NMM 2270 Final Review
Existence and Uniqueness Theorem
If f x y and Ey are continuous there is someinterval I where thesolution y x is unique
f x y XY
1 5
continuous on
continuous on
xy
Example
XE 0,0
8
yeto.nl
i The DEhas a
uniquesolution on x yt 0,0
SeparableDEs
11 9 x hly
114 fgexdx
Linear 1st Order DEs
Pixly x Qix
StandardForm
d
Integration factor µ x
Nointegrationconstant
Solve by multiplying Mex to standardform Mex
Solution y x
Example
1
3 4 6
µx
MIX p
32
2
1 MixPexy x MIXQ X
Fix Mixidexidx
P 1 3 2 Q 162
yx
fe
6 2dx
yex é 2e C
y X 2 CE
solvewithu sub
Exact DEs
0Mex y
Mix.lydy Nex.ydx
fxf x.g
Nix y
Jyf xiglf
x gl
C
Check for exactness 8ᵗyʰ
2xydx x 1 dy O
Example
M 2xy
2x
y
8 24
f x.yj
fzxydxpfyfxytg.ly
21yg
2
f x yl x y
Exact
f xy
x y ytCz
x2 1
y
x2y
C C
C
Cz
ytC C2letC
y
x
What if Ty 2
Ey N x 1
2 M 2xy
NX1
y
x
x2
g'ly x2 1
gy 1
gly fady
g.ly y
yextx.tt
can try to make it exact
To make a non exact DE into an exact DE Multiply alltermsby a functionmlx.gl
Case 1
If 8 3
Nexis
is only a functionof x
ncx.es y Ex Nex.yidx
Case 2
If 8 8
next
Mixis is only a functionof y
Ex mex.gsdy
2x ty dy x y x tx 0
Example
Mix.ly 2x2tyN x.y x2y X
85 1
8 N
Try
8 2xy 1
peexg
et
é2nlx
2xtyldy.tl x2y x dx 0
ldyt xltx o
e
x
211g
x.g 2
y x
c f
yx2dxJ 2x yxtgly y.x
f X.ly 2x yX'tgly
3 X
Exact
Onlyfunctionof x
2 yx
X
flx.ly C 2x yxtEy2tC2c c
C 2x
y
xty.ly y x
glyl
fyd gly
y2 C2
Substitutions Change of Variables
1 F x.gl
Gexis
11 y
8
let ulx.is
y
uxyProductrul 8t8 14
Example
48
e
xcutxuy ux.eu
fe
uxtux uy.ie
191144
He du dx
14
if f tx.ly Ef x.y then f is a homogeneous function of order α
Homogeneous Functions
omo9eneous
11 f x.gs
if
then
11 F E F
f x y ishomogeneous the substitutions y ux or
f x.y in
It
ÉÉÉÉ
e
4 211 4 4
f xy
e
f tx.ly Ete
fltx.ly E
homogenerous
e414
8
fé du
of orderzero
e
In xltc
et In1xltC
x
P x y Qcxyn
but neR sowhat if n 0,1
Use v y
Bernoulli Equations
Example
8
y
1 92
ᵗ
xy2
Tx
1
nz.gg
Mex es
mexte.tn
v
X
Mix
when no DE islinear
whenn 1 DE is separable
substitution
I
yu
vix
s
52
MxQ x dx
xf textdx
I
11 921
yx V
yx
X
x
y x xEx
x vy shouldbemade
LogisticEquation and Populations
Notreallythatimportant read2.8
IF P a GP
P PLE
pct aP t GPCR
P E Gpo ad Poleat
higher Order DEs
An x y
x
Az x y x
x
an x y
Y X Gy x Caya x
or non homogeneous
y
x
do x y x
FX
If F x 0 DE ishomogeneous
If F X 0 DE is non homogeneous
a x and F x are known
Forhomogeneous
a x
where yi x are all linearlyindependent
nyn x
solve for Associated Homogeneous solution
if F x an x y
x
an x y
findHomogeneoussolution Yc for 0 An X y
x
x
an x y
a x y x do x y x
x
a x y x do x y x
ThenFind ParticularSolution yp for F X
x
y x Yc typ
Reduceable DES
Homogeneous DE
DES that can be reduced downto 1st order from 2ndorder
Fyyy x 0
Case 1 No dependance on y x
let P x y
x
Px y
Fly y x
0
y x F PP x
0
Example
P y Ply
xy ty 4x
P P Yx
1storderlinear
P4
x
y x xtc.tn xl
P x 2x
C Case2
letPlyl
Nodependanceonx Fly yiy
y
y
y
y PFy
Fly P.PE
Example
0
yy
1
y3p
1
let y P y P
PIP
SPJP
P
52 4
P Fy2
11 752
1
IEjEitC2
Ay ByltCy o
HigherOrder DEs
Guess y x
er
Ark Brett
0
Ar BrtC O
Case 1 2 Real DistinctRoots
Example
y 2y 159
r
2ndorderlinear
homogeneousODE
2r 15 0
0
Quadratic
y x Cer Czer
11 5 82 3
yex Cés Cze
yes
y é
r 5 r 3 0
Case 2 1 Real RepeatedRoot y x Cer Caxer
Example
re
912 12 4 0
Case 3 2 Complex DistinctRoots
Example y 2y loy O
y x Ge
yz xe
y e
I
axe
yex Ge cos bx Creatsin bx
Quadratic formula
82 2 10 0
rn 1 3i
No factorable
écos3x
y
yz x é sin 3x
y x C é cos 3x Czésin 3x
nthOrder Linear HomogeneousDES same idea as before but with more terms
x am
Any
y
x
azy x a y x doy
azr
a r go 0
Anrntan.ir t
y 3y Yy O
13 32 4 0
Example
r 1 r 212 0
r 1 V2 5
y
e
x
0
y x Get Czé C3xé
yz e
y xé
2
Non Homogeneous Higher Order ODEs
y x Ye YP
Solution to
particularsolution
ssociated Homogeneous
DifferentialEquation
Method of Undetermined Coefficients
Ly x f x
if x in f x
yp Anx An ix Anex
Azx A x Ao
if er in f x
yp Bet
if cos Kx or sinKx in f x
yp Dcos Kx Esin Kx
If
any of theaboveyp terms are in the AHDE multiply the matchingterm byXswhere s is th
smallestpositiveintegersuchthatthereare nomorematchingterms
y Yy Yy 5
Example
2
6 4 22 3es
f x 5 6 4 28 38
r
4 Yr 0
rer 212 0
458 1
1,123
92224
1 t.EE
Bx Des
yp Azx A x Aox Bex
B x Box e
90 1
the
Ye it
yz xe
4
Plug backinto original DE as
y x ye x Yp X
y N C.tl
C3xe
Desx
y to solveforAs Bsand I
Fx3 1x2 3x fxe 3x3e
Variation of Parameters
x
x
y x Pxy Qxyx f
Assume AHDE is solved yo x C y x zyz x
Assume Particular Solution looks like yp x
2nd order DE
f
yp X
Example
AHDE
w
uicx
Y2 ᵗ
dxy.at
1f5jxyax
y Yy Yy extile
82 4 4 0
zxÉÉxl
W 2x
e
Emox
4 x
if
_x
Y
Y
we
W e4x
r 212 0
axe
xlyz x
YfIw
uiex
Yax C e
u.ee y N ua
1 exy
f ej
Determinant
2K
dx
42 4
Eye
fx idx
x X
Yp x 4,41 4242
Y x Yc x Yp x
1 2 3 3 8 x tx xe
13 32
4 2 2 321 27
y x C e Exe 2 28 61
3224
61 32 12224
Cauchy Euler Equations
Anx y
x an
y
Assume y x Xm
Example
ax y x
y
fx
Xm 9k mcm1 m 2
kᵗhterm ofDE akx
AKM M 1 m 2
m k 1 xm
M K 1 Xm
bxy x cy x 0
yx X
mxm I
x
Y
azx y x a xy x doy x
x
axfmemiixmxtbxmxm.ttcxm 0
2
x mem 1 Xm
xMamcm 1 6m c
Case 1 Real and distinct
Case 2 Realand repeated
O
am m 1 bmtc O
im
yoxtcixm.tc.am
ÉYixixm
M α iB
Case 3 Complex roots me α if
yc x Cxm canxlxm
Y XcosBinx1
y xsin Binx1
Yo x Gx cos Bin 1
2x'sin Binx1
Harmonic Motion
K SpringConstant
In
W
β DampingCoefficient
2X
m mass
SimpleHarmonic Motion X E W X E D
Damped Harmonic Motion
x t Cicos wt Casin wt Acos wt α
X E 2Xx t w x t 0
12 2Xr who
r
X R
W
Case 1 Overdamped R w 70 real distinctroots
wTt
xltke.tt GeFwt CzEx
Case 2
t
Underdamped
e
c cost
X w L0 complexroots
t Casin ret
Case 3 Critical Damping X2 w
t E
ᵗ C
Csé cos Ext
α
0 real repeated roots
Cat
Forced Oscilator X t w X t F t
F t Focos rt
Xc t C cos wt Casin wt
Xp t Acos rt Bsin rt
Case 1 wer
E L costat a
wÉ
court
wt
x t Ccos wt Casinwt Eutsin
Case 2 w r
Forced oscillator withdamping
EÉWx
Ft
all termsfrom natural frequency have an et termwhichdecays
to zero onlyleftwithtermsfromforcingfunction
Laplace Transforms
L ft
f t has a Laplace Transform if
a f t is piecewisesmooth for 20
ésᵗf e at
af t 6g t
aL FCE
6 f t is exponential order
6L g t
Notransform it f t blowsup fasterthan an
eat
exponentialfunction
Noinfinitediscontinuities
1
tay.gg 1qf
L eat sta
costwit
sÉw
sinwt
star
specialcase
f t
Mk a
SHEISE
if tn f Ent f tri
ft
if
eatf t
Fs a
LET
L tcos Kt ÉÉÉ
L tsin Kt
f is
shifts gntffol gnzg.jo
F
sFcs
fo
ésᵗʰjttn
piecewisesmoothwithjumpdiscontinuities
finill
511FCS
0
if thislimit isDNE there is no inverseLaplace transforms
F s GCS but f E doesnotalways
glt therecanbepoint discontinuities
Example x lox fCt Fct
UCL
Koko
5
I
f t COSME COSCUE u E e
L UCH
Esf
ult a
X
fct a ult a é F s
16L x
s Xcs sxcolx.co 16Xes
5 52 16 1 521
LEECH
F'est
F'CSI
EFC
1 F
EFCE
E0 E6et's
SITE'S
Xis Ffp isÉE
s
tsin 41
sin 4
t
sincut ftsin 4t
t a sin Yt u t t
45inch ᵗsinUt
O t TL
Ysincut Isincut
Tilt
f t g t e de FCS G s
felde
Foldo
Fs G S
f t
510180
Let f E beperiodic with period P
FCH FCS
t t sin 4 t t ult th
xt
f ᵗf t g t E JT
f g t
s
É é 452
XH L
a
f g t
coscut u E N
COSCH
Fsfésᵗfe at
selt
e.tl
imoSelt to
S t to
S t.to dt
LLSCt
toD e
st tDLTS
1
Linear Systems We can solve systemsofDEs with Laplace Transform
Example
1
y loy 42 0
y.co yzlo 0yilol
y 4y Yy
2
y 10g 4yz 0
54,15 sy.pl yYojt10Ycs 442151 0
Y 4yz Yy
5421s
52 104,15
47215
4415
52 4 4215
D
1
1
52 4
Dif
4114
a
DFI 5 4
D 54 1452 40 16
D 54 1452 24
D 5212 52 2
ji4Y2cs 44,1s 1
s
4
1
is a
D 52 10 52 4
0
52 414215 44 5
52 107,15 442157 1
4
4 I
De
i
D2 1152 10 21 4
Disty y
12 52 10 4
Dis
D2
isiziist
42
1
lyil01
52 6
isiiziis.la
I
4 s
y t
82
52 2
42151 52112
sin Bt
Isin Et
Yz5
52 2
Eosin2BE Essin Et
Fourier Series
Twofunctions f x g x
yet
A
Ancos
fixidx
are orthogonal on interval
ab
if
f x g x dx 0
Fit Basin 7ft
An of f x cos ⁿtx dx Bn
f x sin f x dx
0
