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Problems Collection
Differential Equation II
Meiva Marthaulina4103312019
0
Determine the solution of :
1.d2 ydt 2
−2dydt
−5 y=0
With y(0) = 0 and y’(0) = 1
Solution :
Eq : y2−2 y−5=0
To find the root
y1,2=−b±√b2−4ac
2a
y1,2=2±√4−4.1 .−5
2.1
y1,2=2±2√6
2
y1,2=1±√6
To determine the case
Since y1≠ y2 so we take case 1 that
y=c1 e(1+√6 )t+c2 e
(1−√6)t
To determine y(0) = 0
c1+c2=0
c1=−c2 ... (1)
To determine y’(0)=1
y’(t)=(1+√6 ) c1 e( 1+√6) t+ (1−√6 )c2 e
( 1−√6 ) t
y’(0)=1
(1+√6 ) c1+(1−√6 )c2=1
c1+c2+√6c1−√6c2=1 ... (2)
To subsitute (1) to (2)
c1+c2+√6c1−√6c2=1
−√6c2−√6c2=1
1
−2√6c2=1
c2=−1
2√6=−1
12√6
c1=1
2√6= 1
12√6
Conclusion
So, we get :
y= 112
√6e(1+√6 )t− 112
√6e(1−√6)t
2.d2 ydt 2
−3dydt
=0
With y(0) = 0 and y’(0) = 1
Solution :
Eq : y2−3 y=0
To find the root
y ( y−3 )=0
y=0∨ y=3
To determine the case
Since y1≠ y2 so we take case 1 that
y=c1+c2 e3t
To determine y(0) = 0
c1+c2=0
c1=−c2 ... (1)
To determine y’(0)=1
y’(t)=3c2 e3 t
y’(0)=1
2
3c2=1
c2=13
... (2)
To subsitute (2) to (1)
c1=−c2
c1=−13
Conclusion
So, we get :
y=−13
+13e3t
3.d2 ydt 2
−4dydt
−4 y=0
With y(0) = 0 and y’(0) = 1
Solution :Eq : y2−2 y−5=0
To find the root
y1,2=−b±√b2−4ac
2a
y1,2=4 ±√16−4.1 .−4
2.1
y1,2=4 ±4√6
2
y1,2=2±2√2
To determine the case
Since y1≠ y2 so we take case 1 that
y=c1 e(2+2√2)t+c2e
(2−√2)t
To determine y(0) = 0
c1+c2=0
c1=−c2 ... (1)
3
To determine y’(0)=1
y’(t)=(2+2√2 )c1 e(2+2√2) t+ (2−2√2 )c2 e
(2−2√2) t
y’(0)=1
(2+2√2 )c1+(2−2√2 ) c2=1 ... (2)
To subsitute (1) to (2)
(2+2√2 )c1+(2−2√2 ) c2=1
(2+2√2 )−c2+(2−2√2 )c2=1
−2c2−2√2c2+2c2−2√2c2=1
−4√2c2=1
c2=−1
4 √2=−1
8√2
c1=1
4 √2=1
8√2
Conclusion
So, we get :
y=18
√2e(2+2√2)t−18√2e(2−2√2)t
4.d4 ydx4 +10
d2 yd x2 +9 y=0
Solution :
Eq : γ 4+10 γ2+9=0
(λ2+9 ) (λ2+1 )=0
To find the root
λ1=−3 i
λ2=3 i
4
λ3=−i
λ4=i
To determine the case
Since the root is imaginer so we take case 3 that
y=eax (c¿¿1 cosqx+c2 sinqx)¿
Conclusion
So, we get :
y=C1 cos3 x+C2 sin3 x
+C3 cosx+C4 sinx
5.d4 ydx4 + d
3 yd x3 + d
2 yd x2 +2 y=0
Solution :Eq : λ4+λ3+ λ2+2=0
(λ¿¿2−λ+1)(λ2+2 λ+2)¿
To find the root
λ1,2=−b±√b2−4 ac
2a
λ1,2=1±√1−4.1 .1
2.1
λ1,2=1±√−3
2
λ1=12+ 1
2√3i
λ2=12−1
2√3 i
λ3,4=−2±√4−4.1 .2
2
λ3,4=−2±√−4
2
λ3=−1+i
λ4=−1−i
5
To determine the case
Since the imaginer root so we take case 3 that
y=ea x¿
Conclusion
So, we get :
y=e12x¿¿) + e− x(C3 cos x+C4 sin x )
6. y ' ' '+4 y '=0
Solution :
Eq : λ3+4 λ=0
To find the root
λ ( λ2+4 )=0
λ ( λ−2 i ) ( λ+2i )=0
λ1=0
λ2=2i
λ3=−2 i
To determine the case
Since λ1 is real number so we take case 1 that y=c1 exand λ2 λ3 is imaginer and take
case 3
Conclusion
So, we get :
y=C1+C2cos2 x+C3 cos2 x
7. y(4)+4 y ' '− y '+6 y=0
Solution :
Eq : λ4+4 λ2− λ+6=0
(λ2−λ+2 ) ( λ2+λ+3 )=0
6
To find the root
λ1,2=1±√1−4.1 .2
2
λ1,2=1±√−7
2
λ1=12+ 1
2√7 i
λ2=12−1
2√7 i
To determine the case
Since the imaginer root so we take case 3
Conclusion
So, we get :
y=e12x(C1 cos
12
√3 x+C2sin12√3x )+e− x(C3 cos x+C4sin x )
8.d6 ydx6 −4
d5 yd x5 +16
d4 yd x4 −12
d3 ydx3 +41
d2 ydx2 −8
dydx
+26 y=0
Solution :
Eq : λ6−4 λ5+16 λ4−12 λ3+41 λ2−8 λ+26=0
(λ4+2 λ2+2 ) (λ2−4 λ−13 )=0
(λ2+1 ) ( λ2+2 ) (λ2−4 λ−13 )=0
To find the root
λ1,2=±√−4.1.1
2
λ1=i
λ2=−i
7
λ3,4=±√−4.1.2
2
λ3=¿√2i ¿
λ4=¿−√2 i ¿
λ5,6=±√16−4.1 .13
2
λ5=¿2+3 i ¿
λ6=¿2−3 i ¿
To determine the case
Since the imaginer root so we take case 3
Conclusion
So, we get :
y=(C1 cos x+C2 sin x )(C3 cos√2 x+C4 sin√2x )+e2(C5 cos3 x+C6 cos3 x)
9.d3 ydt 3
−5d2 yd t2
+25dydt
−125 y=−60e7 t
y(0) = 0 , y’(0) = 1 , y’’(0) = 2
solution :Eq :y3−5 y2+25 y−125=0
( y−5 ) ( y2+25 )=0
y= y l+ yr
To find the root
y1=5
y2=5 i
y3=−5 i
To determine the case
y l=c1 e5 t+c2 cos5 t+c3 sin 5 t
yr=a0 e7 t
y 'r=7a0 e7 t
y ' 'r=49a0 e7 t
y ' ' 'r=343a0 e7 t
8
To find the value of a0
d3 ydt 3
−5d2 yd t2
+25dydt
−125 y=−60e7 t
343a0e7 t−245a0 e
7 t+175a0 e7 t−125 a0 e
7 t=−60e7 t
148a0e7 t=−60e7 t
a0=−1537
To find constanta
y=c1 e5 t+c2 cos5 t+c3 sin 5 t−15
37e7 t
y(0) = 0
0 = c1+c2−1537
c1+c2=1537
... (1)
y '=5c1 e5 t−5c2cos5 t+5c3sin 5 t−105
37e7 t
y’(0) = 1
1=5c1+5c3−10537
c1+c3=142185
... (2)
y’’(0) = 2
y ' '=25c1e5 t−25c2 cos5 t−25c3 sin5 t−735
37e7 t
2=25c1−25c2−73537
c1−c2=809925
... (3)
Doing elimination
(1) and (3)
9
c1+c2=1537
c1−c2=809925
-
c2=−217925
Equation 1
c1+c2=1537
c1=592925
Equation 2
c1+c3=142185
c3=142185
−592925
=118925
Conclusion
So, we get :
y=592925
e5t−217925
cos5 t+ 118925
sin5 t−1537e7 t
10. y(4)−6 y(3)+16 y2+54 y '−225 y=100 e−2 x
y(0) = y’(0) = y’’(0) = y’’’(0) = 1
solution :eq : y4−6 y3+16 y2+54 y−225=0
To find the root
By horner method, we get the roots are 3 and -3, so we have :
( x−3 ) ( x+3 ) (x2−6 x+25 )=0
x3,4=6±√36−4.1.25
2
10
x3,4=6±√−64
2
x1=3
x2=−3
x3=3+4 i
x4=3−4 i
To determine the case
y l=c1 e3 t+c2 e
−3 t+e3t ¿¿
yr=a0 e−2 t
y 'r=−2a0 e−2 t
y ' 'r=4 a0 e−2 t
y ' ' 'r=−8a0e−2 t
y ' ' ' 'r=16a0e−2 t
To find the value of a0
y(4)−6 y(3)+16 y2+54 y '−225 y=100 e−2 x
16a0e−2 t+48a0 e
−2 t+64 a0 e−2 t−108a0 e
−2 t−225a0e−2 t=100a0 e
−2 t
−205a0e−2 t=100e−2 t
a0=−2041
To find constantay=c1 e
3 t+c2 e−3t+e3 t ¿¿
y(0) = 0
0 = c1+c2+c3−2041
c1+c2+c3=6141
... (1)
y '=3c1 e3 t−3c2e
−3 t+e3 t ¿
y’(0) = 1
1=3c1−3 c2+3c3+4 c4+4041
11
3c1−3c2+3c3+4c4=141
... (2)
y ' '=9c1 e3 t+9c2 e
−3 t+3e3 t (3c3cos 4 t+3c4 sin 4 t−4c3 sin 4 t+4 c4 cos4 t )+33t (−12c3 sin 4 t+12c4 cos 4 t−16c3cos 4 t−16 c4sin 4 t )−8041e−2 t
y’’(0) = 1
1=9c1+9c2+9c3+12c4+12c4−16 c3−8041
1=9(c¿¿1+c2+c3)+24 c4−16 c3−8041
¿
1=9( 6141
)+24 c4−16c3−8041
16c3−24 c4=42841
2c3−3c4=10782
... (3)
y ' ' '=27 c1 e3 t−27 c2 e
−3 t−21e3 t ¿
y’’’(0) = 1
1=27 c1−27c2−117c3−44c4−16041
27c1−27 c2−117 c3−44c4=20141
... (4)
Doing elimination
(1) and (2)
c1+c2+c3=6141
3c1−3c2+3c3+4c4=141
-
−3c2+2c4=−182
82 ... (5)
(5) and (3)
−9c2+6c4=−546
82
4 c3−6c4=21782
-
12
4 c3−9c2¿−329
82 ... (6)
(2) and (3)
9c1−9 c2+9c3+12c4=3
41
8c3−12c4=42882
-
9c1−9 c2+c3=−422
82
9c1−9 c2+c3=−211
41 ... (7)
(1) and (7)
9c1+9c2+9c3=54941
9c1−9 c2+c3=−211
41 -
18c2+8c3=76041
... (8)
(8) and (6)
8c3−18c2=−658
82
18c2+8c3=76041
-
36c2=−658−1520
82=−2178
82
c2=−21782952
Subtitute to equation (6)
4 c3−9c2=−329
82
4 c3+196022952
=−32982
13
4 c3=−11844−19602
2952
c3=−3144611808
Subtitute to equation (1)
c1+c2+c3=6141
c1−21782952
−3144611808
=1756811808
c1=1756811808
+ 3144611808
+ 871211808
=5772611808
Subtitute to equation (3)
2c3−3c4=10782
6289211808
−3c4=1540811808
3c4=4748411808
c4=4748435424
Conclusion
So, we get :
y=5772611808
e3 t−21782952
e−3 t+e3 t ¿
11.d2qdt2
+1000dqdt
+25000q=24
q(0) = q’(0) = 0
solution :Eq : q2+1000q+25000=0y= y l+ yr
To find the root
14
q1,2=−1000±√1000000−4.1 .(25000)
2
q1,2=−10000±√900000
2
q1,2=−10000±300√10
2
q1,2=−500±150√10
q1=−500+150√10
q2=−500−150√10
To determine the case
y l=c1 e(−500+150 √10 )t+c2 e
(−500−150 √10)t
yr=A0
y 'r=0
To find the value of A0
d2qdt2
+1000dqdt
+25000q=24
25000 A0=24
A0=3
3250
To find constanta
y=c1 e(−500+150 √10)t+c2 e
(−500−150 √10 )t+ 33250
y(0) = 0
0 = c1+c2+3
3250
c1+c2=−3
3250... (1)
c1=−3
3250−c2
y '=(−500+150√10 )c1 e(−500+150√10)t−(−500−150√10)c2 e
(−500−150 √10)t
y’(0) = 0
15
0=(−500+150√10 )c1−(−500−150√10)c2 ... (2)
0=(−500+150√10 )( −33250
−c2)−(−500−150√10)c2
0=15003250
+500c2−450√10
3250−150√10c2−500c2−150√10c2
0= 613
−9√1065
−300√10c2
300√10c2=30−9√10
65
c2=√10−3
6500
Equation (1)
c1+c2=−3
3250
c1+√10−36500
= −33250
c1=−3
3250−( √10−3
6500)
c1=−3−√10
6500
Conclusion
So, we get :
y=−3−√106500
e(−500+150 √10 )t−√10−36500
e(−500−150 √10)t+ 33250
12.d2 ydt 2
−4dydt
+ y=2t 3+3 t2−1
y(0) = y’(0) = 0
solution :Eq : y2−4 y+1=0y= y l+ yr
To find the root
16
y1,2=4 ±√16−4.1 .1
2
y1,2=4 ±√12
2
y1,2=4 ±2√3
2
y1=2+√3
y2=2−√3
To determine the case
y l=c1 e(2+√3 )t+c2 e
(2−√ 3)t
yr=A3 t3+A2 t
2+A1t+A0
y 'r=3 A3 t2+2 A2t+A1
y ' 'r=6 A3 t+2 A2
To find the value of A0
d2 ydt 2
−4dydt
+ y=2t 3+3 t2−1
6 A3 t+2 A2−4 (3 A3t2+2 A2t+A1 )+A3t
3+A2 t2+A1 t+A0=2 t3+3 t 2−1
A3 t3+(−12 A3+A2 )t 2+( A1−8 A2+6 A3 ) t+2 A2−4 A1+A0=2 t3+3 t 2−1
(1)
A3 t3=2t 3
A3=2
(2)
(−12 A3+A2 ) t 2=3 t2
−12 A3+A2=3
−12.2+A2=3
A2=27
(3)
( A1−8 A2+6 A3 ) t=0 t
17
A1−8.27+6.2=0
A1=204
(4)
2 A2−4 A1+A0=−1
A0=761
To find constantay (t )=c1 e
(2+√ 3)t+c2 e(2−√ 3)t+2 t3+27 t 2+204 t+761
y(0) = 1
1 = c1+c2+761
c1+c2=−760... (1)
c1=−760−c2
y '=(2+√3 ) c1e(2+√3 )t+(2−√3 ) c2 e
(2−√3 )t+6 t3+54 t+204
y’(0) = 1
y '=(2+√3 ) c1+ (2−√3 ) c2+204 ... (2)
(2+√3 ) (−760−c2 )+(2−√3 ) c2=−203
−1520−2c2−760 √3−√3 c2+2c2−√3c2=−203
−1317−7600√3+203=2√3c2
c2=−1317−7600√3
2√3
c2=−439√3−7600
2
Equation (1)
c1+c2=−760
c1+(−439√3−76002
)=−760
c1=6080+439√3
2
Conclusion
18
So, we get :
y=6080+439√32
e(2+√ 3)t+−439√3−76002
e(2−√3 )t+2 t3+27 t 2+204 t+761
13.d2 xdt2
+4dxdt
+8x=(20 t 2+16 t−78)e2 t
solution :Eq : y2+4 y+8=0y= y l+ yr
To find the root
y1,2=−4±√16−4.1 .8
2
y1,2=−4±√−16
2
y1,2=−4±4 i
2
y1=−2+2 i
y2=−2−2i
To determine the case
y l=e−2t ¿¿
yr=(A2t2+A1t+A¿¿0)e2t ¿
y 'r=2 A2t2 e2 t+(2 A1+2 A ¿¿2) t e2 t+(2 A0+A1)e
2 t ¿
y ' 'r=4 A2t2 e2 t+( 4 A1+8 A2) t e2 t+(4 A0+4 A1+2 A2)e
2t
To find the value of A0 , A1 , A2
d2 xdt2
+4dxdt
+8x=(20 t 2+16 t−78)e2 t
4 A2t2 e2 t+ (4 A1+8 A2 )t e2 t+ (4 A0+4 A1+2 A2 )e2 t+8 A2 t
2e2 t+(8 A1+8 A ¿¿2) t e2 t+( 8 A0+4 A1 )e2 t+(8 A2 t2+8 A1 t+8 A ¿¿0)e2 t=(20 t 2+16 t−78)e2t ¿¿
20 A2 t2 e2t+(20 A1+16 A2 ) t e2 t+(20 A0+8 A1+2 A2 )e2 t=(20 t 2+16 t−78)e2t
19
(1)
20 A2=20
A2=1
(2)
20 A1+16 A2=16
20 A1+16=16
A1=0
(3)
20 A0+8 A1+2 A2=−78
20 A0+0+2=−78
20 A0=−80
A0=−4
To find constanta
y (t )=e−2 t ¿
y(0) = 0
0 = c1−4
c1=4
y ' (t )=−2e−2 t ¿¿
y ' (0 )=0
0=−2c1+2c2−8
−8+2c2=8
c2=8
Conclusion
So, we get :
20
y (t )=e−2 t ¿
14.d3qdt3
−5d2qdt 2
+25dqdt
−125q=(−522 t2+465 t−387)e2t
q ( 0 )=q ’ (0 )=q' ' (0 )=0
solution :Eq : y3−5 y2+25 y−125=0y= y l+ yr
To find the root
By horner method is gotten ( y−5 ) ( y2+25 )=0
( y−5 ) ( y−5i )( y+5 i)=0
y1=5, y2=5 i and y3=−5 i
To determine the case
y l=c1 e5 t+c2 cos5 t+c3 sin 5 t
yr=(A2t2+A1t+A¿¿0)e2t ¿
y 'r=2 A2t2 e2 t+( 2 A1+2 A2 ) t e2 t+(2 A0+A1)e
2 t
y ' 'r=4 A2t2 e2 t+( 4 A1+8 A2) te2 t+(4 A0+4 A1+2 A2)e
2t
y ' ' 'r=8 A2 t2 e2t+(8 A1+24 A2) te2 t+(8 A0+12 A1+12 A2)e
2t
To find the value of A0 , A1 , A2
d3qdt3
−5d2qdt 2
+25dqdt
−125q=(−500+465t−387)e2 t
8 A2 t2 e2 t+(8 A1+24 A2 )te2 t+ (8 A0+12 A1+12 A2)e2 t−20 A2t
2 e2 t−(20 A1+40 A2 ) te2 t−(20 A0+20 A1+10 A2 )e2t+50 A2t2 e2 t+(50 A1+50 A2 ) t e2 t+(50 A0+25 A1 )e2 t−(125 A2 t
2+125 A1 t+125 A ¿¿0)e2 t=(−522 t2+465 t−387)e2 t ¿
−87 A2t2e2 t−( 87 A1−34 A2 )te2 t−(87 A0+17 A1+2 A2) e2 t=(−500+465 t−387)e2 t
(1)
21
−87 A2=−500
A2=50087
(2)
−87 A1+34 A2=465
−87 A1+17000
87=465
−87 A1=40455−17000
87
A1=−23455
7569
(3)
−87 A0−17 A1−2 A2=−387
−87 A0+398735
7569−1000
87=−387
87 A0=398735
7569−1000
87+387
87 A0=398735−87000+2929203
7569
A0=3240938658503
To find constanta
y (t )=c1e5 t+c2 cos5 t+c3sin 5 t+( 500
87t 2−23455
7569t+ 3240938
658503)e2 t
y(0) = 0
0 = c1+c2+3240938658503
c1+c2=−3240938
658503... (1)
y '=5c1 e5 t−5c2sin 5 t+5 c3 cos5 t+( 1000
87t−23455
7569 )e2 t+( 50087t 2−23455
7569t+ 3240938
658503)2e2 t
y’(0) = 0
22
0=5c1+5c3−234557569
+ 6481876658503
5c1+5c3=2040585−6481876
658503
5c1+5c3=−4441291
658503… (2)
y ' ' ( t )=25c1 e5 t−25c2cos 5t−25c3 sin 5 t+ 1000
87e2 t+(1000
87−23455
7569 )2e2t+4 e2 t (50087t2−23455
7569t+ 3240938
658503 )
y ' ' (0 )=0
0=25c1−25c2+1000
87+ 2000
87−46910
7569+ 12963752
658503
25c1−25c2=−1000
87−2000
87+ 46910
7569−12963752
658503
25c1−25c2=−7569000−15138000+4081170−12963752
658503
c1−c2=−9971058216462575
…(3)
Equation (1) and (3)
c1+c2=−3240938
658503=−81023450
16462575
c1−c2=−9971058216462575
__
2c2=1868713216462575
c2=1868713232925150
Equation (1)
c1+c2=−3240938
658503
c1+1868713232925150
=−3240938658503
c1=−180734032
32925150
Equation (2)
23
5c1+5c3=−4441291
658503
c1+c3=−44412913292515
−18073403232925150
+c3=−44412913292515
c3=1801734032−44412910
32925150
c3=1757321122
32925150
Conclusion
So, we get :
y (t )=−18073403232925150
e5t+1868713232925150
cos5 t+ 175732112232925150
sin 5 t+(50087t2−23455
7569t+ 3240938
658503)e2t
24