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
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