[Math] Largest open interval the solution is defined

ordinary differential equations

I am taking an ODE course, and I have the following problem.

$$y' = y^2 -4y,$$

$$y(0)=8$$

What is the largest open interval containing $t=0$ for which the solution is defined?

I solved the differential equation using separation of variables:

$$
\int\frac{1}{y^2-4y} dy = \int dt
\\
\frac{1}{4}\log\left|\frac{y-4}{4}\right| = t+C
\\
\frac{y-4}{y} = Ce^{4t}
\\
y-4 = Cye^{4t}
\\
y(t) = \frac{4}{1-Ce^{4t}}
$$

How do I find the largest interval for which the solution is defined? Thanks.

Best Answer

I find it useful to plot the function after solving for the constant from the ICs.

We have:

$$y(x) = -\dfrac{8}{e^{4 x}-2}$$

A plot of $y(x)$ shows:

enter image description here

Of course, analytically, we care about where that denominator is $0$, so we have:

$$e^{4 x}-2 = 0 \rightarrow x = \dfrac{\ln 2}{4} \approx 0.173286795139986$$

From the plot, you can also see what happens as $x \rightarrow \pm \infty$, but it is not bad to derive that analytically either.

One last helpful thing you may have not gotten to yet, is a phase plot, which looks like:

enter image description here

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