modul 1 pd linier orde satu
TRANSCRIPT
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MODUL 1
PERSAMAAN
DIFERENSIAL ORDE
SATU
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RUMUS-RUMUS DASAR INTEGRAL
caxa
axdxax
caxa
axdx
caxa
axdxax
caxa
ax
caxa
axdx
caxa
axdx
cx
ncxndxx
nn
csc1
cotcsc).7(
cot1
csc).6(
sec1
tansec).5(
tan1
sec).4(
sin1
cos).3(
cos1
sin).2(
||ln
1,1
1).1(
2
2
1
-1n ,
caxdxax
x
cxxxdx
cx
cxxdx
cxxxdx
cx
cxxdx
22
22).12(
|cotcsc|lncsc)11(
|sin|ln
|csc|lncot)10(
|tansec|lnsec)9(
|cos|ln
|sec|lntan)8(
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c|bax|lna
b
a
xdx
bax
x )14(
c|bax|lna
1
bax
dx )13(
2
cax
axln
a2
1dx
ax
1)16(
c|ax|ln2
1dx
ax
x )15(
22
2222
c|axx|ln2
1
ax
dx )17( 22
22
dxexa
n
a
exdxex)20(
cedue)19(
cea
1dxe)18(
ax1naxn
axn
uu
axax
ca
xcos
ca
xsindx
xa
1 )21(
1
1
22
ca
xcot
a
1
ca
xtan
a
1dx
xa
1 )22(
1
122
ca
xcsc
a
1
ca
xsec
a
1dx
axx
1 )23(
1
1
22
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Rumus-rumus Reduksi
dx xsinn
1nxcosxsin
n
1dx xsin).1 2n1nn
dx xcosn
1nxsinxcos
n
1dx xcos).2 2n1nn
dx xtanxtan1n
1dx xtan).3 2n1nn
dx xcotxcot1n
1dx xcot).4 2n1nn
dx xsec1n
2ntanxsec
1n
1dx xsec).5 2n2nn
dx xcsc1n
2nxcotxcsc
1n
1dx xcsc).6 2n2nn
dxbxcosxb
nbxcos
b
xdxbxsinx).7 1n
nn
dxbxsinxb
nbxsin
b
xdxbxcosx).8 1n
nn
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dx exa
nex
a
1dx ex).9 ax1naxnaxn
dx axaln
nax
aln
1dx ax).10 x1nxnxn
c)bxcosbbxsina(ba
edx bxsine).11
22
axax
c)bxcosabxsinb(ba
edxbxcose).12
22
axax
c)1n(
xxln
1n
xdx xlnx).13
2
1n1nn
cdxxlnnxlnxdxxln).14 1nnn
cdxxlnx1m
n
1m
xlnxdxxlnx).15 1nm
n1mnm
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PENGERTIAN PERSAMAAN DIFERENSIAL
Persamaan diferensial adalah suatu persamaan yang melibatkan satu atau
lebih turunan fungsi yang belum diketahui, dan atau persamaan itu mungkin
juga melibatkan fungsi itu sendiri dan konstanta.
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SOLUSI PERSAMAAN DIFERENSIAL
Solusi persamaan diferensial adalah menentukan suatu fungsi dimana
turunnya, dan disubsitutiskan memenuhi persamaan diferensial yang
diberikan.
Contoh :
Diberikan persamaan diferensial,
dy = (4x + 6 cos 2x)dx
Dengan cara mengintegralkan
diperoleh solusi PD yaitu :
cxx
dxxxy
2sin32
)2cos64(
2
Contoh :
Apakah, y = e2x, solusi persamaan
diferensial,
y” – 4y’ + 4y = 0
Dengan cara mensubstitusikan,
y=e2x, y’ = 2e2x, dan y’’ = 4e2x pada
persamaan dihasilkan,
4e2x - 4(2e2x) + 4e2x = 0
0=0
Jadi 4e2x e2x 4adalah solusi PD.Jadi, y= 2x2 + 3 sin 2x + c adalah
solusi PD
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PD Variabel Terpisah
Bentuk Umum PD Variabel Terpisah,
cdyyg
ygdx
xf
xf
dyyg
ygdx
xf
xf
ygxf
dyygxfdxygxf
)(
)(
)(
)(
0)(
)(
)(
)(
)()(
0)()()()(
1
2
2
1
1
2
2
1
12
2211
PD Solusi
Contoh :Carilah penyelesaian umum PD,
(4x + 6xy2)dx + 3(y + x2y)dy = 0
Tulislah PD menjadi,
2x(2 + 3y2)dx + 3y(1 + x2)dy = 0
---------------------------(1 + x2)(2 + 3y2)
Diperoleh PD,
032
3
1
2
22
dy
y
ydx
x
x
Solusi PD adalah,
cyx
cyx
cdyy
ydx
x
x
)32()1(
ln)32ln(2
1)1ln(
32
3
1
2
222
22
22
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Soal Latihan PD Variabel Terpisah
1. x(1 + y)dx + y(1 + x) dy = 0
2. xydx + (x2 – 1)ln y dy = 0
3. (1 + y2)sin x dx + 2y (1 – cos x)dy= 0
4. (1 + y) (1 + sin x)dx + y cos x dy = 0
5. xy dx + (x – 1)(1 + ln y)dy = 0
6. 2(1 + ey)dx + x(1 + x)dy = 0
7. 2xy(1 – y)dx + (x2 – 4)dy = 0
8. (y2 − 4) dx + x(x – 2)dy = 0
9. y(1 + x2)dx + 2x(1 + ln y)dy = 0
10.ex(1 + ey)dx + (1 + ex )e−y dy = 0
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PD HOMOGEN
Fungsi f(x,y) dikatakan fungsi homogen berderajad n, jika terdapat α,
sedemikian sehingga
f(αx, αy) = αn f(x,y)
Bentuk umum PD :
g(x,y)dx + h(x,y)dy = 0
---------------------------------------------------------------------------------------------------------
Kasus 1. Substitusi, y=ux, dy=udx+xdu
PD menjadi.
[g(1,u)+uh(1,u)]dx + xh(1,u)du=0
cduuuhug
uhdx
x
duuuhug
uhdx
x
),1(),1(
),1(1
0),1(),1(
),1(1
Solusi
Kasus 2. Substitusi, x=vy, dx=vdy + ydv
PD menjadi.
yg(v,1)dv + [vg(v,1)+h(v,1)]dy=0
cdyy
dvvhvvg
vg
dyy
dvvhvvg
vg
1
)1,()1,(
)1,(
01
)1,()1,(
)1,(
Solusi
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Contoh
Carilah penyelesaian umum PD,.
(4x2 – 3y2)dx + 4xy dy = 0Jawab
Substitusikan, y = ux,dy=udx+xdu
ke persamaan , maka dihasilkan :
x2(4–3u2)dx + x24u(udx + x du) = 0
(4–3u2)dx + 4u(udx + x du) = 0
(4–3u2+4u2) dx + 4xu du) = 0
3222
22
2
2
2
)4(
,ln)4(ln
ln)4ln(2ln
ln4
41
04
41
cxyx
x
yucux
cux
cduu
udx
x
duu
udx
x
Carilah penyelesaian umum, PD,
x2ydx – (x3 + y3) dy = 0
Jawab
Substitusi, x=vy, dx=vdy+ydv, ke PD
diperoleh,
v2y3 (vdy +ydv) – y3(v3 + 1)dy = 0
v2(vdy +ydv) – (v3 + 1)dy = 0
v2y dv – dy = 0
3)/(
3
3
2
2
3
3,lnln
lnln3
1
ln1
01
cye
y
xvcye
cyv
cdyy
dvv
dyy
dvv
yx
v
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Reduksi Persamaan Homogen
Kasus khusus PD berbentuk,
(ax+by+d)dx + (px+qy+r)dy=0
---------------------------------------------------------------------------------------------------------
Kasus1, d=0,r=0Jika d=r=0, PD menjadi
(ax+by)dx+(px+qy)dy=0
Substitusikan, y=ux,dy=udx+xdu
diperoleh,
x(a+bu)dx+x(p+qu)(udx+xdu) = 0
atau,
[a+(b+p)u+qu2]dx+x(p+qu) du = 0
cduaupbqu
pqudx
x
duaupbqu
pqudx
x
)(
1
0)(
1
2
2
Solusi,
Contoh :
Carilah penyelesaian umum PD,
(x + 4y)dx + (4x + 2y)dy = 0
Jawab
Substitusi, y=ux,dy=udx+xdu. PD
menjadi,
x(1+4u)dx + x(4+2u)[udx+xdu] = 0
atau,
(1 + 8u + 2u2)dx + x (4 + 2u) du = 0
cduuu
udx
x
duuu
udx
x
182
421
0182
421
2
2
Solusi,
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Kasus 2 aq – bp = 0Bila, aq – bp =0 maka berlaku :
px + qy = k(ax + by)
konstanta tak nol. Substitusikanlah,
z = ax + by , dz = adx + bdy
diperoleh PD,
b
adxdzdy
cdzarbdzakb
rkzdx
dzarbdzakb
rkzdx
dzrkzdxzakbarbd
b
adxdzrkzdxdz
)()(
0)()(
0)(])()[(
0)()(
Solusi,
Contoh
Carilah penyelesaian umum PD,
(2x+5y + 2)dx+(4x+10y + 3)dy=0
Jawab
aq – bp = (2)(10) – (5)(4) = 0.
Subsitusi, 4x + 10y = 2(2x + 5y),
dan z=2x+5y, dz=2dx+5dy. Maka
diperoeh PD
czzx
cdzz
zdx
dzz
zdx
dzzdxz
dxdzzdxz
)4ln(52
4
32
04
32
0)32()4(
05
2)32()2(
Solusi,
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Kasus Ketiga, aq – bp ≠ 0
(ax+by+d)dx + (px+qy+r)dy=0
Substitusi pertama,u=ax+by+d, du=adx+bdy
v=px+qy+r, dv=pdx+qdy
atau,
diperoleh
dv
du
dy
dx
qp
ba
0)()(
0
dvavbudupvqu
bpaq
advpduv
bpaq
bdvqduu
bpaq
advpdudy
bpaq
bdvqdudx
PD diperoleh Sehingga
Substitusi kedua,
v = uz dan dv = udz + zdu
kedalam persamaan homogen,
sehingga dihasilkan :
u(q–pz)du + u(az – b)(udz+zdu) = 0
[az2 – (b+p)z + q]du + u(az–b)dz = 0
cduqzpbaz
bazdu
u
duqzpbaz
bazdu
u
)(
1
0)(
1
2
2
Solusi,
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Contoh
Carilah penyelesaian umum PD,
(2x + 4y + 2)dx + (4x + 3y + 3)dy = 0
Jawab
Substitusi pertama
u=2x + 4y + 2,
v=4x + 3y + 3,
10
24
10
43
dvdudy
dvdudx
dv
du
dy
dx
34
42
Diperoleh PD,
(3u – 4v)du + (–4u + 2v) dv = 0
010
24
10
43
dvduv
dvduu
Substitusi kedua,
v=uz, dv=udz+zdu
diperoleh hasil,
(3u–4uz)du + (–4u+2uz)(udz+zdu)= 0
u(3–4z)du + u(–4+2z)(udz+zdu) = 0
(3 – 4z – 4z + 2z2)du +(2z – 4)d
czzu
czzu
cdzzz
zdu
u
dzzz
zdu
u
)382(
ln)382ln(2
1ln
382
421
0382
421
22
2
2
2
Solusi,
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Soal-soal Latihan PD Homogen
1. (x2 + y2)dx – xydy = 0
2. x2y dx + (x3 + y3)dy = 0
3. y dx – (x−yex/y)dy = 0
4. y(1 + ey/x) dx + (xey/x+ y) dy = 0
5. x2(x+3y)dx + (x3+ y3)dy = 0
6. y(y + xex/y)dx – x2ex/y dy = 0
7. (3x2y + y3)dx + (x3+ 3xy2)dy = 0
8. (2x – 3y)dx + (3x – 8y)dy = 0
9. (2x – 2y + 3)dx + (2x – 8y+4)dy = 0
10. (2x – y)dx + (x – 6y + 2)dy = 0
11. (2x + 5y + 2)dx + (5x + 3y – 2)dy = 0
12. (x – 2y + 3)dx + (2x – 9y – 4)dy = 0
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PD Eksak dan Non Eksak
Persamaan diferensial linier orde satu yang berbentuk,
M(x,y)dx + N(x,y)dy = 0
dikatakan sebagai persamaan diferensial eksak jika hanya jika
--------------------------------------------------------------------------------------------------------x
N
y
M
Solusi, F(x,y)=c dimana
dydxyxMy
yxNyg
yxNygdxyxMy
ygdxyxMyxF
),(),()(
),()(),(
)(),(),(
dimana
Solusi, F(x,y)=c dimana
dxdyyxNx
yxMxf
yxMxfdyyxNx
xfdyyxNyxF
),(),()(
),()(),(
)(),(),(
dimana
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Contoh : Carilah penyelesaian PD,
(1 + yexy) dx +(xexy + 2y) dy = 0
Jawab
PD Eksak, karena :
Solusi, F(x,y)=C, dimana
Contoh : Carilah penyelesaian PD,
Jawab
PD Eksak, karena :
Solusi, F(x,y)=c dimana :
01
sin)]1(ln(cos1[
dy
y
xydxyx
xyxyxy
xyxyxy
exyyexex
N
exyxeyey
M
)1()(
)1()(
cyex
cyydyyg
yxeygexy
ygex
ygdxxeyxF
xy
xyxy
xy
xy
2
2
,
2)(
2)]([
)(
)()1(),(
Solusi
y
x
x
N
y
x
y
M
1
cos
1
cos
)1()1(,
)1ln(1
)(
1
sin)]()1ln(sin[
)()1ln(sin(
)()1ln(cos1(),(
sin ycye
yydyy
yyg
y
xyygyxx
y
ygyxx
ygdxyxyxF
xyx
Solusi
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PD Non Eksak dan Faktor Integrasi
Persamaan diferensial linier orde satu yang berbentuk,
M(x,y)dx + N(x,y)dy = 0
dikatakan sebagai persamaan diferensial non eksak jika hanya jika
PD Non eksak diubah menjadi PD eksak dengan mengalikan faktor integrasi
u, sehingga PD berbentuk.
uM(x,y)dx+uN(x,y)dy = 0
--------------------------------------------------------------------------------------------------------
0
x
N
y
M
x
N
y
M atau
Kasus Pertama, u = u(x)Faktor integrasi u diberikan oleh,
N
x
N
y
M
xp
eudxxp
)(
)(
Kasus Kedua, u = u(y)Faktor integrasi u diberikan oleh,
M
x
N
y
M
yq
eudyyq
)(
)(
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Contoh :Carilah penyelesaian umum PD,
(4x3 + x2 – y2)dx + 2xy dy = 0
Jawab
PD Eksak or Non eksak
xy
y
N
x
N
y
M
pyx
N
y
M
yx
Ny
y
M
xyNyxxM
2
4,4
2,2
2,4 223
Faktor Integrasi u
2
1ln
ln2
2
12
xe
eeu
x
xdx
x
PD menjadi,
02
14
0)2(1
)4(1
2
2
2223
2
dyx
ydx
x
yx
xydyx
dxyxxx
PD Eksak,
Solusi PD Eksak, F(x,y) = c, dimana :
cygygx
yxx
ygdxx
yxyxF
)()(2
)(14),(
22
2
2
Solusi.
2x3 + x2 + y2 = cx
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Contoh :Carilah penyelesaian umum PD,
Jawab
PD Eksak or Non eksak
0ln32
34)(
22
dyy
xyedxxyey xx
yxyey
xye
M
x
N
y
M
q
xyex
N
y
M
xyex
Nxye
y
M
yx
yeNxyeyM
x
x
x
xx
xx
2
)(
)(2
)(2
342
ln32
34,
22
,
Faktor Integrasi u
2ln
ln2
2
2ye
eeu
y
ydy
y
PD Non Eksak
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PD menjadi,
0ln32
34)(
0ln32
34)(
222
334
2222
dyyyyx
eydxxyey
dyyx
yeydxxyeyy
xx
xx
PD
Eksak
Solusi PD Eksak, F(x,y)=c dimana :
332
222
3324
324
34
3
1lnln3)(
ln32
34)(
2
1
)(
)(2
1
)()(),(
yyyydyyyg
yyyx
eyygyxeyy
yg
ygyxey
ygdxxyeyyxF
xx
x
x
: dari diperoleh
cyyyyxey x 33324
3
1ln
2
1
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Soal Latihan PD Eksak Non Eksak
1. (x3 + y2)dx + (2xy − y3)dy = 0
2. (x + y sin 2x)dx + (sin2 x + 3y2)dy = 0
3. [2x + y cos(xy)] dx + [x cos(xy) – 2y]dy = 0
4. (x + y)2 dx + (x2 + 2xy + yey)dy = 0
5. (xex + yexy)dx + (1 + xexy )dy = 0
6. (xex − ey)dx + ey(y − x)dy = 0
7. 3x2(y − 1)2dx + 2x3 (y − 1)dy = 0
0)1()33(.10
0)1ln(1
)1(.9
0lnln.8
3/2
22
2
dyxdxyey
dyxdxx
yx
dyyy
xydxx
x
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PD Linier Orde Satu
Persamaan diferensial biasa linier orde
satu adalah suatu persamaan yang
berbentuk,
y′ + P(x)y = Q(x)
Tulislah PD menjadi,
[P(x)y – Q(x)]dx + dy = 0
Persamaan diatas adalah non eksak
faktor integrasinya adalah,
dxxP
eu)(
Solusi PD adalah,
cdxexQyedxxPdxxP
)()(
)(
ContohCarilah penyelesaian umum PD,
xy′ + (1 – x)y = 4xex ln x
Jawab
Tulis PD menjadi,
cxxxyxe
dxxx
dxxexeyxe
xeeeeu
xxdxx
xdxxp
xeyx
xy
x
xxx
xxxxx
x
22
lnln
ln2
ln4
))(ln4(
ln1
)(
ln41
Solusi,
integrasi, Faktor
![Page 25: Modul 1 pd linier orde satu](https://reader034.vdocuments.net/reader034/viewer/2022042503/5597c81d1a28ab7a468b459c/html5/thumbnails/25.jpg)
PD Bernoulli
Bentuk umum PD Bernoulli,
y′ + P(x)y = Q(x)yn
Tulislah PD menjadi,
yny′ + P(x)y1–n = Q(x)
Substitusi, z = y1–n,dan z′=(1–n)y–n y′,
PD menjadi
z′ + (1 – n)P(x)z = (1 – n)Q(x)
PD adalah linier orde satu,
cdxexQnze
eu
dxxPndxxPn
dxxPn
Solusi,
integrasi, Faktor
)()1()()1(
)()1(
)()1(
ContohCarilah pernyelesaian umum PD,
xy′ + y = y3 x3 ln x
Jawab
Tulislah PD menjadi,
Substitusi, z = y–2, z′=–2y–3 y′, PD
menjadi,
xxyx
yy ln1 223
cdxx
xxx
z
xeeu
xxzx
z
xdx
x
22
2
2ln2
2
2
1ln2
1
ln22
PD, Solusi
integrasi, Faktor
![Page 26: Modul 1 pd linier orde satu](https://reader034.vdocuments.net/reader034/viewer/2022042503/5597c81d1a28ab7a468b459c/html5/thumbnails/26.jpg)
PD Bernoulli
Bentuk umum PD Bernoulli - lain
yn–1 y′ + P(x)yn = Q(x)
Substitusi, z = yn,dan z′=nyn – 1y′, PD
menjadi
z′ + n P(x)z = nQ(x)
PD adalah linier orde satu,
cdxexnQze
eu
dxxnPdxxnP
dxxnP
Solusi,
integrasi, Faktor
)()(
)(
)(
ContohCarilah penyelesaian PD,
Jawab,
Substitusi , z = y3,dan z′= 3y2y′,
PD menjadi
xxyx
xyy 2sin
13 32
cdxxex
ey
cdxex
xxex
z
ex
eeu
xxzx
xz
xx
xx
xxxdx
x
x
PD Solusi
integrasi, Faktor
2sin
12sin
1
1
2sin1
3
ln1
![Page 27: Modul 1 pd linier orde satu](https://reader034.vdocuments.net/reader034/viewer/2022042503/5597c81d1a28ab7a468b459c/html5/thumbnails/27.jpg)
Reduksi Orde PD
Bantuk Umum PD adalah,
y(n) + P(x)y(n–1) = Q(x)
Substitusi, z = y(n–1),dan z′= y(n), PD
menjadi
z′ + P(x)z = Q(x)
PD adalah linier orde satu,
cdxexQey
cdxexQze
eu
dxxPdxxPn
dxxPdxxP
dxxP
Solusi,
integrasi, Faktor
)()()1(
)()(
)(
)(
)(
Contoh
carilah penyelesaian khusus dari,
y′′′ – y′′ = xex
y(0) = 1, y′(0) = 2 dan y′′(0) = 4
Jawab
Substitusi, z = y′′,dan z′= y′′′, PD
menjadi,
z′ – z = xex
Faktor integrasi,
dxcecexe
y
dxceexey
cexeey
cdxexeze
eeu
xxx
xxx
xxx
xxx
xdx
1
22
22
2
4
3
2
])[(
])([
)(
Solusi,
![Page 28: Modul 1 pd linier orde satu](https://reader034.vdocuments.net/reader034/viewer/2022042503/5597c81d1a28ab7a468b459c/html5/thumbnails/28.jpg)
Soal-soal Latihan1. y′ + y tan x = 2 x cos x
2. y′ – xy = 6xe2x
3. (1 + x2)y′ + 2xy = x2
4. x2 ln x y′ + xy = 1
5. x y′ + 2 y = 4 ln x
6. sin x y′ + y cos x = sin x – x cos x
7. (1 + x2) y′ + 2xy = x ln x
8. (x – 1) y′ – 2y = x(x − 1)4
9. (1 + ex)y′ + ex y= xex
10.x ln x y′ + y = x3 ln x
11.3y′ + y = (1 − 2x)y4
12.x ln x y′ – y = x3 y2
13.x y′′′ – y′′ = x4 ln x
14.y′′′ – 2y′′ = x e2x