# Document 13559582

```18.034 ;
Feb 6, 2004
Lecture 2
1. Set-up model for a mixing problem
Rate of mass of chemical in
=(conc. in) &times; (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
&times; (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 &gt; 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
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