y[t]) + 
     2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[
       t] == (-4 (1 + C/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == C}, 
  y, t]

where $m$ is a nonnegative integer and $C$ is a real number.

I want to show that there exists a $C$ such that the solution is $0$ at infinity. When I try that code:

Limit[y[t] /. solution[[1]], t -> Infinity, m \[Element] Integers]

it just spits out the same thing.

What should I do? (Note that I don't need to find that value of $C$; I just need to show that for every $m$, there is a number $C$ in which the solution vanishes at infinity. )

EDIT:

I amanged to get that it's true for many values of $m$. Here is the code I used for $m=10$.

solutionm = 
 DSolve[{-((10 (10 + 1) + 4/(9 (-2/3 + t) t)) y[t]) + 
     2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[
       t] == (-4 (1 + C/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == C}, 
  y, t]

Limit[FullSimplify[Re[y[t] /. solutionm[[1]]]], t -> Infinity, 
 Assumptions -> C \[Element] Reals]

Which spits out:

DirectedInfinity[360 (1036 - 943 Log[3]) + C (-59572 + 54225 Log[3])]

Then I choose the $C$ that makes that number in "DirectedInfinity" zero:

{a} = Solve[360 (1036 - 943 Log[3]) + C (-59572 + 54225 Log[3]) == 0, 
  C]
a = C /. a[[1]]

Then when $C=a$, the limit is 0:

Limit[Re[y[t] /. solutionm[[1]]], t -> Infinity, 
 Assumptions -> C == a]
0

I tried for many values of $m$, and I get the same thing. When I try to make $m$ arbitrary, something weird happens:

$Assumptions={m \[Element] Integers, C \[Element] Reals}
solutionm = 
 DSolve[{-((m (m + 1) + 4/(9 (-2/3 + t) t)) y[t]) + 
     2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[
       t] == (-4 (1 + C/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == C}, 
  y, t]

When I run y[t] /. solutionm[[1]] /. {m -> 1}, it gives me an error: Power::infy: Infinite expression 1/0 encountered. I get the same error with any $m$. I am not sure why this happens.

Also, when I repeat the same thing as above, and run

Limit[FullSimplify[Re[y[t] /. solutionm[[1]]]], t -> Infinity, 
 Assumptions -> C \[Element] Reals]

it doesn't compute the limit. It just spits out the same thing. Is there a way around this? Or, as Nasser suggested, is this too complicated for Mathematica? Also, I don't need to find that $C$. I just want to show that there exists a $C$ in which the solution vanishes at infinity.

Answer 1

Here is an answer for m==2 in version 13.1 on Windows 10 (C is reserved in WL.)

ClearAll[m, c]; $Assumptions = {m \[Element] Integers, c \[Element] Reals}; m = 2;
solution = AsymptoticDSolveValue[{-((m (m + 1) + 4/(9 (-2/3 + t) t)) y[t]) + 
  2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[t] ==
 (-4 (1 + c/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == c}, y[t], {t, Infinity, 1}];
Limit[solution, t -> Infinity]

\[Infinity] (48-36 Log[3]+c (-68+63 Log[3]))

Now

Solve[(48 - 36 Log[3] + c (-68 + 63 Log[3])) == 0, c]

{{c -> (-48 + 36 Log[3])/(-68 + 63 Log[3])}}

and then

ClearAll[m, c]; m = 2; c = (-48 + 36 Log[3])/(-68 + 63 Log[3]);
solution = AsymptoticDSolveValue[{-((m (m + 1) + 4/(9 (-2/3 + t) t)) y[t]) + 
  2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[
    t] == (-4 (1 + c/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == c},
y[t], {t, Infinity, 1}];
Limit[solution, t -> Infinity]

0

A similar behavior for other natural m I tried. In the general case AsymptoticDSolveValue fails.

Answer 2

We can have an insight about the solutions behavior for each m by doing

tmax = 1000;
solution = ParametricNDSolve[{-((m (1 + m) + 4/(9 (-2/3 + t) t)) y[t]) + 2 (-1/3 + t) y'[t] + (-2/3 + t) t y''[t] == (-4 (1 + c/2))/(9 (-2/3 + t) t), y[1] == 1, y'[1] == c}, y, {t, 1, tmax}, {m, c}]

m = 1;

Plot3D[Evaluate[y[m, c][t] /. solution], {t, 1, tmax}, {c, -6, 0}]