Re ˇsi diferencialno ena ˇcb o z nastavk om y ( x ) = P

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Matematik a 4

3.vaja B.JurˇciˇcZlobec1 1UniverzavLjubljani, FakultetazaElektrotehniko 1000Ljubljana,Trˇzaˇska25,Slovenija MatematikaFE,Ljubljana,Slovenija15.april2013

Re ˇsi diferencialno ena ˇcb o z nastavk om y ( x ) = P

∞ i=0

a

i

x in zapi ˇsi prvih p et ˇclenov vrste razli ˇcnih o d ni ˇc. y

00

( x ) + x

2

y ( x ) = 0 , y (0) = 1 , in y

0

(0) = − 1 / 2 .

Iy(x)=P∞ i=0aixi iny00 (x)=P∞ i=2i(i−1)aixi−2 . IP ∞ i=2i(i−1)aixi−2 +P ∞ i=0aixi+2 =0, I(i+1)(i+2)ai+2+ai−2=0,i=2,3,...a2=a3=0, Iai+2=−ai−2 (i+1)(i+2),i=2,3,..., Iy(x)=1−x 2−x4 12+x5 40+x8 672....

Re ˇsi diferencialno ena ˇcb o z nastavk om y ( x ) = x

r

P

∞ i=0

a

i

x

i

. 4 xy

00

( x ) + 2 y

0

( x ) + y ( x ) = 0, y (0) = 1 in lim

x→0

y

0

( x ) = −

1 2

.

IP∞ i=0(4(i+r)(i+r−1)ai+2ai(i+r)+aix)xi+r−1 =0. IPrvipogoj(4r(r−1)+2r)a0=0 I(4(i+r)(i+r−1)+2(i+r))ai+ai−1=0,i=1,3,... Iai=−ai−1 2(i+r)(2i+2r−1),i=1,3,... I

ˇ Ce jea06=0,potemjer(2r−1)=0,r=0alir=

1 2. a−1ain0IZar=0,a=−=(−1),n2i(2i−1)(2i)! Izar=

1 2ai0,a=(−1),n(2i+1)! √√√22(1)(2)xxxxxxIa1−+−...+ax−+−....002!4!3!5! √ Iy(x)=cosx.

Uvedi novo o dvisno sp remenljivk o v diferencialno ena ˇcb o. ( x

2

− 1) y

00

( x ) + 4 x −

2 x

y

0

( x ) − 10 y ( x ) , y =

z x

.

Iy0 (x)=z0(x) x−z x2,y00 (x)=z00(x) x−2z0(x) x2+2z(x) x3. I(x2 −1) z00 x−2z0 x2+2z x3 +4x−2 x z0 x−z x2 −10

z x. 2000I(1−x)z(x)−2xz(x)+12z(x)=0.

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Besslova di ferencialna ena ˇcba

x2 y00 (x)+xy0 (x)+(x2 −ν2 )y(x)=0 IParameterνniceloˇstevilo,y(x)=AJν(x)+BJ−ν(x). IParameterν=njeceloˇstevilo,f(x)=AJn(x)+BNn(x). IParameterν=

q 1 22 ,y(x)=(Asin(x)+Bcos(x)) πx IRekurzivneformule. 2ν J+J=J(x)ν−1ν+1ν x IOdvodiBesslovihfunkcij. dJ(x)νν =−J(x)+J(x)νν−1 dxx

P oi ˇsˇci re ˇsitev diferencialne ena ˇcb e x

2

y

00

( x ) − xy

0

( x ) + (1 + x

2

) y ( x ) = 0, y (0) = 0, y

0

(0) = 1, z uvedb o y ( x ) = xz ( x ).

Iy0 (x)=z(x)+xz0 (x),y00 (x)=2z0 (x)+xz00 (x). Ix2 (2z0 (x)+xz00 (x))−x(z(x)+xz0 (x))+(1+x2 )xz(x)=0 I2xz0 (x)+x2 z00 (x)−z(x)−xz0 (x)+z(x)+x2 z(x)=0 Ix2 z00 (x)+xz0 (x)+x2 z(x)=0 Iy(x)=xz(x)=AxJ0(x)+BxN0(x),z(0)=0,z0 (0)=1. Iy(x)=xJ0(x).

P oi ˇsˇci re ˇsitev diferencialne ena ˇcb e x

2

y

00

( x ) + xy

0

( x ) + (4 x

4

− 1) y ( x ) = 0, lim

x→0

y ( x ) = 0, lim

x→0

y

0

( x ) = 1, z uvedb o ξ = x

2

.

I2xdx=dξ,dξ dx=2√ ξ,^{d dx}=

√ dξd d^{d d}=2ξ. ξdxξ √√dy0Iy(x)=2ξ=2ξy.ξdξ √√√√ 100I√y(x)=2ξ2ξy=2ξy+2ξy=→ξξξξξξ 22I2y+4ξy,4ξy(ξ)+4ξy(ξ)+(4ξ−1)y(ξ)=0.ξξξξξξ 122Iξy(ξ)+ξy(ξ)+(ξ−)y(ξ)=0ξξξ 4 sinξcosξ00I√√y(ξ)=AJ(ξ)+BJ(ξ)=A+B.1/2−1/2ξξ 12Iy(x)=sin(x). x

Legendrovi p olinomi

IDiferencialnaenaˇcba: (1−x2 )y00 (x)−2xy0 (x)=−n(n+1)y(x). IInterval:[−1,1].Uteˇz:ρ(x)=1. IRodriguesovaformula:Pn(x)=1 2nn!dn dxnx2 −1n . IOrtogonalnost:(Pm,Pn)=Z1 −1Pm(x)Pn(x)dx=2 2n+1δmn.

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P oi ˇsˇci re ˇsitev diferencialne ena ˇcb e y

00

( x ) − tan( x ) y

0

( x ) + 2 y ( x ) = 0, y ( x ) = 0, y

0

( x ) = 1, z uvedb o ξ = sin( x ).

Idξ=cos(x)dx,^{d dx}=

dξd d^{d d}=cos(x). ξdxξ dy0Iy(x)=cos(x). dξ dy00^{d d}Iy(x)=cos(x)cos(x) ξdξ 2dydydcos(x)002Iy(x)=cosx+cos(x).2dξdξdξ dcos(x)^{dx d}I=−sin(x)=−tan(x) dξξ 2dydy002Iy(x)=cos(x)−sin(x).2dξdξ 2000I(1−ξ)y(ξ)−2ξy(ξ)+2y(ξ)=0. Iy(ξ)=ξ→y(x)=sin(x).

P oi ˇsˇci p olinomsk o re ˇsitev diferencialne ena ˇcb e ( x

2

− 1) y

00

( x ) +

2 x

y

0

( x ) − (2 +

2 x2

) y ( x ) = 0, z uvedb o y ( x ) = xz ( x ).

Iy0 (x)=z(x)+xz0 (x),y00 (x)=2z0 (x)+xz00 (x). I(x2 −1)(2z0 (x)+xz00 (x))+2 x(z(x)+xz0 (x))− 2+2 x2 xz(x)=0. Ix(x2 −1)z00 (x)+(2x2 −2+2)z0 (x)+(2 x−2x−2 x)z(x). I(1−x2 )z00 (x)−2xz0 (x)+2z(x)=0,2=l(l+1),l=1o todz(x)=P1(x)=xiny(x)=x2 . Iy(x)=x2 .

Hermitovi p olinomi

IDiferencialnaenaˇcba:y00 (x)−xy0 (x)=−ny(x). IInterval:(−∞,∞).Uteˇz:ρ(x)=e−x2/2 . IRodriguesovaformula:Hn(x)=(−1)n ex2/2dn dxne−x2/2 . IOrtogonalnost: (Hm,Hn)=Z∞ −∞Hm(x)Hn(x)e−x2/2 dx=√ πn!δmn.

Laguerrovi p olinomi

IDiferencialnaenaˇcba:xy00 (x)+(1−x)y0 (x)=−ny(x). IInterval:[0,∞).Uteˇz:ρ(x)=e−x . IRodriguesovaformula:Ln(x)=1 n!exdn dxnxn e−x . IOrtogonalnost:(Lm,Ln)=Z∞ 0Lm(x)Ln(x)e−x dx=δmn.

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P olinomi ˇ Cebi

ˇseva

IDiferencialnaenaˇcba:(1−x2 )y00 (x)−xy0 (x)=−n2 y(x). IInterval:[−1,1].Uteˇz:ρ(x)=1√ 1−x2. ITrigonometriˇcnaformula:Tn(x)=cos(narccos(x)). IRekurzivnaformula: T0(x)=1,T1(x)=x,Tn+1=2xTn(x)−Tn−1(x). IOrtogonalnost:(m|n>0),(Tm,Tn)= Z1 −1Tm(x)Tn(x)dx √ 1−x2=π 2δmn,(T0,T0)=π. ILastnosti.Medvsemipolinomiistestopnjezistimvodilnim koeficientomimapolinom

ˇ Cebi

ˇsevanaintervalu[−1,1] najmanjˇseabsolutnevrednostiekstremov.

Dolo ˇci konstante α

ij

, tak o, da b o do funk cije:

f0(x)=α00,f1(x)=α10+α11xinf2(x)=α20+α21x+α22x2 ortonormiranenaintervalu[−1,1]. Iˆf0(x)=1,kˆf0(x)k2 =R 1 −1ˆf^{2 0}(x)dx=2. Iˆf1(x)=αˆf0(x)+x→ I ˆf1(x),ˆf0(x) =α ˆf0(x),ˆf0(x) + x,ˆf0(x) =0. Iα=−(x,ˆf0(x)) (ˆf0(x),ˆf0(x))=−R1 −1xdx R1 −1dx=0,ˆf1(x)=x,kˆf1(x)k2 =

2 3. 2ˆf(x),x()i2ˆˆˆIf(x)=αf+αf(x)+x,α=−.20011iˆˆf(x),f(x)()ii ^{1 3}22ˆˆIf(x)=−+x,kf(x)k=22

8 45.Normirani→ q 1I√f(x)=,f(x)=01 2

q 3 2x,f(x)=2

5 82 (3x−1).

Dolo ˇci konstante α , tak o, da b o do funk cije:

ij 2 f(x)=α,f(x)=α+αxinf(x)=α+αx+αx000110112202122 ortonormiranenaintervalu[0,2]. R22^{2 0}ˆˆÎf(x)=1,kf(x)k=f(x)dx=2.000 ˆÎf(x)=αf(x)+x→10 ˆˆˆˆÎf(x),f(x)=αf(x),f(x)+x,f(x)=0.10000 R2ˆx,f(x)()xdx00ÎRα=−=−=−1,f(x)=−1+x,21ˆˆf(x),f(x)()dx000 2ˆkf(x)k=1

2 3. 2ˆf(x),x()i2ˆˆˆIf(x)=αf+αf(x)+x,α=−.20011iˆˆf(x),f(x)()ii ˆIf(x)=2 2 322ˆ−2x+x,kf(x)k=2 8 45.Normirani→ q 1I√f(x)=,f(x)=01 2

q 3 2(x−1),f(x)=2 5 82 (3(x−1)−1).

Dolo ˇci konstante α

ij

, tak o, da b o do funk cije:

f0(x)=α00,f1(x)=α10+α11xinf2(x)=α20+α21x+α22x2 ortonormiranenaintervalu[0,2],zuteˇzjoρ(x)=x. Iˆf0(x)=1,kˆf0(x)k2 =R2 0xˆf^{2 0}(x)dx=2. Iˆf1(x)=αˆf0(x)+x→ I ˆf1(x),ˆf0(x) =α ˆf0(x),ˆf0(x) + x,ˆf0(x) =0. Iα=−(x,ˆf0(x)) (ˆf0(x),ˆf0(x))=−R2 0x2dx R2 0xdx=−^{4 3},ˆf1(x)=−^{4 3}+x, kˆf1(x)k2 =4 9. Iˆf2(x)=α0ˆf0+α1ˆf1(x)+x2 ,αi=−(ˆfi(x),x2 ) (ˆfi(x),ˆfi(x)). Iˆf2(x)=

6 5−

12 5

x+x2 ,kˆf2(x)k2 =

8 45.Normirani→ q 13xI√f(x)=,f(x)=−2+,f(x)=01222 5 82 (6−12x+5x).

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Preizkusi ortogonalnost La ge ndr ejevih p olinomov P

2

( x ) in P

3

( x )

IP2(x)=1 23x2 −1 , IP3(x)=1 25x3 −3x . I(P2(x),P3(x)=R1 −1P2(x)P3(x)dx=→ IR1 −115x5 4−7x3 2+3x 4=0. I(P2(x),P3(x))=0.

P oi ˇsˇci no rmo Legendrovih p olinomov P

1

( x ) in P

3

( x

IP1(x)=x→ IkP1(x)k2 =R1 −1|P1(x)|2 dx=

2 3. IP(x)=3

1 2

5x3 −3x → IkP3(x)k2 =

R211 43 5x−3xdx= −1

2 7. rr 22 IkP(x)k=,kP(x)k=.13 37

Dolo ˇci ko eficienta a in a v razvoju funk cije f ( x )

13 poLegendrovihpolinomihnaintervalu[−1,1].Funkcija   −1,−1≤x<0 f(x)=1,0<x≤1   0,x=0

. If(x)=∞X n=0anPn(x),zax∈[−1,1]. Ia1=(f(x),P1(x))/kP1(x)k2 ina3=(f(x),P3(x))/kP3(x)k2 . Ia1=3 2Z 1 −1xf(x)dx=3 2Z 1 0xdx=3 4. Ia3=7 2Z1 −11 25x3 −3x f(x)=7 8Z1 0x3 −3xdx=−7 16 Ia1=3 4,a3=−7 16

Ap roksimacija f ( x ) z Legendrovimi p olynomi

-1.0-0.50.5 -1.5

-1.0

-0.5

0.5

1.0

1.5

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Ena izmed re ˇsitev difere ncia lne ena ˇcb e je p ol inom. Dolo ˇci njegovo stopnjo.

y00 (x)−xy0 (x)+3y=0 IHermitovadiferencialnaenaˇcbazan=3. Iy(x)=H3(x). IStopnjaje3.

Ena izmed re ˇsitev difere ncia lne ena ˇcb e je p ol inom. Dolo ˇci njegovo stopnjo.

y00 (x)−2xy0 (x)+6y=0 Ian+2=2(n−3)an (n+2)(n+1). IStopnjaje3.

P oi ˇsˇci Laplaceovo transfo rmacijo Besslove funk cije J

0

( x ).

IFunkcijay=J0(x)jereˇsitevdiferencialneenaˇcbe xy00 (x)+y0 (x)+xy(x)=0,y(0)=1iny0 (0)=0. ILaplaceovatransformacijadiferencialneenaˇcbejeenaˇcba −d dss2 Y(s)−s +sX(s)−1−Y0 (s)=0, Y0 (s)(s2 +1)+sY(s)=0. IRezultatjediferencialnaenaˇcbaprvegaredazloˇcljivimi spremenljivkami. dY Y=−sds 1+s2→lnY(s)=−1 2ln(1+s2 )→. IY(s)=1 √ 1+s2.