Differential equations: error on founding solution

Question

I have to found the solution of this differential equation: y'=-2-y^2, it look simple but I didn't think so. After solving the differential eqaution I have to plot the solution.

What I do:

function DY=solution(t,y)

DY=$-2-y^2$;

end

$[t,y]=ode45(@solution, [0, 4], 0)$; %% I have this message

Warning: Failure at $t=1.110695e+00$. Unable to meet integration tolerances without reducing the step size below the smallest value allowed $(3.552714e-15)$ at time $t$.

In $ode45$ at $309$

%% $[t,y]=ode45(@solution, [0, 4], 0)$ I get an array of numerical solution

%% in both cases the solution plotted give something not good

plot$(t,y)$[solution figure]1

Everything works fine here and the warning is appropriate since the solution does explode in finite time. Furthermore, for the initial condition y(0)=0 the time of explosion is t=pi/(2sqrt(2))=1.11072..., in accordance with the warning message. — Did, Aug 21 '15 at 09:21

Narasimham · Answer 1 · 2015-08-21T09:37:21.177

0

Using Matlab? Before attempting numerical solution, first attempt a symbolic closed form solution.

Try the $t$ interval [0,1.1] because of vertical asymptote for $ \tan$ function.

edited Aug 21 '15 at 09:37

answered Aug 21 '15 at 09:23

Narasimham

40,495

score -1 · Accepted Answer · answered Aug 21 '15 at 09:21

-1

Notice, we have $$y'=-2-y^2$$ $$\frac{dy}{dt}=-(2+y^2)$$ $$\frac{dy}{2+y^2}=-dt$$ integrating both the sides, we get $$\int \frac{dy}{2+y^2}=-\int dt$$ $$\int \frac{dy}{(\sqrt2)^2+y^2}=-\int dt$$ $$\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{y}{\sqrt 2}\right)=-t+c$$

Can you simplify further?

answered Aug 21 '15 at 09:21

Harish Chandra Rajpoot

37,450

Could someone please tell what is wrong with the answer? – Harish Chandra Rajpoot Aug 21 '15 at 09:24
Your answer is correct, indeed. What I suppose is that someone does not understand that the problem happens if $t=something$. – Claude Leibovici Aug 21 '15 at 09:29
No, kindly recheck the derivative it's $y'=-2-y^2$. – Harish Chandra Rajpoot Aug 21 '15 at 09:32
@HarishChandraRajpoot Sorry I erred seeing and so cancelled my comment – Narasimham Aug 21 '15 at 09:34
At the very least, this lacks an explanation of the explosion of the solution witnessed by the OP (who, as happens on the site, seems to be happy with an answer to a different question than the one they asked). – Did Aug 21 '15 at 09:40
Yes, you are absolutely right. But perhaps it could help him. – Harish Chandra Rajpoot Aug 21 '15 at 09:43

score -2 · Answer 3 · answered Aug 21 '15 at 09:38

-2

It work good, so now how solve this problem whene I wnat to plot the solution on a bigger interval conataining O for example [-5,5]?

answered Aug 21 '15 at 09:38

Aws Ben Hamed

1

?? Do you realize that the solution is NOT defined on the whole interval [-5,5]? – Did Aug 21 '15 at 09:40
I know, but the problem that I found some figure of the solution of this differential equation in some courses, how did they do it? – Aws Ben Hamed Aug 21 '15 at 09:47
They plotted the solution on its interval of definition, probably. – Did Aug 21 '15 at 12:25

Differential equations: error on founding solution

3 Answers3