Task compilation - Differential Equation II
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Transcript of Task compilation - Differential Equation II
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DIFFERENTIAL EQUATION II
TASK COMPILATION
3/27/2014
MARIA PRISCILLYA PASARIBUIDN. 4103312018
TASK I (February, 13th 2014)
Determine the solution of::
1.d2 yd t2
β2dydt
β5 y=0, y(0)=0; yβ(0)=1
2.d2 yd t2
β3dydt
=0, y(0)=0; yβ(0)=1
3.d2 yd t2
β4dydt
β4 y=0, y(0)=0; yβ(0)=1
Solution:
1.d2 yd t2
β2dydt
β5 y=0, y(0)=0; yβ(0)=1
Characteristic Equation: Ξ»2β2 Ξ»β5=0
Ξ»1,2=βbΒ±βb2β4 ac
2a=
β(β2 )Β±β (β2 )2β4 (1 ) (β5 )2 (1 )
ΒΏ 2Β±β4+202
=2Β±β242
=2Β±2β62
=1Β±β6
Ξ»1and Ξ»2 are real and distinct, then y=c1 em1x+c2 e
m2 x
y (0 )=0
0=c1+c2
c1=βc2 .................... (1)
y ' (0 )=1
y ' (x )= (1+β6 )c1e(1+β6 )x+(1ββ6 )c1 e
(1ββ6) x
1=(1+β6 ) c1+(1ββ6 )c2
1=(1+β6 ) c1+(1ββ6 )(βcΒΏΒΏ1)ΒΏ
1=c1 (1+β6β1+β6 )
1=2β6c1
c1=1
12β6
c1=βc2 .................... (1)
c2=β112
β6
Maka, y=1
12β6e (1+β6 ) xβ 1
12β6e (1ββ6) x
2.d2 yd t2
β3dydt
=0, y(0)=0; yβ(0)=1
Characteristic Equation:Ξ»2β3 Ξ»=0
Ξ» (Ξ»β3)=0
Ξ»1=0β¨ Ξ»2=3
Ξ»1and Ξ»2 are real and distinct, then y=c1 em1x+c2 e
m2 x
y=c1 e0x+c2 e
3 x
y (0 )=0
0=c1+c2
c1=βc2 .................... (1)
y ' (0 )=1
y ' (x )=3c23 x
1=3c2
c2=13
c1=β13
Then, y=β13em1 x+ 1
3em2 x
3.d2 yd t2
β4dydt
β4 y=0, y(0)=0; yβ(0)=1
Characteristic Equation:Ξ»2β4 Ξ»β4=0
Ξ»1,2=βbΒ±βb2β4 ac
2a=
β(β4 )Β±β (β4 )2β4 (1 ) (β4 )2 (1 )
ΒΏ 4Β±β16+162
=4 Β±4β22
=2Β±2β2
Ξ»1and Ξ»2 are real and distinct, then y=c1 em1x+c2 e
m2 x
y=c1 e(2+2β2) x+c2 e
(2β2β2 )x
y (0 )=0
0=c1+c2
c1=βc2 .................... (1)
y ' (0 )=1
y ' (x )= (2+2β2 )c1 e( 2+2β2)x+(2β2β2)c2e
(2β2β2)x
1=( 2+2β2 ) c1+(2β2β2)c2
1=( 2+2β2 ) c1+(2β2β2)(βc1)
1=c1(2+2β2β2+2β2)
1=c1(4β2)
c1=18
β2
c1=βc2 .................... (1)
c2=β18
β2
Maka, y=18
β2e(2+2β2) xβ18β2e(2β2β2) x
TASK II (February, 20th 2014)
Solve these equation:
1.d4 yd x4 +10
d2 yd x2 +9 y=0
2.d4 yd x4 + d
3 yd x3 + d
2 yd x2 +2 y=0
3. y ' ' '+4 y '=0
4. y ( 4)+4 y ' 'β y '+6 y=0
5.d6 yd x6 β4
d5 yd x5 +16
d4 yd x4 β12
d3 yd x3 +41
d2 yd x2 β8
dydx
+26 y=0
Solution:
1.d4 yd x4 +10
d2 yd x2 +9 y=0
Characteristic equation: Ξ»4+10 Ξ»2+9=0
(Ξ»2+9 ) (Ξ»2+1 )=0
( Ξ»+3i ) ( Ξ»β3 i) ( Ξ»+i ) ( Ξ»βi )=0
Ξ»1=β3 iβ Ξ»2=β3 iβ Ξ»3=βiβ Ξ»3=i
So, the solution is y=C1 cos3 x+C2 sin3 x+C3cos x+C4 sin x
2.d4 yd x4 + d
3 yd x3 + d
2 yd x2 +2 y=0
Characteristic equation: Ξ»4+Ξ»3+ Ξ»2+2=0
(Ξ»2βΞ»+1 ) ( Ξ»2+2 Ξ»+2 )=0
(Ξ»2βΞ»+1 )=0
Ξ»1,2=βbΒ±βb2β4 ac
2a
Ξ»1,2=β(β1)Β±β(β1)2β4(1)(1)
2(1)
Ξ»1,2=1Β±β1β4
2
Ξ»1,2=1Β±β3i
2
Ξ»1=12+ 1
2β3i and Ξ»2=
12β1
2β3 i
(Ξ»2+2Ξ»+2 )=0
Ξ»3,4=βbΒ±βb2β4ac
2a
Ξ»3,4=β(2)Β±β(2)2β4 (1)(2)
2(1)
Ξ»3,4=β2Β±β4β8
2
Ξ»3,4=β2Β±2 i
2
Ξ»3=β1+i and Ξ»4=β1βi
So, the solution is y=e12x(C1 cos
12
β3 x+C2sin12β3x )+eβx (C3 cos x+C4 sin x )
3. y ' ' '+4 y '=0
Characteristic equation: Ξ»3+4 Ξ»=0
Ξ» ( Ξ»2+4 )=0
Ξ» ( Ξ»+2i ) ( Ξ»β2i )=0
Ξ»1=0β Ξ»2=β2 iβ Ξ»3=2 i
So, the solution is y=C1+C2cos2 x+C3 sin 2x
4. y ( 4)+4 y ' 'β y '+6 y=0
Characteristics equation is Ξ»4+4 Ξ»2β Ξ»+6=0
(Ξ»2βΞ»+2 ) ( Ξ»2+Ξ»+3 )=0
(Ξ»2βΞ»+2 )=0
Ξ»1,2=βbΒ±βb2β4 ac
2a
Ξ»1,2=β(β1)Β±β(β1)2β4(1)(2)
2(1)
Ξ»1,2=1Β±β1β8
2
Ξ»1,2=1Β±β7 i
2
Ξ»1=12+ 1
2β7 i and Ξ»2=
12β1
2β7 i
(Ξ»2+ Ξ»+3 )=0
Ξ»3,4=βbΒ±βb2β4ac
2a
Ξ»3,4=β(1 )Β±β (1 )2β4 (1 ) (3 )
2 (1 )
Ξ»3,4=β1Β±β1β12
2
Ξ»3,4=β1Β±β11i
2
Ξ»3=β1
2+ 1
2β11 i and Ξ»4=
β12
β12β11 i
So, the equation is:
y=e12x(C1 cos
12
β7 x+C2sin12β7 x)+e
β12x(C3 cos
12
β11 x+C4 sin12
β11 x )
5.d6 yd x6 β4
d5 yd x5 +16
d4 yd x4 β12
d3 yd x3 +41
d2 yd x2 β8
dydx
+26 y=0
Characteristic equation is Ξ»6β4 Ξ»5+16 Ξ»4β12 Ξ»3+41 Ξ»2β8 Ξ»+26=0
(Ξ»4+3 Ξ»2+2 ) ( Ξ»2β4 Ξ»+13 )=0
(Ξ»2+1 ) ( Ξ»2+2 ) (Ξ»2β4 Ξ»+13 )=0
(Ξ»2+1 )=0
Ξ»1,2=βbΒ±βb2β4 ac
2a
Ξ»1,2=0Β±β0β4 (1)(1)
2 (1)
Ξ»1,2=Β±ββ4
2
Ξ»1,2=Β±2 i
2
Ξ»1=i and Ξ»2=βi
(Ξ»2+2 )=0
Ξ»3,4=βbΒ±βb2β4ac
2a
Ξ»3,4=0Β±β0β4 (1 ) (2 )
2 (1 )
Ξ»3,4=Β±ββ8
2
Ξ»3,4=Β±2β2 i
2
Ξ»3=β2i and Ξ»4=ββ2 i
(Ξ»2β4 Ξ»+13 )=0
Ξ»5,6=βbΒ±βb2β4ac
2a
Ξ»5,6=β(β4)Β±β(β4)2β4 (1 ) (13 )
2 (1 )
Ξ»5,6=4Β±β16β52
2
Ξ»5,6=4Β±6 i
2
Ξ»5=2+3 i and Ξ»6=2β3 i
So, the solution is
y=C1 cos x+C1 sin x+e2 x (C3cos 3x+C4 sin 3 x )+C5 cosβ2x+C5sin β2 x
TASK III (March 7th, 2014)
1.d3qd t 3
β5d2qd t2
+25dqdt
β125q=β60 e7 t
y(0)=0, yβ(0)=1, yβ(0)=2
2. y(IV )β6 y ' ' '+16 y ' '+54 y 'β225 y=100eβ2x
y (0 )= y ' (0 )= y ' ' (0 )= y ' ' ' (0 )=1
Solution:
1.d3qd t 3
β5d2qd t2
+25dqdt
β125q=β60 e7 t
y(0)=0, yβ(0)=1, yβ(0)=2
Quadratic equation: Ξ»3β5 Ξ»2+25 Ξ»β125=β60e7 t
Y= yl + yr
Ξ»3β5 Ξ»2+25 Ξ»β125=0
(Ξ»β5ΒΏ(Ξ»+5 i)(Ξ»β5 i)=0
y l=c1 e5 t+c2 cos5 t+c3 sin 5 t
yr=A0 e7 t
yr'=7 A0 e
7t
yr' '=49 A0 e
7 t
yr' ' '=343 A0e
7 t
y ' ' 'β5 y ' '+25 yβ125 y=β60e7 t
343 A0 e7 tβ245 A0e
7 t+175 A0e7 tβ125 A0 e
7 t=β60e7 t
148 A0 e7 t=β60 e7 t
A0=β60148
=β1537
yr=β1537
e7 t
Then, we got the equation is equal to
Y= yl + yr
y (t )=c1 e5 t+c2 cos5 t+c3sin 5 tβ15
37e7 t
Now, we are going to find the value of c1, c2, and c3.
y(0)=0
0=c1+c2β1537
c1+c2=1537
........................................ (1)
yβ(0)=1
y ' (t)=5 c1 e5tβ5c2 sin5 t+5c3cos 5tβ105
37e7 t
1=5c1+5c3β10537
c1+c3=142185
.................................... (2)
yβ(0)=2
y ' '( t)=25c1 e5 tβ25c2cos 5tβ25 c3 sin 5 tβ735
37e7 t
2=25c1β25c2β73537
25c1β25c2=80937
c1βc2=809925
....................................... (3)
Elimination of c2 from (1) and (3)
c1βc2=809925
....................................... (3)
c1+c2=1537
........................................ (1)
2c1=809+375
925
c1=592925
From (3), we can get the value of c2 by substituting the value of c1.
c1βc2=809925
....................................... (3)
592925
βc2=809925
c2=β217925
From (2), we can get the value of c3 by substituting the value of c1.
c1+c3=142185
.................................... (2)
592925
+c3=142185
c3=118925
After getting the value of c1, c2, and c3, then the equation is:
y (t )=592925
e5 tβ217925
cos5 t+ 118925
sin 5 tβ1537e7t
2. y ( IV )β6 y ' ' '+16 y ' '+54 y 'β225 y=100eβ2x
y (0 )= y ' (0 )= y ' ' (0 )= y ' ' ' (0 )=1
Quadratic equation: Ξ»4β6 Ξ»3+16 Ξ»2+54 Ξ»β225=100eβ2x
+
y= y l+ yr
Ξ»4β6 Ξ»3+16 Ξ»2+54 Ξ»β225=0
By using Horner method and ABC formula, so that the roots are gotten:
Ξ»1=3 Ξ»3=3+4 i
Ξ»2=β3 Ξ»4=3β4 i
y l=c1 e3 t+c2 e
β3 t+e3t (c3 cos4 t+c4 sin 4 t)
yr=A0 eβ2x
yr'=β2 A0 e
β2t
yr' '=4 A0 e
β2 t
yr' ' '=β8 A0 e
β2t
yr(IV )=16 A0 e
β2 t
y ( IV )β6 y ' ' '+16 y ' '+54 y 'β225 y=100eβ2 t
16 A0 eβ2 t+48 A0 e
β2t+64 A0 eβ2 tβ108 A0 e
β2 tβ225 A0 eβ2 t=100eβ2 t
β205 A0 eβ2 t=100eβ2 t
A0=β100205
=β2041
So, the value of yr is
yr=β20
41eβ2 t
y=c1 e3 t+c2 e
β3t+e3 t(c3 cos4 t+c4sin 4 t)β2041eβ2 t
Now, we are going to find the value of c1, c2, and c3.
y(0) = 1
1=c1+c2+c3β2041
c1+c2+c3=6141
....................................... (1)
yβ(0)=1
y '=3c1 e3 tβ3c2e
β3 t+3e3 t (c3 cos 4 t+c4 sin 4 t )+e3 t (β4 c3sin 4 t+4c4 cos 4 t )+ 4041eβ2 t
1=3c1β3 c2+3c3+4 c4+4041
3c1β3c2+3c3+4c4=141
........................ (2)
yββ(0)=1
y ' '=9c1 e3 t+9c2 e
β3 t+9e3 t (c3 cos4 t+c 4sin 4 t )+3e3 t (β4c3 sin 4 t+4 c4 cos 4 t )+3e3 t (β4c3 sin 4 t+4 c4 cos4 t )β16e3 t (c3cos 4 t+c4 sin 4 t )+ 8041eβ2t
1=9c1+9c2+9c3+12c4+12c4β16 c3β8041
1=9(c1+c2+c3)+24 c4β16c3β8041
12141
=9( 6141
)+24 c4β16 c3
β42841
=24 c4β16 c3
2c3β3c4=10782
.......................................... (3)
yβββ(0)=1
y ' ' '=27 c1 e3 tβ27 c2 e
β3 t+75e3 t (c3 cos4 t+c4 sin 4 t )β100e3t (βc3sin 4 t+c4 cos4 t )+ 16041eβ2t
1=27 c1β27c2+75 c3β100c4+16041
27c1β27 c2+75c3β100c4=β119
41 ........ (4)
To get the value of c1, c2, c3, and c4, we can use elimination and substituion method from the
above equations.
From (4) and (2), we can get the new equation:
27c1β27 c2+75c3β100c4=β119
41 ........ (4)
27c1β27 c2+27c3+36c4=9
41 ................ (2)
43 c3β136c4=β128
41
6c3β176 c4=β1641
..................................... (5)
From (5) and (3), we can get the value of c4.
6c3β176 c4=β1641
..................................... (5)
6c3β9c4=32182
... ....................................... (3)
β8c4=β32β321
82
c4=β353656
From (3), we can get the value of c3.
2c3β3(β353656
)=10782
.......................................... (3)
2c3=1915656
c3=19151315
+
+
From (1), we get:
c1+c2+19151315
=6141
....................................... (1)
c1+c2=37
1312 ................................................ (6)
From (2), we get:
3c1β3c2+3( 19151315
)+4 (β353656
)= 141
........................ (2)
3c1β3c2=32β5745β2824
1312
3c1β3c2=β85371312
c1βc2=β85373936
................................................ (7)
From (6) and (7), we can get the value of c1 and c2 by elimination method.
c1+c2=37
1312 ................................................ (6)
c1βc2=β85373936
............................................. (7)
2c1=β8537+111
3936
2c1=β84263936
c1=β42133936
From (7), we can get the value of c2.
β42133936
βc2=β85373936
............................................. (7)
c2=8537β4213
3936
+
c2=1081656
After getting the value of c1, c2, c3, and c4, we get the characteristic value.
y=β42133936
e3 t+ 1081656
eβ3 t+e3 t( 19151315
cos4 tβ353656
sin 4 t )β2041eβ2 t
TASK IV (March 13th, 2014)
1.d2qd t 2
+1000dqdt
+25000q=24
q ( 0 )=q ' (0 )=0
2.d2 yd t2
β4dydt
+ y=2t 3+3 t2β1
y (0 )= y ' (0 )=1
Solution:
1. q ' '+1000q '+25000q=24
Quadratic equation:
t 2+1000 t+25000=0
t 1,2=βbΒ±βb2β4 ac
2a
t 1,2=β1000Β±β(1000)2β4 (1 ) (25000 )
2 (1 )
t 1,2=β1000Β±β900000
2
t 1,2=β1000Β±300β10
2
t 1,2=β500Β±150β10
t 1=β500+150β10 and t 2=β500β150β10
q l=c1 e(β500+150 β10 )t+c2e
(β500β150 β10 )t
qr=A0
q ' r=0
q ' ' r=0
q ' '+1000q '+25000q=24
0+1000 (0)+25000(A0)=24
A0=3
3125, then
qr=3
3125
q=q l+qr
q (t)=c1 e(β500+150 β10) t+c2 e
(β500β150β10) t+ 33125
q ( 0 )=0
0=c1+c2+3
3125
c1+c2=β3
3125 ...................................(1)
q ' (t)= (β500+150β10 ) c1 e(β500+150 β10) t+(β500β150β10 )c2 e
(β500β150β10) t
q ' (0 )=0
0=(β500+150β10 )c1+(β500β150β10 )c2 ........................... (2)
By using elimination method, we can find the value of c1 from (1) and (2).
(β500β150β10 )c1+(β500β150β10 )c2=β3
3125(β500β150β10 )
(β500+150β10 )c1+ (β500β150 β10 )c2=0
β300β10c1=β3
3125(β500β150β10 )
c1=β5β10β15
3125
By using (1), we can find the value of c2.
c1+c2=β3
3125
β5β10β153125
+c2=β3
3125
c2=5β10+12
3125
After finding the value of c1 and c2, we get the equation.
q (t)=(β5β10β153125 )e (β500+150 β10) t+( 5β10+12
3125 )e (β500β150β10) t+ 33125
2. y ' 'β4 y '+ y=2t 3+3 t2β1
Quadratic equation:
t 2β4 t+1=0
t 1,2=βbΒ±βb2β4 ac
2a
t 1,2=β(β4)Β±β(β4)2β4 (1 ) (1 )
2 (1 )
t 1,2=4Β±β12
2
β
t 1,2=4Β±2β3
2
t 1,2=2Β±β3
t 1=2+β3 and t 2=2ββ3
y l=c1 e(2+β3 ) t+c2 e
( 2ββ3 ) t
yr=A3 t3+A2 t
2+A1t+A0
y 'r=3 A3 t2+2 A2t+A1
y ' 'r=6 A3 t+2 A2
y ' 'β4 y '+ y=2t 3+3 t2β1
(6 A3t+2 A2 )β4 (3 A3 t2+2 A2 t+A1)+(A3 t
3+A2t2+A1t+A0 )=2 t3+3 t 2β1
A3 t3+(β12 A3+A2 )t 2+(6 A3β8 A2+A1 ) t+A1+A0=2 t3+3 t 2β1
Equation similarity:
A3=2
β12 A3+A2=3
β12(2)+A2=3
A2=27
6 A3β8 A2+A1=0
6 (2)β8(27)+A1=0
A1=204
A1+A0=β1
204+A0=β1
A0=β205
yr=2t 3+27 t 2+204 tβ205
y= y l+ yr
y (t )=c1 e( 2+β3) t+c2 e
(2ββ3) t+2 t 3+27 t2+204 tβ205
y (0)=1
1=c1+c2β205
c1+c2=206 .........................(1)
y '(t )=(2+β3 ) c1e(2+β3) t+(2ββ3 )c2 e
(2ββ3) t+6 t2+54 t+204
y '(0)=1
1=( 2+β3 )c1+(2ββ3 )c2+204
(2+β3 )c1+ (2ββ3 )c2=β203 ........................... (2)
By using elimination and substitution method, the value of c1 and c2 can be obtained
from (1) and (2).
β203=(2+β3 )c1+( 2ββ3 )c2........................... (2)
206=(2ββ3 )c1+(2ββ3 )c2 ...................................(1)
2β3 c1=β203β206 (2ββ3 )
2β3 c1=β715+206β3
c1=β715β3+618
6
c1+c2=206
(β715β3+6186 )+c2=206
c2=715β3+618
6
y (t )=(β715β3+6186 )e (2+β3) t+(715β3+618
6 )e( 2ββ3 )t+2 t3+27 t 2+204 tβ205
TASK V (March 20th, 2014)
1.d2 xd t 2
+4dxdt
+8x=(20 t 2+16 tβ78)e2 t
y(0)=yβ(0)=0
2.d3qd t 3
β5d2qd t2
+25dqdt
β125q=(β500 t2+465 tβ387)e2 t
q(0)=qβ(0)= qββ(0)=0
Solution:
β
1.d2 xd t 2
+4dxdt
+8x=(20 t 2+16 tβ78)e2 t
y(0)=yβ(0)=0
quadratic equation is
Ξ»2+4 Ξ»+8=0
Ξ»1,2=βbΒ±βb2β4 ac
2a
Ξ»1,2=β4Β±β42β4 (1)(8)
2(1)
Ξ»1,2=β4Β±ββ16
2
Ξ»1,2=β2Β±2 i
y l=eβ2t (c1cos 2t+c2 sin2 t)
yr=(A2t2+A1t+A0)e
2 t
y 'r=(2 A2t+A1 )e2 t+(A2t2+A1t+A0) (2e2 t )
y ' 'r=2 A2 e2 t+(2 A2 t+A1 ) ( 2e2 t )+( 2 A2t+A1 ) (2e2 t )+(A2 t
2+A1t+A0)( 4e2 t )
y ' '+4 y '+8 y=(20 t 2+16 tβ78)e2t
2 A2 e2 t+( 2 A2t+A1 ) (2e2 t )+ (2 A2t+A1 ) (2e2 t )+(A2 t
2+A1t+A0 ) ( 4e2t )+4 {( 2 A2 t+A1 )e2 t+(A2 t2+A1 t+A0) (2e2 t )}+8 {(A2t
2+A1 t+A0)e2 t }=(20 t 2+16 tβ78)e2 t
(2 A2+8 A1+20 A0 )e2 t+(16 A2+20 A1 ) t e2 t+(20 A2)t2e2 t=(20 t 2+16 tβ78)e2t
20 A2=20
A2=1
16 A2+20 A1=16
16(1)+20 A1=16
20 A1=0
A1=0
2 A2+8 A1+20 A0=β78
2 (1 )+8 (0 )+20 A0=β78
20 A0=β78β2
20 A0=β80
A0=β4
yr=(t 2β4)e2 t
y = yl + yr
y (t )=eβ2 t (c1cos 2t+c2sin 2 t )+(t 2β4)e2 t
y ' (t )=(β2eβ2 t ) (c1 cos2 t+c2sin 2 t )+eβ2t (β2c1 sin 2t+2c2cos2 t )+2 t e2 t+(t 2β4) (2e2 t )
y (0 )=0
0=c1β4
c1=4
y ' (0 )=0
0=β2c1+2c2β8
8=β2(4)+2c2
c2=8
y (t )=eβ2 t (4cos 2t+8 sin 2 t )+(t2β4)e2t
2.d3qd t 3
β5d2qd t2
+25dqdt
β125q=(β500 t 2+465 tβ387 )e2 t
q(0)=qβ(0)= qββ(0)=0
Quadratic equation is:
Ξ»3β5 Ξ»2+25 Ξ»β125=0
Ξ»1=5 Ξ»2=5 i
Ξ»3=β5 i
y l=c1 e5 t+c2 cos5 t+c3 sin 5 t
yr=(A2t2+A1t+A0)e
2 t
y ' r=( 2 A2t+A1 )e2 t+(A2 t2+A1t+A0)(2eΒΏΒΏ2 t)ΒΏ
ΒΏ2 A2t e2t+A1 e
2 t+2 A2 t2 e2 t+2 A1t e
2 t+2 A0e2 t
y ' 'r=2 A2 e2 t+8 A2t e
2 t+2 A1 e2 t+4 A2t e
2 t+4 A2t2e2 t+2 A1 e
2 t+4 A1 t e2 t+4 A0 e
2t
y ' ' ' r=12 A2e2 t+12 A1 e
2 t+8 A0 e2 t+24 A2t e
2 t+8 A1t e2 t+8 A2t
2 e2 t
y ' ' 'β5 y ' '+25 y 'β125 y=(β500 t 2+465 tβ387 )e2 t
12 A2 e2 t+12 A1 e
2t+8 A0 e2 t+24 A2 t e
2t+8 A1 t e2t+8 A2 t
2 e2 tβ5 {2 A2 e2 t+8 A2t e
2 t+2 A1e2 t+4 A2t e
2 t+4 A2t2e2 t+2 A1 e
2 t+4 A1 t e2 t+4 A0 e
2t }+25 {2 A2 t e2 t+A1 e
2t+2 A2t2 e2 t+2 A1 t e
2 t+2 A0 e2 t }β125 {(A2 t
2+A1t+A0 )e2 t }=(β500 t 2+465 tβ387 )e2 t
(2 A2+17 A1β87 A0 )e2t+ (34 A2β87 A1 ) t e2 t+(β87 A2) t 2 e2 t=β500 t2 e2t+465 t e2 tβ387e2t
β87 A2=β500
A2=50087
34 A2β87 A1=465
34 (50087 )β87 A1=465
A1=β23455
7569
2 A2+17 A1β87 A0=β387
2( 50087 )+17 (β23455
7569 )β87 A0=β387
A0=2617468658503
yr=( 50087t 2β23455
7569t+ 2617468
658503 )e2 t
y= y l+ yr
y (t )=c1e5 t+c2 cos5 t+c3sin 5 t+( 500
87t 2β23455
7569t+ 2617468
658503 )e2 t
y ' (t)=5 c1 e5tβ5c2 sin5 t+5c3cos 5t+( 1000
87tβ23455
7569 )e2 t+(50087t 2β23455
7569t+ 2617468
658503 ) (2e2 t )
y ' ' (t )=25c1 e5 tβ25c2cos 5tβ25c3 sin 5 t+ 1000
87e2 t+(1000
87tβ23455
7569 )( 2e2 t )+( 100087
tβ234557569 ) (2e2 t )+( 500
87t 2β23455
7569t+ 2617468
658503 ) (4 e2 t )
y (0 )=0
0=c1+c2+2617468658503
c1+c2=β2617468
658503 ................................... (1)
y ' (0 )=0
0=5c1+5c3β234557569
+5234936658503
5c1+5c3=β3194351
658503
c1+c3=β31943513292515
.................................... (2)
y ' ' (0 )=0
0=25c1β25c2+1000
87β46910
7569β 46910
7569+ 10469872
658503
25c1β25c2=β9876532
658503 .................................... (3)
25c1+25c2=β65436700
658503 ................................... (1)
50c1=β75313232
658503
c1=β7531323232925150
From equation (1), we get the value of c2.
c2=β2617468
658503+75313232
32925150
c2=β5556016832925150
From equation (2), we get the value of c3.
c3=β31943513292515
+ 7531323232925150
c3=4336972232925150
+
y (t )=β7531323232925150
e5 tβ5556016832925150
cos5 t+ 4336972232925150
sin 5 t+( 50087t 2β23455
7569t+ 2617468
658503 )e2 t