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differential equations - DSolve not finding solution I expected


Try to solve the following ODE via DSolve


{y′(x)+2y(x)ex−y(x)2=e2x+exy′(0)=1


The expected solution is y(x)=ex, 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)=ex 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⇔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


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