EDIT: Solution can be found by hand quite easily. Don't know how to get Mathematica to do it though. Solution found by hand will give the curve, but with other curves too, which are not part of the actual solution.
I'm having a problem solving a 1st order nonlinear ODE analytically. DSolve
finds all the solutions to the equation, however the actual solution -- which can be seen using NDSolve
-- is made up of two of the solutions that were given by DSolve
. When 'glued' together, these two solutions form a smooth curve which is the actual solution. I need to how to tell Mathematica to somehow obtain this solution.
I've simplified the equation a lot to remove unnecessary variables. Here's the equation now:
$$ y'(t) = \sqrt{\frac{1+y(t)}{y(t)^2}}$$
$$y(0)=1$$
eqn = y'[t] == Sqrt[(1 + y[t])/y[t]^2];
Here's my code for analytically solving and plotting the equation:
solE = DSolve[{eqn, y[0] == c1}, y, t];
Plot[Evaluate[y[t] /. solE /. c1 -> 1], {t, -3, 3}]
Here's the code for numerically solving and plotting the equation (Used ListLinePlot
to avoid extrapolation due to division by 0 at around t = -0.4):
solN = NDSolve[{eqn, y[0] == 1}, y, {t, -5, 5}];
Show[ListLinePlot[y /. solN], PlotRange -> {{-3, 3}, {-1, 4}}]
As you can see from the plots below, the correct analytical solution (the positive curve that starts between $t=-1$ and $t=0$ at $y = 0$) is made up of two separate solutions (starts with the green curve, then transitions to red), glued together. How do I get Mathematica to return this 'composite curve' as the final solution?
Analytical solutions (left) and numerical solution (right):
I would prefer a mathematical method of doing this, instead of just visually inspecting the two plots like I have. Previously, when I had multiple solutions, I successfully found the correct solution by verifying it using {eqn, initial condition} /. sol
. However, it doesn't seem to work on this equation. For the original equation, the first solution returned true only for the initial condition, but the second solution returned false for both, so my method must be wrong.
DSolve
offers more solutions than admissable, I think. Assumingy>-1
(and real solutions y[t]) the ode should givey'[t]>0
. Together with the conditiony[0]==1
only the branch remains which NDSolve returns! $\endgroup$y > -1
as an assumption toDSolve
, but thenDSolve
ran for over 5 minutes without any results. I thought this was because I had left out the conditiony[0]==1
earlier on (I substituted 1 with 'c', or it would give missing solutions), but putting it back still yields the same result. I also tried assumingy[t] > -1
instead, but that doesn't have any effect, with or without specifyingy[0]==1
insideDSolve
. I also tried adding the assumption when evaluation the equation inside plot, but it didn't have any effect either? $\endgroup$y[t] > -1
with the initial condition asy[0] == c
, solutions are returned, but then when I try to plot it, nothing shows up. I also forgot to explicitly say in the question that y(0) = 1, edited now. $\endgroup$