Finding the interval of definition of an ODE solution: $x'(t)=\frac{x^2-1}{2}$.

Question

This is the first exercise of my ODE notes, so I'd like you to check if my approach is correct or not.

Find the maximum interval of definition of the solutions: $\dot{x}=\frac{x^2-1}{2}, \ \ x(t_0)=x_0.$

My approach:

I found the solution of this ODE. They're of the form:

$$x(t)=\frac{-e^{c_1+t}-1}{e^{c_1+t}-1}.$$

Now we take the denominator and equal it to $0$ to see what points are not in the domain. We see that $${e^{c_1+t}-1}=0 \implies t=-c_1.$$

So we have a vertical asymptote on $x(c_1)$ in every solution considered.

Therefore, the interval is $$(c_1,\infty)\ \ if \ \ c_1>0$$

$$(-\infty,c_1)\ \ if \ \ c_1<0$$

$$(c_1,\infty)\ \ or \ \ (-\infty,c_1) \ \ if \ \ c_1=0.$$

Is my approach correct?

Thanks for your time.

Answer 1

A more careful computation would have the integration of the partial fraction decomposition as $$ \ln|x-1|-\ln|x+1|=t+c $$ so that in the exponential $$ \frac{x(t)-1}{x(t)+1}=Ce^t $$ where $\frac{x_0-1}{x_0+1}=C=\pm e^c$ incorporates the signs of the expressions under the absolute value. By uniqueness the signs are constant.

Considering the sign of the constant $C$ one finds that for $x_0\in [-1,1]$ the solution $$ x(t)=\frac{1+Ce^t}{1-Ce^t} $$ exists at all times, outside that interval it has poles in finite time.

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Alternatively to confirm that solution use $$ u(t)=\frac{x(t)-1}{x(t)+1} $$ which leads to $$ u'(x)=\frac{2x'}{(x+1)^2}=u $$ which has the solution $$ u(t)=e^tu(0)\implies \frac{x(t)-1}{x(t)+1}=e^t\frac{x_0-1}{x_0+1} \\~\\ \implies x(t)=\frac{(x_0+1)-e^t(1-x_0)}{(x_0+1)+e^t(1-x_0)} $$

Answer 2

In general at first you should find $c_1$ in terms of by the initial condition $x(t_0)=x_0$ and then find the interval of definition.

Note: the general solution should be

$$x(t) = \frac{- e^{c_1 + t}+1}{e^{c_1 + t} + 1}$$

In this particular case, since it is an autonomous differential equation, it easy to see that the interval of definition is $\mathbb{R}$.