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I need help figuring out why the fisr order ODE : $x'(t)=x^2(t)$ with the initial condition $x(0)=x_{0}\neq 0$ doesn't have a solution defined for all t.

Does anyone have an idea ?

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Hm? It has a solution, namely $x(t) = x_0/(1-tx_0)$. – martini Apr 24 '12 at 14:16
@martini - sorry, I will correct the question in a second – Belgi Apr 24 '12 at 14:17
As the solution is $x(t) = x_0/(1-tx_0)$ and solutions are (locally) unique. – martini Apr 24 '12 at 14:19
@martini - I am at the stage of my ODE course that we don't know this yet. The proof should have other considirations to it according to the notes am reading. – Belgi Apr 24 '12 at 14:20
up vote 2 down vote accepted




$\int\frac{dx}{x^2}=\int dt$



if $x(0)=x_0$ then

$x(0)=\frac{-1}{0+c}=\frac{-1}{c}=x_0$ ($x_0\neq 0$)



Let's find the values that $x(t)$ is not defined:



if $t=\frac{1}{x_0}$ then $x(t)$ is not defined. $x(t)$ is defined for all other values except $t=\frac{1}{x_0}$ ($x_0\neq 0$).

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What is the meaning of multiplying in $dt$ ? what justification can you give for this ? – Belgi Apr 24 '12 at 15:39
I do not know your knowledge about differential equations . You can write $x'(t)=\frac{dx}{dt}$ to solve differential equations. You can check the link about Leibniz's – Mathlover Apr 24 '12 at 19:36

Nobody has said that any initial value problem you cook up will have a solution for all $t$ – in the same vein that you cannot expect a real solution for all equations of the form $x^2=c\in{\mathbb R}$.

On the positive side you can say the following: When the hypotheses of the existence and uniqueness theorem are satisfied throughout the domain $\Omega\subset{\mathbb R}^2$ of the given ODE $x'=f(x,t)$ then the graph of any "complete" solution will finally leave any given compact $K\subset\Omega$.

In the case at hand the graph of the solution $y(t)=x_0/(1-t x_0)$ will go off to $\infty$ when $t x_0\to 1-$.

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