advertisement

18.034 ; Feb 6, 2004 Lecture 2 1. Set-up model for a mixing problem Rate of mass of chemical in =(conc. in) × (rate of flow of liquid)= a.c(t) Rate of mass out = (conc. out) So y ' = a ⋅ c (t ) − a y v V=volume, a=rate of flow of solution × (rate of flow) = q ⋅ y' + y (t ) . v C(t)= concentration in a y = a.c(t ) v y(t)= mass of chemical in tank at time t where a, V > 0 are constants y '+ p (t ) y = q (t ) 2. Discussed method of integrating factors: (a) Guess there exists u(t ) s.t. u(t )y '+u(t )p(t )y equals [u(t )y ] ' . (b) This leads to assorted separable equations, u' = p(t )u which has a solution U (t ) = e ∫ p(t ) , which is evacuated. =e p (t ) p(t ) (c) Define x = e y , y = e x . Then y '+ p(t )y = q(t ) p(t ) iff x' = e q(t ) . Moreover, choosing P (0) = 0, y (0) = y 0 iff x (0) = y 0 . So have existence/ uniqueness of original IVP p (t ) iff existence/ uniqueness of IVP x' = e q(t ), x (0) = y 0 . But this follows from F.T. of calculus. (d) Conclusion: If p(t ), q(t ) are defined and cts. on (a, b ) ⊂ IR , then there exists a pdt solution y (t ) of y '+ p (t ) y = q (t ) defined on all of (a, b ) , the solution is unique, and it has the form. y (t ) = e − p (t ) t ∫ e p (s ) q(s )ds + y 0 e − p(t ) where p (t ) = p (t ) ' , p(0) = 0 0 3. Used this method to solve the mixing problem: y (t ) = e −α t t v ∫ α t ae v c (s )ds + y 0 e −α t v 0 (a) If c (t ) = c is constant, get y (t ) = cV − (cV − y 0 ) (cV −α t ev − y (t )) = (cV − y 0 ) 18.034, Honors Differential Equations Prof. Jason Starr , i.e. −α t ev . Page 1 of 2 So equil. solution is y = cV (which makes physical sense), and the “half-life”, to V 1 of equilibrium from initial value is τ = ln(2) (increasing V or 2 a decreasing a increases the half-life). come with (b) Consider the case that c (t ) = A ⋅ sin (ωt ) . Let to integral t ∫ α aAe v s sin (ωs ) ds . 0 a Set λ = , get v aA 2 λ +ω 2 e λt sin (ωt − φ ), tan (φ ) = ω . Didn’t have time to really λ analyze the solution. y (t ) = aA 2 λ +ω 2 sin (ωt − φ ) + be − λt for some b 4. Particular solution method. To find the general solution of y '+ p(t )y = q(t ) , (i) Find general solution of undriven/ homo system y 0' + p(t )y 0 = 0 . (ii) Find a particular solution y p of original equation. (iii) General solution is y 0 + y p . 18.034, Honors Differential Equations Prof. Jason Starr Page 2 of 2