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Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex · 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ NINFORMATION NEWS MENU PHOTO STAFF MAP BLOG RECRUIT LINKSoluciones del examen de sept05 de Ecuaciones Diferenciales I 1 Sea la ecuación y0= 1 2ty3 3t2y2 a Hallar su solución general por dos caminos diferentes b Precisar cuántas soluciones cumplen y(1)=1 y escribirlas explícitamente a Bernouilli 3y2y0= 2 t y 3 1 2 z=y3!z0= 2z t 1!z = C 1 R t2!y =Ct t2 1/3;
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