Skip to main content

differential equations - DSolve not finding solution I expected


Try to solve the following ODE via DSolve


$$ \left\{\begin{aligned} y'(x)+2 y(x) e^x-y(x)^2 &= e^{2 x}+e^x \\ y'(0) &=1 \end{aligned}\right. $$


The expected solution is $y(x)=e^x$, but Mathematica produces the message:



DSolve::bvfail: For some branches of the general solution, the given boundary conditions lead to an empty solution




Note: $y(x)=e^x$ is obviously a special solution to the original ODE and satisfies the initial/boundary value condition which can be easily verified by substituting it into the ODE.


Additionally, it is easy to deduct that: $y'(0)=1\Leftrightarrow y(0)=1$ for this ODE;


How to handle it?



Answer



The problem - Genericity of solutions


The problem we encounter here is that DSolve can return only a generic solution however that general solution cannot satisfy such an initial condition as y'[0] == 1. The issue is related to an arbitrary choice of constants of integration i.e. such constants that are specific to certain types of a differential equations DSolve tries to solve during the integration process.


DSolve[{y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x, y'[0] == 1}, y[x], x]


DSolve::bvnul: For some branches of the general solution, the given boundary conditions 

 lead to an empty solution. >>

{}

Solving the ODE without the initial condition we will see that the general solution excludes exceptional ones:


y[x_, c_] = y[x] /. First @ DSolve[{ y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x},
                                    y[x], x] /. C[1] -> c


 E^x + 1/(c - x)


Namely there is no constant c reducing the general solution to E^x however putting e.g. y'[0] == 2 we will find a solution:


DSolve[{y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x, y'[0] == 2}, y[x], x]// First


 {y[x] -> (-1 - E^x + E^x x)/(-1 + x)}

This is quite similar issue to one we can find here: Solving a differential equation with initial conditions.


The problem cannot be fixed changing the variables in the ODE to :


y'[x] + 2 E^x y[x] - y[x]^2 == E^(2 x) + E^x /. 

{y[x] -> z[x] + E^x, y'[x] -> z'[x] + E^x} // Simplify


z[x]^2 == z'[x]

DSolve[{z'[x] == z[x]^2, z'[0] == 0}, z[x], x]


DSolve::bvnul: For some branches of the general solution, the given boundary conditions 
lead to an empty solution. >>


{}

Remedy


Introducing an initial condition in a general symbolic way we can find also the special solution:


f[x_, c_] = z[x] /. DSolve[{z'[x] == z[x]^2, z'[0] == c}, z[x], x]


 { -(Sqrt[c]/(-1 + Sqrt[c] x)), -(Sqrt[c]/(1 + Sqrt[c] x))}


f[x, c] /. Solve[D[f[x, c] == 0], c]


{ 0, 0}

I.e. we get the special solution y[x] == E^x.


we can see that the same trick can work also in the original equation.


Alternatively the general solution can be reparametrized in a different way:


E^x + c/(1 - x)


yields also the special solution although this excludes e.g. E^x - 1/x


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more