Why is DSolve unable to solve this second order ode with initial conditions? Any workaround?

Question

Fyi, report to WRI as suggedted. [CASE:4956902]

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This ode is similar to one here but for some reason DSolve could not able to solve this. This ode is from a textbook

My question is: What is the reason it can't solve this since it is similar to the one in the above link, and any workaround to make DSolve solve this one?

I solved by hand below as well. I used the Trace commands by Michael E2 from the above answer, but it did not help me figure where it went wrong and I am not good at internal tracing of Mathematica functions.

Code

ClearAll[y, x];
ode = y''[x] == y'[x]*Exp[y[x]];
ic = {y[3] == 0, y'[3] == 1};
DSolve[{ode, ic}, y[x], x]
(* {} *)

The solution it should have given is y[x]->-Log[4-x] as shown below.

Verification of solution

mysol = y -> Function[{x}, -Log[4 - x]]
ode /. mysol

ic/.mysol

Hand solution

\begin{align*} y^{\prime\prime} & =y^{\prime}e^{y}\\ y\left( 3\right) & =0\\ y^{\prime}\left( 3\right) & =1 \end{align*} Let $p=y^{\prime}$, then $y^{\prime\prime}=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=\frac{dp}{dy}p$. The ode becomes \begin{align*} \frac{dp}{dy}p & =pe^{y}\\ \frac{dp}{dy} & =e^{y} \end{align*} Which has the solution \begin{equation} p=e^{y}+c_{1}\tag{1}% \end{equation} Using initial conditions given by $p\left( 3\right) =1,y\left( 3\right) =0$ the above becomes $$ 1=1+c_{1} $$ Hence $c_{1}=0$ and the solution (1) simplifies to \begin{align*} p & =e^{y}\\ \frac{dy}{dx} & =e^{y} \end{align*} This is separable. $$ e^{-y}dy=dx $$ Integrating gives \begin{equation} -e^{-y}=x+c_{2}\tag{2} \end{equation} But $y\left( 3\right) =0$, hence $$ -1=3+c_{2} $$ Therefore $c_{2}=-4$ and (2) becomes \begin{align*} -e^{-y} & =x-4\\ e^{-y} & =4-x\\ -y & =\ln\left( 4-x\right) \\ y & =-\ln\left( 4-x\right) \end{align*}

Attempt at Tracing using Michael E2's code

ClearAll[y,x];
ode=y''[x]==y'[x]*Exp[y[x]];
ic={y[3]==0,y'[3]==1};
Trace[DSolve[{ode,ic},y[x],x],s_Solve:>HoldForm[s]->s,TraceInternal->True]//Flatten//Column

Answer 1

I'd report this to WRI. If buried in the DSolve code base is a way to solve this, then the decision tree misses it. Otherwise, they should implement a way. DSolve has a method to solve y''[x] == F[y[x]]. Here's is a modification of it to solve y''[x] == F[y[x], y'[x]] (tested only on this problem!).

I snatched the code for y''[x] == F[y[x]] and modified by hand. The modifications are surround by (**) on either side.

mySpecialOrder2BVP // ClearAll;
First@DownValues@DSolve`DSolveExtendedLibraryDump`SpecialOrder2BVP /. 
  s_Symbol :> 
   With[{v = Symbol@SymbolName[s]}, 
    RuleCondition[v, ! MemberQ[Context[s], $ContextPath]]] /. 
 SpecialOrder2BVP -> mySpecialOrder2BVP

Edit the output above to be the following:

(*"
 * Solves y''[x] == F[y[x], y'[x]] if LHS is integrable w.r.t x
"*)
DownValues@mySpecialOrder2BVP = {
   HoldPattern[
     mySpecialOrder2BVP[eqn_, ycond_, yprimecond_, a_, y_, x_]] :> 
    Block[{rhs, w, res1, c, res},
     rhs = eqn[[2]] /.  (**) {y[x] | y'[x] -> w}(**);
     (res1 = 
        Quiet[Solve[ (**)
          Derivative[1][y][x] == Integrate[eqn[[2]], x](**)  + c /. 
            x -> a /. {ycond, yprimecond}, c]];
       (res1 = (**)
          Derivative[1][y][x] == 
           Integrate[eqn[[2]], x](**)  + (c /. res1[[1]]);
         res = (**)
          DSolve[res1 && ycond[[1]] == ycond[[2]], {y}, {x}]; (**)
         res /; FreeQ[res, (**)DSolve(**)]) /;
        MatchQ[res1, {{c -> m_}}] && FreeQ[res1, Integrate]) /;
      FreeQ[rhs, y | x]]
   };


mySpecialOrder2BVP[ode, Sequence @@ First@Solve[ic], 3, y, x]

(*  {{y -> Function[{x}, -Log[4 - x]]}}  *)

I don't have a way to hook this into DSolve automatically.

Edit notice: I neglected to mark the change from the internal DSolve`DSolveParser to DSolve. DSolveParser[] threw a couple of Part::partw errors that could be ignored. Calling the top-level DSolve seemed the best fix for this answer. The Condition should therefore be FreeQ[res, DSolve]. It seems to me that the errors are but that arise because the internal SpecialOrder2BVP[] hasn't been updated to pass DSolveParser[] complete option value data (with the new options like IncludeSingularSolutions). I haven't noticed anyone reporting such a bug yet, though. (Also, I don't know why Integrate is called twice. It seems wasteful.)

Answer 2

With a bit of assistance, DSolve can obtain the desired solution. ode has a first integral, which is obtained by

Integrate[Subtract @@ ode, x, GeneratedParameters -> C] == 0
(* -E^y[x] + C[1] + Derivative[1][y][x] == 0 *)

Applying the initial conditions then yields

Solve[% /. x -> 3 /. Rule @@@ ic, C[1]] // Flatten
(* {C[1] -> 0} *)

DSolve now obtains the desired solution

DSolve[{%% /. %, ic}, y[x], x] // Flatten
(* {y[x] -> -Log[4 - x]} *)

Addendum

To answer why DSolve fails in the question, consider

s = DSolveValue[{ode, First@ic}, y[x], x, , Assumptions -> C[1] ∈ Reals]
(* -Log[-(1/C[1]) + (E^(3 C[1] - x C[1]) (1 + C[1]))/C[1]] *)

Applying the second boundary condition then yields

Solve[1 == D[s, x] /. x -> 3, C[1]] // Flatten
{C[1] -> 0}

Of course, inserting this value of C[1] into s fails.

Power::infy: Infinite expression 1/0 encountered.

On the other hand, as suggested by Mariusz Iwaniuk in a comment,

Limit[s, %]
(* -Log[4 - x] *)

does work. Apparently, DSolve does not try this.