Solutions that Blow Up: The Domain of a Solution . Example 1. Solve the IVP y = y2 , y(0) = 1. Solution. We can solve this using separation of variables. Separate: dy = dx. y2 Integrate: −1/y = x + C. Solve for y: y = −1/( x + C ). Find C using the IC: y(0) = 1 = −1/C, therefore C = −1. Solution: y = −1/( x − 1) = 1/(1 − x ). The graph has a vertical asympote at x = 1. 5 y 4 3 2 1 −2 −1 −1 1 2 t −2 −3 −4 −5 Fig. 1. Graph of y = 1/(1 − x ). Starting at x = 0 the graph goes to inﬁnity as x → 1. Informally, we say y blows up at x = 1. The graph has two pieces. One is deﬁned on (−∞, 1) and the other is deﬁned on (1, ∞). For technical reasons we prefer to say that we actually have two solutions to the DE. We indicate this by carefully specifying the domain of each. y(x) = 1/(1 − x) y(x) = 1/(1 − x) x in the interval (−∞, 1) (1) x in the interval (1, ∞). (2) Thus, the solution to the IVP in this example is solution (1). � The rule being followed here is that solutions to ODE’s have domain con sisting of a single interval. The example shows one reason for this: starting at (0, 1) on solution (1) there is no way to follow the solution continuously to solution (2).

Solutions That Blow Up: The Domain of a Solution

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