lecture_12 (dragged) 2

lecture_12 (dragged) 2 - MA 36600 LECTURE NOTES: MONDAY,...

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MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9 3 We explain the relationship with solutions to these equations. Recall how to solve the diferential equation when the equation is “homogeneous” i.e., when g ( t ) is the zero Function: dy dt + p ( t ) y =0 dy dt = p ( t ) y 1 y dy dt = p ( t ) d dt ln | y | = p ( t ) ln | y | = ° t p ( τ ) = y ( t )= y 0 exp ± ° t t 0 p ( τ ) ² . We can solve the diference equation by considering its iterates: y 1 = ρ 0 y 0 = y 0 ³ 1 p ( t 0 ) h ´ y 2 = ρ 1 y 1 = y 0 ³ 1 p ( t 0 ) h ´³ 1 p ( t 1 ) h ´ y 3 = ρ 2 y 2 = y 0 ³ 1 p ( t 0 ) h ´³ 1 p ( t 1 ) h ´³ 1 p ( t 2 ) h ´ = y n = y 0 n 1 µ k =0 ± 1 p ( t k )∆ t k ² where we have set ∆ t k = h . Note the similarity with the exponential and the product. Autonomous Equations. Say that we have a linear autonomous equation in the Form dy dt + py = g where p and g are constants. The associated diference equation is in the Form y n +1 = ρy n + b where ρ =1 ph, b = gh.
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