# Lesson #6 October 21st 2011

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Publication Date
Source
Legacy
Keywords
• Calculus (1: Modulo Generico)
• 37258
• Economia
• 0054
• Quantitative Finance
• 8409
• 2011
• 6
Disciplines
• Mathematics

## Abstract

1/16 P�i? 22333ML232 Universita` di Bologna Calculus @ QFinance Lesson 2.5 Friday October 21st 2011 Ordinary differential equations professor Daniele Ritelli www.unibo.it/docenti/daniele.ritelli 2/16 P�i? 22333ML232 Separable equation: Exampley′(x) = a(x) y(x)y(x0) = y0 (E1) being a a continuous function. 2/16 P�i? 22333ML232 Separable equation: Exampley′(x) = a(x) y(x)y(x0) = y0 (E1) being a a continuous function. here b(y) = y so using (S)∫ y y0 1 z dz = ∫ x x0 a(s) ds =⇒ ln y y0 = ∫ x x0 a(s) ds 2/16 P�i? 22333ML232 Separable equation: Exampley′(x) = a(x) y(x)y(x0) = y0 (E1) being a a continuous function. here b(y) = y so using (S)∫ y y0 1 z dz = ∫ x x0 a(s) ds =⇒ ln y y0 = ∫ x x0 a(s) ds y = y0 exp (∫ x x0 a(s) ds ) 3/16 P�i? 22333ML232 For instance if a(x) = −x 2 the initial value problemy ′(x) = −x 2 y(x) y(0) = 1 (E1es) ha solution y(x) = e− x2 4 4/16 P�i? 22333ML232 Example y′(x) = a(x) y2(x)y(x0) = y0 (E2) being a a continuous function. 4/16 P�i? 22333ML232 Example y′(x) = a(x) y2(x)y(x0) = y0 (E2) being a a continuous function. here b(y) = y2 so using (S)∫ y y0 1 z2 dz = ∫ x x0 a(s) ds 4/16 P�i? 22333ML232 Example y′(x) = a(x) y2(x)y(x0) = y0 (E2) being a a continuous function. here b(y) = y2 so using (S)∫ y y0 1 z2 dz = ∫ x x0 a(s) ds −1 y + 1 y0 = ∫ x x0 a(s) ds 5/16 P�i? 22333ML232 so that y = 1 1 y0 − ∫ x x0 a(s) ds 5/16 P�i? 22333ML232 so that y = 1 1 y0 − ∫ x x0 a(s) ds For instance if a(x) = −2x the initial value problemy′(x) = −2x y2(x)y(0) = 1 (E2ap) has solution y = 1 1 + x2 6/16 P�i? 22333ML232 Exercises Solve the equations  y′′(x) = 1 y′(x) y(0) = 0 y′(0) = 1 (a) 7/16 P�i? 22333ML232 First order equations Consider the differential equationy′(x) = α(x)y(x) + β(x)y(x0) = y0 (L) 7/16 P�i? 22333ML232 First order equati

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