lecture_11

# lecture_11 - MA 36600 LECTURE NOTES FRIDAY FEBRUARY 6...

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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 6 Numerical Solutions Euler’s Method. Consider an initial value problem dy dt = G ( t,y ) , y ( t ) = y ; in terms of some function G ( t,y ). In general it is too difficult to solve such an equation. Indeed, we have seen that we can solve this equation only if we have some type of trick – such as using integrating factors or studying separable equations. Instead we will discuss how to compute an approximate solution. Say that we have a solution y = y ( t ) to the initial value problem above. Choose a value t 1 which is “near” t . We explain how to choose a value y 1 which is “near” y ( t 1 ). Consider a secant line through the two points ( t ,y ) and ( t 1 ,y 1 ). Also consider the line tangent to the curve y = y ( t ) at the point ( t ,y ). We want the slope of the secant line to be equal to the slope of the tangent line: Δ y Δ t = dy dt ( t ,y ) that is, y 1- y t 1- t = G ( t ,y ) . Solving for y 1 gives the expression y 1 = y + G ( t ,y ) · ( t 1- t ) ≈ y ( t 1 ) . We may proceed in a recursive fashion. Choose a value t n +1 which is “near” t n . Consider the secant line through two points ( t...
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## This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue.

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lecture_11 - MA 36600 LECTURE NOTES FRIDAY FEBRUARY 6...

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