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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9 Existence and Uniqueness (cont’d) Example. We give an example of the proof of the theorem. Consider the initial value problem dy dt = 2 t (1 + y ) , y (0) = 0 . First we find an exact solution y = y ( t ), then we find a sequence of approximate solutions y n = y n ( t ) via Picard’s Method. To find the exact solution, note that this differential equation is a separable equation: dy dt = 2 t (1 + y ) 1 y + 1 dy dt = 2 t d dt ln  y + 1  = 2 t ln  y + 1  = t 2 + C = ⇒ y + 1 = C 1 e t 2 for some constant C 1 = ± e C . For the initial condition, set t = 0: C 1 = 1 = ⇒ y ( t ) = e t 2 1 . For a sequence of approximate solutions, define the following: y n +1 = Z t G ( τ, y n ( τ ) ) dτ where G ( t,y ) = 2 t (1 + y ) . The initial function is a constant, so we have y ( t ) = 0; y 1 ( t ) = Z t 2 τ (1 + 0) dτ = t 2 ; y 2 ( t ) = Z t 2 τ ( 1 + τ 2 ) dτ = t 2 + t 4 2 ; y 3 ( t ) = Z t 2 τ 1 + τ 2 + τ 4 2 dτ = t 2 + t 4 2 + t 6 6 . It is easy to show in general that y n ( t ) = t 2 + t 4 2! + t 6 3! + ··· + t 2 n n ! = n X k =0 t 2 k k ! . Now we take the limit as n increases without bound: lim n →∞ y n ( t ) = ∞ X k =1 t 2 k k ! = ∞ X k =0 x k k ! ! 1 where x = t 2 . Recall the Taylor Series expansion for the exponential function: e x = ∞ X k =0 x k k ! = ⇒ lim n →∞ y n ( t ) = e t 2 1 = y ( t ) . Hence the functions y n = y n ( t ) do indeed give a good approximation of the exact solution y = y ( t ). 1 2 MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9 Difference Equations First Order Equations. Recall that every first order differential equation is in the form dy dt = G ( t,y ) , y ( t ) = y ; for some function G = G ( t,y ). As discussed in the previous lecture, often it is too difficult to find an exact solution, so we seek approximate solutions. We review the ideas....
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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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