problems collection of differential equation ii

qwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmrtyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqw

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



y=c1+c2 e3t


c1+c2=0

c1=−c2 ... (1)


y’(t)=3c2 e3 t

y’(0)=1

2

3c2=1

c2=13

... (2)


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



y=c1 e(2+2√2)t+c2e

(2−√2)t


c1+c2=0

c1=−c2 ... (1)

3


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)


(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


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


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


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


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 ¿


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


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


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


c1+c2+c3=6141

c1−21782952

−3144611808

=1756811808

c1=1756811808

+ 3144611808

+ 871211808

=5772611808


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


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


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


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


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

problems collection of differential equation ii

Science