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Department
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Course Code
MAT244H1
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MAT244H1a.doc
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Introduction
INTRODUCTION TO DIFFERENTIAL EQUATIONS
A differential equation is an equation involving some hypothetical function and its derivatives.
Example
xyy =
+
2 is an differential equation. As such, the differential equation is a description of some function
(exists or not).
A solution to a differential equation is a function that satisfies the differential equation.
Example
xxy 5
3+= is a solution to
3
xyxy +=
.
Some differential equations are famous/important:
yy =
,
x
ey =.
ayy =
,
ax
ey =.
0=+
yy ,
x
y
cos
=
.
0=+
ayy , xaycos=.
Recall that a differential equation describes a phenomenon in terms of changes. For example, if mvP
=
, then
vdmFdt= or
F
dt
dP=
.
Example
A pool contains V liters of water which contains M kg of salt. Pure water enters the pool at a constant rate of v
liters per minute, and after mixing, exits at the same rate. Write a differential equation that describes the
density of salt in the pool at an arbitrary time t.
Let
(
)
t
ρ
be the density at time t. Then
( )
(
)
V
tM
t=
ρ
.
To model change in
(
)
t
ρ
, let
(
)
11 t
ρρ
= and
(
)
22 t
ρρ
=. Then
(
)
(
)
12112 ttvV
ρρρ
, so
V
v
V
tt 1
12
12
ρ
ρρ
or
(
)
(
)
( )
V
v
tV
ttt
ttt
ρ
ρρ
+
+
.
Now, as 0
t,
( )
V
v
t
dt
d
ρ
ρ
=
or V
v
ρρ
=
.
Without solving this equation, we can predict facts about this system. As
, 0
ρ
.
To solve this differential equation, write
dt
V
v
d=
ρ
ρ
. Integrating both sides, we get
( ) ( )
t
V
v
t
V
v
C
Ct
V
v
AeeeetCt
V
v
t
+
===+=
ρρ
ln
.
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MAT244H1a.doc
Page 2 of 32
If we add
(
)
ρρ
=0 to V
v
ρρ
=
, then we have an IVP (initial value problem).
ISSUES ABOUT THE USE OF DIFFERENTIAL EQUATIONS
1) How to translate a real problem to a differential equation. Keep your eyes open!
2) Some patterns of nature are ill-defined. Use different points of view and different mental differential equation
models to reformulate them.
3) Differential equations have infinitely many solutions. Which one is yours? The initial value are extremely
important.
4) There may be no analytic solution found.
Is there a solution?
Is this solution unique?
If a numeric answer is required, i.e. the value of the solution at one particular point, then use numerical
approximations. It does not give any feelings for the pattern, nor does it give elbow room.
Use theoretical analysis if you need the behavior of the solution. This does not give any values.
To know the behavior locally/in a neighborhood, solve in series.
5) The data does not fit you solution. You need to repeat (as in a feedback/controlled system).
NOTATIONS WITH REGARD TO THE INPUT/OUTPUT SYSTEMS
Example
xyxyyx tansin2 =
+
can be written as
[
]
xyLtan=. Solve it, and the answer is the output.
[
]
yL is the black box system.
xtan is the input.
For theoretical purposes, mathematicians use these equivalents: x
y
y
x
x
x
x
y
+=
2
sintan is the same as
(
)
yyxfy
=
,, or
(
)
0,,,, =
yyyyxF.
LINEAR VS. NON-LINEAR DIFFERENTIAL EQUATIONS
x
yey
x
yx x
tan1
11
tan+
=+
+
is linear.
0
2=+
+
yyy is non-linear.
First Order Differential Equations
LINEAR EQUATIONS
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MAT244H1a.doc
Page 3 of 32
First order linear equations have the form
(
)
(
)
tgytpy=+
.
Derivation
Suppose I can find
(
)
t
µ
so that
(
)
(
)
(
)
ttpt
µµ
=.
Multiply both sides of
(
)
(
)
tgytpy=+
by
(
)
t
µ
:
(
)
(
)
(
)
(
)
(
)
tgtytptyt
µµµ
=+
which is
(
)
(
)
(
)
(
)
tgtytyt
µµµ
=
+
, i.e.
( )( ) ( ) ( )
tgtyt
µµ
=
.
Integrate both sides:
( ) ( ) ( )
Cdttgtyt+=
µµ
, so
( ) ( ) ( ) ( )
Cdttgt
t
ty+=
µ
µ
1.
But what is
(
)
t
µ
? Since
( ) ( )
(
)
( ) ( )
tp
t
t
tpt=
=
µ
µ
µµ
, therefore
( ) ( )
=dttpt
µ
ln. So
(
)
(
)
=dttp
et
µ
.
General Solution
To solve
(
)
(
)
tgytpy=+
,
1) Let
(
)
(
)
=dttp
et
µ
(no constant needed).
2) The solution is
( ) ( ) ( ) ( )
[
]
Cdttgt
t
ty+=
µ
µ
1.
Example
Solve 2
22 t
tetyy
=+
.
Here,
(
)
ttp2=,
( )
2
2t
tetg
=.
Let
( )
2
2t
tdt
eet==
µ
.
The solution is
( )
[
]
[
]
(
)
22
22
22
2
22
1
2
1
2
1tt
tt
tt
tCeetCt
e
Ctdt
e
Cdttee
e
ty+=+=+=+= .
Note that as
, 0
y.
Importance of Analysis
One needs to have an understand of the solution before (even after) solving it, with respect to:
1) Behavior of the solution as
.
2) The nature and behavior of the solution within a family (depends on y0).
Variation of Parameter
Recall that the family of solutions of a first order linear differential equation
(
)
(
)
tgytpy=+
,
(
)
(
)
=dttp
et
µ
,
( ) ( ) ( ) ( )
[
]
( ) ( ) ( ) ( )
t
C
dttgt
t
Cdttgt
t
ty
µ
µ
µ
µ
µ
+=+= 11 . Notice that the family of solutions is generated by
(
)
t
µ
. This leads to the technique of variation of parameter.
Recall a differential equation
[
]
(
)
xgyL=. If
(
)
0=xg, then we have a zero-input system, or a homogeneous
differential equation
[
]
0=yL which describes the solutions to a great extent.
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Description
MAT244H1a.doc Introduction INTRODUCTION TO D IFFERENTIAL E QUATIONS A differential equation is an equation involving some hypothetical function and its derivatives. Example y + 2y = x is an differential equation. As such, the differential equation is a description of some function (exists or not). A solution to a differential equation is a function that satisfies the differential equation. Example y = x +5x y = x+ y x 3 is a solution to . Some differential equations are famousimportant: y = y y = ex , . y = ay y = eax , . y+y= 0 y = cos x , . y+ ay = 0 y = cos ax , . P = mv Recall that a differential equation describes a phenomenon in terms of changes. For example, if , then dP = F Fdt = dmv or dt . Example A pool contains V liters of water which contains M kg of salt. Pure water enters the pool at a constant rate of v liters per minute, and after mixing, exits at the same rate. Write a differential equation that describes the density of salt in the pool at an arbitrary time t. M t) () (t = Let be the density at time t. Then V . (t) = (t) = t ) ( )V v t t ) To model change in , let 1 1 and 2 2 . Then 2 1 1 2 1 , so 2 1 v (t+ t) () v V 1 V (t) t2 1 V or t+ t t V . d = (t) v = v t 0 dt V V Now, as , or . Without solving this equation, we can predict facts about this system. As , 0 . d = v dt V To solve this differential equation, write . Integrating both sides, we get vt+C vt vt ln (t)= v t +C t = e V = e e V = Ae V V . Page 1 of 32 www.notesolution.com MAT244H1a.doc v (0)= = If we add to V , then we have an IVP (initial value problem). ISSUES A BOUT THE U SE OF D IFFERENTIAL EQUATIONS 1) How to translate a real problem to a differential equation. Keep your eyes open! 2) Some patterns of nature are ill-defined. Use different points of view and different mental differential equation models to reformulate them. 3) Differential equations have infinitely many solutions. Which one is yours? The initial value are extremely important. 4) There may be no analytic solution found. Is there a solution? Is this solution unique? If a numeric answer is required, i.e. the value of the solution at one particular point, then use numerical approximations. It does not give any feelings for the pattern, nor does it give elbow room. Use theoretical analysis if you need the behavior of the solution. This does not give any values. To know the behavior locallyin a neighborhood, solve in series. 5) The data does not fit you solution. You need to repeat (as in a feedbackcontrolled system). N OTATIONS W ITH R EGARD TO THE INPUT OUTPUT S YSTEMS Example xy+ 2y sin x y = tan xan be written aL[y]= tan x. Solve it, and the answer is the output. L[y] is the black box system. tan xis the input. For theoretical purposes, mathematicians use these equivayen=s:anx + sinx y 2y is the same as x x x y = f (x,y,y)or F (x,y,y, y,y)= 0 . L INEAR VS . N ON -LINEAR DIFFERENTIAL EQUATIONS tanx y +1 y +e y = 1 is linear. x 1+ tanx 2 y + y + y = 0 is non-linear. First Order Differential Equations L INEAR EQUATIONS Page 2 of 32 www.notesolution.com MAT244H1a.doc First order linear equations have the formy + p ()y = g t) . Derivation Suppose I can find t) so that t)p(t)= t). Multiply both sides of y + p t)y = g(t) by (t): (t)y + t)p t)y = t)g(t) which is t)y + t)y = t)g(t), i.e( t y ) t g ( ) ( ) 1 Integrate both sides:(t y = t g (tdt (C) , so yt )= (tg t(d)+C . t ) (t p t dt But what is t)? Since (t p t = t )= p t , therefore ln t )= p(t)dt. So t)= e . General Solution To solve y + p (t)y = g(t), p t dt 1) Let t)= e (no constant needed). 1 2) The solution is yt )= [ (tg t(d) + C]. (t ) Example t Solve y + t2 = t2 . t2 Here, p t) = 2t, g( )= 2e . 2tdt t2 Let t = e = e . 1 t2 t2 1 1 2 2 t2 t2 The solution is y t = t2 e 2te dt +C = [ t 2tdt +C = t2 t + C )= t e + Ce . e e e Note that as t , y 0 . Importance of Analysis One needs to have an understand of the solution before (even after) solving it, with respect to: 1) Behavior of the solution as t . 2) The nature and behavior of the solution within a family (depends on y 0. Variation of Parameter p t dt Recall that the family of solutions of a first order linear differential equation (t)y = g(t), (t)= e , 1 1 C y( )= ( ) ( )t+ C ]= ( ) ( )dt+ . Notice that the family of solutions is generated by ( ) ( ) ( ) t). This leads to the technique of variation of parameter. Recall a differential equationL[y]= g x ). If (x )= 0 , then we have a zero-input system, or a homogeneous differential equation [y]= 0 which describes the solutions to a great extent. Page 3 of 32 www.notesolution.com
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