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This document explores Euler's Method to approximate solutions for the differential equation y' = x + y with an initial condition of y(0) = 2. It details step-by-step calculations for finding the approximate values of y at various points, including y(1), y(2), y(3), and intermediate points like y(0.5) and y(1.5) using a step size of ∆x = 0.1 and ∆x = 0.5. By analyzing slopes at given points, the approximations yield values such as y(1) ≈ 4, y(2) ≈ 9, y(3) ≈ 20, and others, illustrating the efficacy of Euler's Method in numerical analysis.
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Section 10.2.2Euler's Method Diff. Equ. February 24, 2009 Berkley High School
Consider a Differential Equation: y'=x+y • We would like an approximation of the curve of y(x) that goes through y(0)=2. • What is the slope at y(0)=2? • slope is 0+2=2 • Let ∆x = 1. What is our best guess as to the value of y(1)? • 2+2=4. We think that a point on that curve is approximately y(1)=4
Consider a Differential Equation: y'=x+y • We would like an approximation of the curve of y(x) that goes through y(1)=4. • What is the slope at y(1)=4? • slope is 1+4=5 • Let ∆x = 1. What is our best guess as to the value of y(2)? • 4+5=9. We think that a point on that curve is approximately y(2)=9
Consider a Differential Equation: y'=x+y • We would like an approximation of the curve of y(x) that goes through y(2)=9. • What is the slope at y(2)=9? • slope is 2+9=11 • Let ∆x = 1. What is our best guess as to the value of y(3)? • 9+11=20. We think that a point on that curve is approximately y(3)=20
Consider a Differential Equation: y'=x+y • We would like an approximation of the curve of y(x) that goes through y(0)=2. • What is the slope at y(0)=2? • slope is 0+2=2 • Let ∆x = .5. What is our best guess as to the value of y(.5)? • 2+.5*2=3. We think that a point on that curve is approximately y(.5)=3
Consider a Differential Equation: y'=x+y • We would like an approximation of the curve of y(x) that goes through y(.5)=3. • What is the slope at y(.5)=3? • slope is .5+3=3.5 • Let ∆x = .5. What is our best guess as to the value of y(1)? • 3+.5*3.5=4.75. We think that a point on that curve is approximately y(1)=4.75
Consider a Differential Equation: y'=x+y • We would like an approximation of the curve of y(x) that goes through y(1)=4.75. • What is the slope at y(1)=4.75? • slope is 1+4.75=5.75 • Let ∆x = .5. What is our best guess as to the value of y(1.5)? • 4.75+.5*5.75=7.625. We think that a point on that curve is approximately y(1.5)=7.625
Exercises • Section 10.2: 19-24 all