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Use Euler's method with step size 0.5 to compute the approximate y-values $ y_1, y_2, y_3, $ and $ y_4 $ of the solution of the initial-value problem $ y' = y - 2x, y(1) = 0. $
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Calculus 2 / BC
Chapter 9
Differential Equations
Section 2
Direction Fields and Euler's Method
Campbell University
Oregon State University
Idaho State University
Boston College
Lectures
13:37
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.
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Use Euler's method wi…
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you're given an initial value problem and were asked to use whether given steps to approximate some points on the solution. The initial value problem is the differential equation y prime equals y minus two x with initial value. Why have one people zero and the steps eyes were asked to use is h equals 0.5. Since our initial value is why have 10? We have that X zero is equal to one and why zero is equal to zero. Therefore we have that X one is equal to x zero plus 00.5 or 1.5 and why one is equal to y zero just zero plus our step size 0.5 times are function Why minus two x evaluated at point x zero y zero point 10 This is simply equal to negative too. Times one is equal to 0.5 times negative to his native born This is negative one again There step sizes 0.5 we have X two is equal to 2.0 and therefore why to is equal to why one which is a negative one plus her step size 0.5 times are function Why minus two x evaluated at X one y one were at 1.5 negative one. This is equal to negative one minus two times 1.5, which is equal to negative one plus 0.5 times negative one minus three or negative four is equal to negative one plus negative, too, or negative three. For 1/3 step, we have that X three is equal to two plus 20.5 or 2.5 in that wide three is equal to why to which is negative. Three plus our step size 25 times are function y minus two x evaluated at the point to negative three. So negative three minus two times two, which is equal to negative three plus 0.5 times negative three minus form or native seven is equal to negative three plus negative 3.5 or negative 6.5. Finally, we have our next step. X two x four is equal to 2.5 plus 0.5 or three point out, and why for is equal to why three. His native 6.5 plus our step size 0.5 times their function. Why minus two X evaluated at the point 2.5 Negative 6.5 This sequel to negative 6.5 unless 0.5 times negative. 6.5 minus five or negative 11.5 This is equal to negative 6.5 plus negative 5.25 which is equal to negative. 11 point. Sorry, this is negative. 5.7. Yeah, this is equal to negative 12 0.25
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