Euler's Method for Numerical Solution to Differential Equations
"Learn how Euler's method utilizes local linearity to approximate solutions to differential equations iteratively. Explore the step-by-step process and implications of choosing step sizes for accuracy. Discover how this method helps build solutions graphically with examples."
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Presentation Transcript
A Numerical Technique for Building a Solution to a DE or system of DE s
Eulers Method Suppose we know the value of the derivative of a function at every point and we know the value of the function at one point. We can build an approximate graph of the function using local linearity to approximate over and over again. This iterative procedure is called Euler s Method.
Eulers Method Here s how it works. We start with a point on our solution. . . . . . and a fixed small step size t. . . . then we project a small distance along the tangent line to compute the next point, . . . t . . . and repeat!
Projecting Along a Little Arrow ( , t y ) 1 1 y t ( , t y ) 0 0
Projecting Along a Little Arrow ( , t y ) 1 1 dy dt= ( , f t y ) 0 0 slope y = slope t = f (t0,y0) t t ( , t y ) 0 0
Projecting Along a Little Arrow ( , t y ) 1 1 dy dt= ( , f t y ) 0 0 slope = (t0 + t , y0+ y) = (t0 + t , y0+f (t0,y0) t) t ( , t y ) 0 0
Projecting Along a Little Arrow ( , ) t y new new = slope ( , ) f t y old old = (told + t, yold+ y) = (told + t, y0+ f (told , yold) t) ( , ) t y old old
Summarizing Eulers Method dy dt= You need a differential equation of the form , ( , ) f t y an initial condition (t0,y0), A smaller step size will lead to more accuracy, but will also take more computations. and a fixed step size t. tnew = told + t ynew = y0+ f (told , yold) t
For instance, if dy dt= 2 sin( ) t and (1,1) lies on the graph of y, then 1000 steps of length .01 yield the following graph of the function y. This graph is the anti-derivative of sin(t 2); a function which has no elementary formula!
Exercise dy dt= 2 t y Start with the differential equation , the initial condition , and a step size of t = 0.5. ( ) 0 0 , (2,1) t y = Compute the next two (Euler) points on the graph of the solution function.
Exercise tnew = told + t ynew = y0+ f (told , yold) t dy dt= 2 t y Start with the differential equation , the initial condition , and a step size of t = 0.5. ( ) 0 0 , (2,1) t y = ( , ( , t y ) = = (2 (2,1) + t y 0 0 .5,1 + 2(.5)) = (2.5, 2) ) 1 1 ( , t y ) =(2.5 + .5, 2 + 10(.5)) = (3, 7) 2 2
