Hi everyone,

I wanted to post a quick followup to our work from Tuesday, regarding intervals of validity – this may be of use as you work on WeBWorK #3.

Let’s look at an example – this is my version of Problem 6:

Separate the following differential equation and integrate to find the general solution:
y'=(5-2x)y^2

Then give the particular solution that satisfies the initial condition y(0)=\frac{1}{-6} and state the interval on x for which this solution is valid.

Let’s look at the particular solution satisfying the initial condition – it is:  y=\frac{1}{x^2-5x-6}.

What is the interval of validity?  It is the largest interval of the x-axis which a) contains the initial condition, and b) on which the solution is defined and continuous.

By setting the denominator equal to zero, we find that the solution is undefined at x=6 and x=-1. These two values break the entire x-axis up into three intervals on which the solution is defined (and continuous): (-\infty,-1) \cup (-1,6) \cup (6,\infty).  We now simply have to choose the correct interval – here, we use the fact that the initial condition y(0)=\frac{1}{-6} refers to the point (0,\frac{1}{-6}), so we must choose the interval containing x=0.

Thus the interval of validity is (-1,6).

Please comment below if you have questions!

Best of luck with the WeBWorK, and enjoy the snow.

Regards,
Prof. Reitz