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First-Order Differential Equations 1.1 Terminology and Separable Equations 1. For x > 1, 2ÕÕ = 2√ x − 1 1 2 √x − 1 = 1, so Õ is a solution. 2. With Õ(x) = Ce−x, Õ + Õ = −Ce−x + Ce−x = 0, so Õ is a solution. 3. For x > 0, rewrite the equation as 2xy + 2y = ex. With y = Õ(x) = 1 2x−1(C − ex), compute y = 1 2 −x−2(C − ex) − x−1ex . Then 2xy + 2y = x −x−2(C − ex) − x−1ex + x−1(C − ex) = ex. Therefore Õ(x) is a solution. 4. For x = ± √2, Õ = −2cx (x2 − 2)2 = 2x 2 − x2 c x2 − 2 = 2xÕ 2 − x2 , so Õ is a solution.
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