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A solution to a differential equation at an interval α < t < b is any function y (t) that satisfies the differential equation in question on the interval α < t < b. It is significant to note that solutions are frequently accompanied through intervals and these intervals can impart several of significant information regarding the solution. Consider the subsequent illustration.
Illustration 1: Show that y(x) = x-3/2 is a solution to 4x2 y′′ + 12xy′ + 3 y = 0 for x > 0.
Solution: We'll require the first and second derivative to do this.
y'(x) = (-3/2) x-5/2 y''(x) = (15/4) x-7/2
Plug these in addition to the function in the differential equation.
4x2 ((15/4) x-7/2) + 12x((-3/2) x-5/2) + 3(x-3/2) = 0
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Continuity requirement : Let's discuss the continuity requirement a little. Nowhere in the above description did the continuity requirement clearly come into play. We need that t
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