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

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...