Differential equation of a mixing problem

18.034, Honors Differential Equations Page 1 of 2 Prof. Jason Starr 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) × (rate of flow) = ( ) vt y q ⋅ . So ( ) y va t c a y − ⋅ = ' where a, 0 > V are constants 2. Discussed method of integrating factors: (a) Guess there exists ( ) t u s.t. ( ) ( ) ( )y t p t u y t u + ' equals ( ) [ ] ' y t u . (b) This leads to assorted separable equations, ( )u t p u = ' which has a solution ( ) ( ) t p pdt e e t U = = ∫ , which is evacuated. (c) Define ( )y e x t p = , ( )x e y t p = . Then ( ) ( ) t q y t p y = + ' iff ( ) ( ) t q e x t p = ' . Moreover, choosing ( ) ( ) 0 0 , 0 0 y y P = = iff ( ) 0 0 y x = . So have existence/uniqueness of original IVP iff existence/uniqueness of IVP ( ) ( ) ( ) 0 0 , ' y x t q e x t p = = . But this follows from F.T. of calculus. (d) Conclusion: If ( ) ( ) t q t p , are defined and cts. on ( ) b a, ⊂ IR , then there exists a solution ( ) t y of ( ) ( ) t q y t p y = + ' defined on all of ( ) b a, , the solution is unique, and it has the form. where ( ) ( ) t p t p = ' ( ) 0 0 = p 3. Used this method to solve the mixing problem: ( ) ( ) t v t t v t v e y ds s c ae e t y α α α − − + = ∫ 0 0 (a) If ( ) c t c = is constant, get ( ) ( ) t v e y cV cV t y α − − − = 0 , i.e. ( ) ( ) ( ) t v e y cV t y cV α − − = − 0 . ( ) t c a y va y . ' = + ( ) ( ) t q y t p y = + ' ( ) ( ) ( ) ( ) , ) ( 0 0 t p t s p t p e y ds s q e e t y − − + = ∫ V=volume, a=rate of flow of solution C(t)= concentration in y(t)= mass of chemical in tank at time t18.034, Honors Differential Equations Page 2 of 2 Prof. Jason Starr So equil. solution is cV y = (which makes physical sense), and the “half-life”, to come with 21 of equilibrium from initial value is aV = τ ln(2) (increasing V or decreasing a increases the half-life). (b) Consider the case that ( ) ( ) t A t c ω sin ⋅ = . Let to integral ( ) ds s aAe t s v ω α sin 0 ∫. Set va = λ , get ( ) ( )λ ω φ φ ω ω λ λ = − + tan , sin 2 2 t e aA t . Didn’t have time to really analyze the solution. ( ) ( ) t be t aA t y λ φ ω ω λ − + − + = sin 2 2 for some b 4. Particular solution method. To find the general solution of ( ) ( ) t q y t p y = + ' , (i) Find general solution of undriven/homo system ( ) 0 0 '0 = + y t p y . (ii) Find a particular solution p y of original equation. (iii) General solution is p y y + 0 .