Use the Cauchy Picard to solve it : Let D ⊆ R^(2), take initial (t_0,x_0) in D. f : D -> R. Consider the i.v.p, x' = f(t,x(0)), x(0) = z_0. Suppose that 1. f is continuous, 2. f is bounded : Sup_D ll f(t,x) ll = M, 3. Lipschitz in x, Uniformly in t. R = { (t,x) such that l t- t_0 l < h, l x - x_0 l < k }, h and k chosen t. R ⊆ D and Mh < k. Then there exists a unique solution to i.v.p x(t) in R ≡ in (t_0-h, t_0 + h). For finding the interval of existence : Find R = { (t,x) such that l t l < h, l x l < k}. Use those to find the interval of existence to the i.v.p x' = x^(2) + e^(-t^(2)), x(0) = 0.