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