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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

−2dydt

−5 y=0, y(0)=0; y’(0)=1

2.d2 yd t2

−3dydt

=0, y(0)=0; y’(0)=1

3.d2 yd t2

−4dydt

−4 y=0, y(0)=0; y’(0)=1

Solution:

1.d2 yd t2

−2dydt

−5 y=0, y(0)=0; y’(0)=1

Characteristic Equation: λ2−2 λ−5=0

λ1,2=−b±√b2−4 ac

2a=

−(−2 )±√ (−2 )2−4 (1 ) (−5 )2 (1 )

¿ 2±√4+202

=2±√242

=2±2√62

=1±√6

λ1and λ2 are real and distinct, then y=c1 em1x+c2 e

m2 x

y (0 )=0

0=c1+c2

c1=−c2 .................... (1)

y ' (0 )=1

y ' (x )= (1+√6 )c1e(1+√6 )x+(1−√6 )c1 e

(1−√6) x

1=(1+√6 ) c1+(1−√6 )c2

1=(1+√6 ) c1+(1−√6 )(−c¿¿1)¿

1=c1 (1+√6−1+√6 )

1=2√6c1

c1=1

12√6

c1=−c2 .................... (1)

c2=−112

√6

Maka, y=1

12√6e (1+√6 ) x− 1

12√6e (1−√6) x

2.d2 yd t2

−3dydt

=0, y(0)=0; y’(0)=1

Characteristic Equation:λ2−3 λ=0

λ (λ−3)=0

λ1=0∨ λ2=3

λ1and λ2 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=−c2 .................... (1)

y ' (0 )=1

y ' (x )=3c23 x

1=3c2

c2=13

c1=−13

Then, y=−13em1 x+ 1

3em2 x

3.d2 yd t2

−4dydt

−4 y=0, y(0)=0; y’(0)=1

Characteristic Equation:λ2−4 λ−4=0

λ1,2=−b±√b2−4 ac

2a=

−(−4 )±√ (−4 )2−4 (1 ) (−4 )2 (1 )

¿ 4±√16+162

=4 ±4√22

=2±2√2

λ1and λ2 are real and distinct, then y=c1 em1x+c2 e

m2 x

y=c1 e(2+2√2) x+c2 e

(2−2√2 )x

y (0 )=0

0=c1+c2

c1=−c2 .................... (1)

y ' (0 )=1

y ' (x )= (2+2√2 )c1 e( 2+2√2)x+(2−2√2)c2e

(2−2√2)x

1=( 2+2√2 ) c1+(2−2√2)c2

1=( 2+2√2 ) c1+(2−2√2)(−c1)

1=c1(2+2√2−2+2√2)

1=c1(4√2)

c1=18

√2

c1=−c2 .................... (1)

c2=−18

√2

Maka, y=18

√2e(2+2√2) x−18√2e(2−2√2) 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 ' '− y '+6 y=0

5.d6 yd x6 −4

d5 yd x5 +16

d4 yd x4 −12

d3 yd x3 +41

d2 yd x2 −8

dydx

+26 y=0

Solution:

1.d4 yd x4 +10

d2 yd x2 +9 y=0

Characteristic equation: λ4+10 λ2+9=0

(λ2+9 ) (λ2+1 )=0

( λ+3i ) ( λ−3 i) ( λ+i ) ( λ−i )=0

λ1=−3 i⋁ λ2=−3 i⋁ λ3=−i⋁ λ3=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: λ4+λ3+ λ2+2=0

(λ2−λ+1 ) ( λ2+2 λ+2 )=0

(λ2−λ+1 )=0

λ1,2=−b±√b2−4 ac

2a

λ1,2=−(−1)±√(−1)2−4(1)(1)

2(1)

λ1,2=1±√1−4

2

λ1,2=1±√3i

2

λ1=12+ 1

2√3i and λ2=

12−1

2√3 i

(λ2+2λ+2 )=0

λ3,4=−b±√b2−4ac

2a

λ3,4=−(2)±√(2)2−4 (1)(2)

2(1)

λ3,4=−2±√4−8

2

λ3,4=−2±2 i

2

λ3=−1+i and λ4=−1−i

So, the solution is y=e12x(C1 cos

12

√3 x+C2sin12√3x )+e−x (C3 cos x+C4 sin x )

3. y ' ' '+4 y '=0

Characteristic equation: λ3+4 λ=0

λ ( λ2+4 )=0

λ ( λ+2i ) ( λ−2i )=0

λ1=0⋁ λ2=−2 i⋁ λ3=2 i

So, the solution is y=C1+C2cos2 x+C3 sin 2x

4. y ( 4)+4 y ' '− y '+6 y=0

Characteristics equation is λ4+4 λ2− λ+6=0

(λ2−λ+2 ) ( λ2+λ+3 )=0

(λ2−λ+2 )=0

λ1,2=−b±√b2−4 ac

2a

λ1,2=−(−1)±√(−1)2−4(1)(2)

2(1)

λ1,2=1±√1−8

2

λ1,2=1±√7 i

2

λ1=12+ 1

2√7 i and λ2=

12−1

2√7 i

(λ2+ λ+3 )=0

λ3,4=−b±√b2−4ac

2a

λ3,4=−(1 )±√ (1 )2−4 (1 ) (3 )

2 (1 )

λ3,4=−1±√1−12

2

λ3,4=−1±√11i

2

λ3=−1

2+ 1

2√11 i and λ4=

−12

−12√11 i

So, the equation is:

y=e12x(C1 cos

12

√7 x+C2sin12√7 x)+e

−12x(C3 cos

12

√11 x+C4 sin12

√11 x )

5.d6 yd x6 −4

d5 yd x5 +16

d4 yd x4 −12

d3 yd x3 +41

d2 yd x2 −8

dydx

+26 y=0

Characteristic equation is λ6−4 λ5+16 λ4−12 λ3+41 λ2−8 λ+26=0

(λ4+3 λ2+2 ) ( λ2−4 λ+13 )=0

(λ2+1 ) ( λ2+2 ) (λ2−4 λ+13 )=0

(λ2+1 )=0

λ1,2=−b±√b2−4 ac

2a

λ1,2=0±√0−4 (1)(1)

2 (1)

λ1,2=±√−4

2

λ1,2=±2 i

2

λ1=i and λ2=−i

(λ2+2 )=0

λ3,4=−b±√b2−4ac

2a

λ3,4=0±√0−4 (1 ) (2 )

2 (1 )

λ3,4=±√−8

2

λ3,4=±2√2 i

2

λ3=√2i and λ4=−√2 i

(λ2−4 λ+13 )=0

λ5,6=−b±√b2−4ac

2a

λ5,6=−(−4)±√(−4)2−4 (1 ) (13 )

2 (1 )

λ5,6=4±√16−52

2

λ5,6=4±6 i

2

λ5=2+3 i and λ6=2−3 i

So, the solution is

y=C1 cos x+C1 sin x+e2 x (C3cos 3x+C4 sin 3 x )+C5 cos√2x+C5sin √2 x

TASK III (March 7th, 2014)

1.d3qd t 3

−5d2qd t2

+25dqdt

−125q=−60 e7 t

y(0)=0, y’(0)=1, y”(0)=2

2. y(IV )−6 y ' ' '+16 y ' '+54 y '−225 y=100e−2x

y (0 )= y ' (0 )= y ' ' (0 )= y ' ' ' (0 )=1

Solution:

1.d3qd t 3

−5d2qd t2

+25dqdt

−125q=−60 e7 t

y(0)=0, y’(0)=1, y”(0)=2

Quadratic equation: λ3−5 λ2+25 λ−125=−60e7 t

Y= yl + yr

λ3−5 λ2+25 λ−125=0

(λ−5¿(λ+5 i)(λ−5 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 ' ' '−5 y ' '+25 y−125 y=−60e7 t

343 A0 e7 t−245 A0e

7 t+175 A0e7 t−125 A0 e

7 t=−60e7 t

148 A0 e7 t=−60 e7 t

A0=−60148

=−1537

yr=−1537

e7 t

Then, we got the equation is equal to

Y= yl + yr

y (t )=c1 e5 t+c2 cos5 t+c3sin 5 t−15

37e7 t

Now, we are going to find the value of c1, c2, and c3.

y(0)=0

0=c1+c2−1537

c1+c2=1537

........................................ (1)

y’(0)=1

y ' (t)=5 c1 e5t−5c2 sin5 t+5c3cos 5t−105

37e7 t

1=5c1+5c3−10537

c1+c3=142185

.................................... (2)

y”(0)=2

y ' '( t)=25c1 e5 t−25c2cos 5t−25 c3 sin 5 t−735

37e7 t

2=25c1−25c2−73537

25c1−25c2=80937

c1−c2=809925

....................................... (3)

Elimination of c2 from (1) and (3)

c1−c2=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.

c1−c2=809925

....................................... (3)

592925

−c2=809925

c2=−217925

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 t−217925

cos5 t+ 118925

sin 5 t−1537e7t

2. y ( IV )−6 y ' ' '+16 y ' '+54 y '−225 y=100e−2x

y (0 )= y ' (0 )= y ' ' (0 )= y ' ' ' (0 )=1

Quadratic equation: λ4−6 λ3+16 λ2+54 λ−225=100e−2x

+

y= y l+ yr

λ4−6 λ3+16 λ2+54 λ−225=0

By using Horner method and ABC formula, so that the roots are gotten:

λ1=3 λ3=3+4 i

λ2=−3 λ4=3−4 i

y l=c1 e3 t+c2 e

−3 t+e3t (c3 cos4 t+c4 sin 4 t)

yr=A0 e−2x

yr'=−2 A0 e

−2t

yr' '=4 A0 e

−2 t

yr' ' '=−8 A0 e

−2t

yr(IV )=16 A0 e

−2 t

y ( IV )−6 y ' ' '+16 y ' '+54 y '−225 y=100e−2 t

16 A0 e−2 t+48 A0 e

−2t+64 A0 e−2 t−108 A0 e

−2 t−225 A0 e−2 t=100e−2 t

−205 A0 e−2 t=100e−2 t

A0=−100205

=−2041

So, the value of yr is

yr=−20

41e−2 t

y=c1 e3 t+c2 e

−3t+e3 t(c3 cos4 t+c4sin 4 t)−2041e−2 t

Now, we are going to find the value of c1, c2, and c3.

y(0) = 1

1=c1+c2+c3−2041

c1+c2+c3=6141

....................................... (1)

y’(0)=1

y '=3c1 e3 t−3c2e

−3 t+3e3 t (c3 cos 4 t+c4 sin 4 t )+e3 t (−4 c3sin 4 t+4c4 cos 4 t )+ 4041e−2 t

1=3c1−3 c2+3c3+4 c4+4041

3c1−3c2+3c3+4c4=141

........................ (2)

y’’(0)=1

y ' '=9c1 e3 t+9c2 e

−3 t+9e3 t (c3 cos4 t+c 4sin 4 t )+3e3 t (−4c3 sin 4 t+4 c4 cos 4 t )+3e3 t (−4c3 sin 4 t+4 c4 cos4 t )−16e3 t (c3cos 4 t+c4 sin 4 t )+ 8041e−2t

1=9c1+9c2+9c3+12c4+12c4−16 c3−8041

1=9(c1+c2+c3)+24 c4−16c3−8041

12141

=9( 6141

)+24 c4−16 c3

−42841

=24 c4−16 c3

2c3−3c4=10782

.......................................... (3)

y’’’(0)=1

y ' ' '=27 c1 e3 t−27 c2 e

−3 t+75e3 t (c3 cos4 t+c4 sin 4 t )−100e3t (−c3sin 4 t+c4 cos4 t )+ 16041e−2t

1=27 c1−27c2+75 c3−100c4+16041

27c1−27 c2+75c3−100c4=−119

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:

27c1−27 c2+75c3−100c4=−119

41 ........ (4)

27c1−27 c2+27c3+36c4=9

41 ................ (2)

43 c3−136c4=−128

41

6c3−176 c4=−1641

..................................... (5)

From (5) and (3), we can get the value of c4.

6c3−176 c4=−1641

..................................... (5)

6c3−9c4=32182

... ....................................... (3)

−8c4=−32−321

82

c4=−353656

From (3), we can get the value of c3.

2c3−3(−353656

)=10782

.......................................... (3)

2c3=1915656

c3=19151315

+

+

From (1), we get:

c1+c2+19151315

=6141

....................................... (1)

c1+c2=37

1312 ................................................ (6)

From (2), we get:

3c1−3c2+3( 19151315

)+4 (−353656

)= 141

........................ (2)

3c1−3c2=32−5745−2824

1312

3c1−3c2=−85371312

c1−c2=−85373936

................................................ (7)

From (6) and (7), we can get the value of c1 and c2 by elimination method.

c1+c2=37

1312 ................................................ (6)

c1−c2=−85373936

............................................. (7)

2c1=−8537+111

3936

2c1=−84263936

c1=−42133936

From (7), we can get the value of c2.

−42133936

−c2=−85373936

............................................. (7)

c2=8537−4213

3936

+

c2=1081656

After getting the value of c1, c2, c3, and c4, we get the characteristic value.

y=−42133936

e3 t+ 1081656

e−3 t+e3 t( 19151315

cos4 t−353656

sin 4 t )−2041e−2 t

TASK IV (March 13th, 2014)

1.d2qd t 2

+1000dqdt

+25000q=24

q ( 0 )=q ' (0 )=0

2.d2 yd t2

−4dydt

+ y=2t 3+3 t2−1

y (0 )= y ' (0 )=1

Solution:

1. q ' '+1000q '+25000q=24

Quadratic equation:

t 2+1000 t+25000=0

t 1,2=−b±√b2−4 ac

2a

t 1,2=−1000±√(1000)2−4 (1 ) (25000 )

2 (1 )

t 1,2=−1000±√900000

2

t 1,2=−1000±300√10

2

t 1,2=−500±150√10

t 1=−500+150√10 and t 2=−500−150√10

q l=c1 e(−500+150 √10 )t+c2e

(−500−150 √10 )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(−500+150 √10) t+c2 e

(−500−150√10) t+ 33125

q ( 0 )=0

0=c1+c2+3

3125

c1+c2=−3

3125 ...................................(1)

q ' (t)= (−500+150√10 ) c1 e(−500+150 √10) t+(−500−150√10 )c2 e

(−500−150√10) t

q ' (0 )=0

0=(−500+150√10 )c1+(−500−150√10 )c2 ........................... (2)

By using elimination method, we can find the value of c1 from (1) and (2).

(−500−150√10 )c1+(−500−150√10 )c2=−3

3125(−500−150√10 )

(−500+150√10 )c1+ (−500−150 √10 )c2=0

−300√10c1=−3

3125(−500−150√10 )

c1=−5√10−15

3125

By using (1), we can find the value of c2.

c1+c2=−3

3125

−5√10−153125

+c2=−3

3125

c2=5√10+12

3125

After finding the value of c1 and c2, we get the equation.

q (t)=(−5√10−153125 )e (−500+150 √10) t+( 5√10+12

3125 )e (−500−150√10) t+ 33125

2. y ' '−4 y '+ y=2t 3+3 t2−1

Quadratic equation:

t 2−4 t+1=0

t 1,2=−b±√b2−4 ac

2a

t 1,2=−(−4)±√(−4)2−4 (1 ) (1 )

2 (1 )

t 1,2=4±√12

2

–

t 1,2=4±2√3

2

t 1,2=2±√3

t 1=2+√3 and t 2=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

y ' '−4 y '+ y=2t 3+3 t2−1

(6 A3t+2 A2 )−4 (3 A3 t2+2 A2 t+A1)+(A3 t

3+A2t2+A1t+A0 )=2 t3+3 t 2−1

A3 t3+(−12 A3+A2 )t 2+(6 A3−8 A2+A1 ) t+A1+A0=2 t3+3 t 2−1

Equation similarity:

A3=2

−12 A3+A2=3

−12(2)+A2=3

A2=27

6 A3−8 A2+A1=0

6 (2)−8(27)+A1=0

A1=204

A1+A0=−1

204+A0=−1

A0=−205

yr=2t 3+27 t 2+204 t−205

y= y l+ yr

y (t )=c1 e( 2+√3) t+c2 e

(2−√3) t+2 t 3+27 t2+204 t−205

y (0)=1

1=c1+c2−205

c1+c2=206 .........................(1)

y '(t )=(2+√3 ) c1e(2+√3) t+(2−√3 )c2 e

(2−√3) t+6 t2+54 t+204

y '(0)=1

1=( 2+√3 )c1+(2−√3 )c2+204

(2+√3 )c1+ (2−√3 )c2=−203 ........................... (2)

By using elimination and substitution method, the value of c1 and c2 can be obtained

from (1) and (2).

−203=(2+√3 )c1+( 2−√3 )c2........................... (2)

206=(2−√3 )c1+(2−√3 )c2 ...................................(1)

2√3 c1=−203−206 (2−√3 )

2√3 c1=−715+206√3

c1=−715√3+618

6

c1+c2=206

(−715√3+6186 )+c2=206

c2=715√3+618

6

y (t )=(−715√3+6186 )e (2+√3) t+(715√3+618

6 )e( 2−√3 )t+2 t3+27 t 2+204 t−205

TASK V (March 20th, 2014)

1.d2 xd t 2

+4dxdt

+8x=(20 t 2+16 t−78)e2 t

y(0)=y’(0)=0

2.d3qd t 3

−5d2qd t2

+25dqdt

−125q=(−500 t2+465 t−387)e2 t

q(0)=q’(0)= q’’(0)=0

Solution:

–

1.d2 xd t 2

+4dxdt

+8x=(20 t 2+16 t−78)e2 t

y(0)=y’(0)=0

quadratic equation is

λ2+4 λ+8=0

λ1,2=−b±√b2−4 ac

2a

λ1,2=−4±√42−4 (1)(8)

2(1)

λ1,2=−4±√−16

2

λ1,2=−2±2 i

y l=e−2t (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 t−78)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 t−78)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 t−78)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=−78

2 (1 )+8 (0 )+20 A0=−78

20 A0=−78−2

20 A0=−80

A0=−4

yr=(t 2−4)e2 t

y = yl + yr

y (t )=e−2 t (c1cos 2t+c2sin 2 t )+(t 2−4)e2 t

y ' (t )=(−2e−2 t ) (c1 cos2 t+c2sin 2 t )+e−2t (−2c1 sin 2t+2c2cos2 t )+2 t e2 t+(t 2−4) (2e2 t )

y (0 )=0

0=c1−4

c1=4

y ' (0 )=0

0=−2c1+2c2−8

8=−2(4)+2c2

c2=8

y (t )=e−2 t (4cos 2t+8 sin 2 t )+(t2−4)e2t

2.d3qd t 3

−5d2qd t2

+25dqdt

−125q=(−500 t 2+465 t−387 )e2 t

q(0)=q’(0)= q’’(0)=0

Quadratic equation is:

λ3−5 λ2+25 λ−125=0

λ1=5 λ2=5 i

λ3=−5 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)(2e¿¿2 t)¿

¿2 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 ' ' '−5 y ' '+25 y '−125 y=(−500 t 2+465 t−387 )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 t−5 {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 }−125 {(A2 t

2+A1t+A0 )e2 t }=(−500 t 2+465 t−387 )e2 t

(2 A2+17 A1−87 A0 )e2t+ (34 A2−87 A1 ) t e2 t+(−87 A2) t 2 e2 t=−500 t2 e2t+465 t e2 t−387e2t

−87 A2=−500

A2=50087

34 A2−87 A1=465

34 (50087 )−87 A1=465

A1=−23455

7569

2 A2+17 A1−87 A0=−387

2( 50087 )+17 (−23455

7569 )−87 A0=−387

A0=2617468658503

yr=( 50087t 2−23455

7569t+ 2617468

658503 )e2 t

y= y l+ yr

y (t )=c1e5 t+c2 cos5 t+c3sin 5 t+( 500

87t 2−23455

7569t+ 2617468

658503 )e2 t

y ' (t)=5 c1 e5t−5c2 sin5 t+5c3cos 5t+( 1000

87t−23455

7569 )e2 t+(50087t 2−23455

7569t+ 2617468

658503 ) (2e2 t )

y ' ' (t )=25c1 e5 t−25c2cos 5t−25c3 sin 5 t+ 1000

87e2 t+(1000

87t−23455

7569 )( 2e2 t )+( 100087

t−234557569 ) (2e2 t )+( 500

87t 2−23455

7569t+ 2617468

658503 ) (4 e2 t )

y (0 )=0

0=c1+c2+2617468658503

c1+c2=−2617468

658503 ................................... (1)

y ' (0 )=0

0=5c1+5c3−234557569

+5234936658503

5c1+5c3=−3194351

658503

c1+c3=−31943513292515

.................................... (2)

y ' ' (0 )=0

0=25c1−25c2+1000

87−46910

7569− 46910

7569+ 10469872

658503

25c1−25c2=−9876532

658503 .................................... (3)

25c1+25c2=−65436700

658503 ................................... (1)

50c1=−75313232

658503

c1=−7531323232925150

From equation (1), we get the value of c2.

c2=−2617468

658503+75313232

32925150

c2=−5556016832925150

From equation (2), we get the value of c3.

c3=−31943513292515

+ 7531323232925150

c3=4336972232925150

+

y (t )=−7531323232925150

e5 t−5556016832925150

cos5 t+ 4336972232925150

sin 5 t+( 50087t 2−23455

7569t+ 2617468

658503 )e2 t