It seems normal to solve a differential equation in order to find its interval of validity but it's not always the case. The following theorem will allow us to find the interval of validity without solving the differential equation. An interval of validity guarantees the existence of a solution and its uniqueness.
Here is the theorem:
Let's consider the following differential equation y' + p(t)y = g(t) with the initial value y(t₀) = y₀. If p(t) and g(t) are continuous on the interval α<t<β and the interval contains t₀, then there is a unique solution of the differential equation on that interval. If the interval is the largest one in which p(t) and g(t) are continuous, then this interval is the interval of validity of the equation.
Example. Determine the interval of validity of the initial value problem without solving it:
(t² - 9)y' + 2y =ln❘20 - 4t❘ y(4) = -3
Let's write the equation under its explicit form i.e y' + p(t)y = g(t). In order to do so, we divide the equation by the coefficient of y' which is t² - 9:
(t² - 9)y'/(t² - 9) + 2y/(t² - 9) = ln❘20 - 4t❘/t² - 9
y' + 2y/(t² - 9) = ln❘20 - 4t❘/t² - 9
Let's study the continuity of the functions p(t) = 2/ t² - 9 and g(t) = ln❘20 - 4t❘/t² - 9.
p(t) being a rational function is not defined for the value of t that make its denominator equal to zero.
t² - 9 = 0 for t = -3 or t = 3. Therefore p(t) is continuous in the intervals ]-∞ -3[ ]-3 3[ ]3 +∞[.
The function g(t) is continuous if its numerator is defined and its denominator are not equal to zero. The numerator is undefined if 20 - 4t = 0 because in this case we would have ln0. The numerator is defined in the intervals ]-∞ 5[ 5[ +∞[. The denominator is defined in the intervals ]-∞ --3[]-3 3[ ]3 +∞[. The function g(t) is continuous in the union of the intervals in which both the numerator and the denominator are defined. The function g(t) is continuous in ]-∞ -3[]-3 3[ ]3 5[ ]5 +∞[. These are also intervals in which both p(t) and g(t) are continuous. The interval of validity is the interval that has the initial value t = 4. In this case this is the interval ]3 5[.
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