Without solving, find out the interval of validity for the subsequent initial value problem.

(t² - 9) y' + 2y = In |20 - 4t|,   y(4) = -3

Solution

First, in order to use the theorem to determine the interval of validity we should write the differential equation in the exact form given in the theorem. Thus we will require dividing out through the coefficient of the derivative.

y' + (2/(t² - 9))y = In |20 - 4t|/(t² - 9)

Subsequently, we need to recognize where the two functions are not continuous. It will allow us to determine all possible intervals of validity for the differential equation. Thus, p(t) will be discontinuous at t = +3 since these points will provide a division by zero. Similarly, g(t) will also be discontinuous at t = + 3 and also t = 5 as at this point we will have the natural logarithm of zero. Remember that in this case we won't have to worry regarding to natural log of negative numbers due to the absolute values.

Here, with these points in hand we can break-up the real number line in four intervals and here both p(t) and g(t) will be continuous. These four intervals are as:

- ∞ < t < -3,     -3< t < 3,          3< t < 5,           5< t <

The endpoints of each of the intervals are points where as a minimum one of the two functions is discontinuous. It will guarantee that both functions are continuous everywhere in all intervals.

At last, let's identify the actual interval of validity for the initial value problem. The real interval of validity is the interval which will include t_o = 4. So, the interval of validity for the initial value problem is:

3 < t < 5

In this last illustration we require to be careful to not jump to the conclusion as another three intervals cannot be intervals of validity. Through changing the initial condition, in specific value of t_o, we can create any of the four intervals the interval of validity.

The first theorem needed a linear differential equation. There is a same theorem for non-linear first order differential equations. This theorem is not as useful for determining intervals of validity like the first theorem was thus we won't be liability all that much along with it.