2 1. Differential Equations has exactly the growth property required. For any interesting differential equation, there will be some func- tions that do satisfy the equation and others that do not. Those which do satisfy it, as the function P (t) defined by (1.2) satisfies (1.1), are called solutions to the equation. The most elementary skill needed for working with differential equations is just the ability to plug some functions into an equation to find out whether they solve it or not. Example 1.1 Determine whether the functions f1(x) = xex (1.3) and f2(x) = x2 (1.4) are solutions to the differential equation (1 + x)f(x) = xf (x). (1.5) Solution Note that while equations (1.3) and (1.4) are merely def- initions specifying what is meant by f1 and f2, equation (1.5) is a differential equation. We expect that some functions, but not all func- tions, would have this property that multiplying their derivative by x always results in the same value as multiplying the original function by 1 + x. For instance, f1 is a solution to (1.5) because (1 + x)f1(x) = (1 + x)xex = (x + x2)ex and xf 1 (x) = x(ex + xex) = xex + x2ex = (x + x2)ex. We therefore see that if f = f1, then the left and right sides of (1.5) would give the exact same value no matter what we choose for x. Note, however, that f2 is not a solution. Some students might mistakenly think that it is a solution and argue as follows: (x + 1)f2(x) = x3 + x2 xf2(x) = x(2x) = 2x2. So, is it true that x3 + x2 = 2x2? It is when x = 0 or x = 1. Then, isn’t it a solution? Remember, however, that we are looking for an equality of functions here. In other words, here x3 +x = 2x2 is asking

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