Differential Equations 2 Equations of the form d2y dx

2



f x 

d2y

dy a  b  cy  0 2 dx dx

a

d2y dx

2

b

dy  cy  hx  dx

are Second Order Differential Equations.

Type 1 d2y dx 2

 f x 

can be solved by integrating twice (assuming that the function f x  and its integral can be integrated).

d2y  3 1  x   0 subject to the conditions that dx 2 dy  1 ). y(0)  0 , y '(0)  1 (ie when x  0 , y  0 and dx d2y d2y 5  3 1  x   3  3 x 5 2  3 1  x   0  dx 2 dx dy 3x 2 5  3 x  A Integrating gives dx 2 dy 5 1  0  0  A  A5 1  y '(0)  1 ie x  0 , dx dy 3x 2 5  3x  5 giving dx 2 3x 2 3x3 5y    5x  B integrating again gives 2 3 2 5 0  0  0  0  B  B0 y(0)  0 ie x  0 , y  0  3x 2 3x3 5y    5x 2 6 hence solution is 3x 2 3x3 y  x 10 30 or

Example Solve the equation 5

Type 2 Linear equations with constant coefficients of the form: d2y dy a  b  cy  0 dx dx 2 dy  cy  0 has solution y  kecx so it seems logical The first order, linear equation dx to try y  emx as the solution for a second order Linear Differential equation... ye

mx

dy d2y mx   me   m2emx 2 dx dx

am2emx  bmemx  ce mx  0

am2  bm  c emx  0

substituting into the equation gives

am2  bm  c  0.

but as emx  0 this gives

This equation is often called the auxiliary equation and there are 3 cases to consider depending on whether the roots of the quadratic equation are (i) real and different, (ii) real and equal or (iii) complex.

Case (i) Two distinct roots Let the roots of the auxiliary equation

am2  bm  c  0 be m1  p and m2  q

then y  e px , y  eqx are solutions of the differential equation and it looks as though the general solution could be y  Ae px  Be qx

( A and B are arbitrary constants), Example Find the general solution of the equation

d2y dx

2

5

dy  6y  0 dx

2

m  5m  6  0

The auxiliary equation is which factorises to give with roots

m  1m  6  0

General solution is

y  Ae  x  Be 6 x

m1  1, m2 6

Case (ii) Roots real and equal Experience shows that if the auxiliary equation has equal roots

am2  bm  c  0 m1  p and m2  p

then the general solution is

y   Ax  B e px 1

Example d2y

Find the general solution of the equation

dx 2

6

dy  9y  0 dx

The auxiliary equation is which gives

m 2  6m  9  0 m1  3 , m2  3

and general solution

y   Ax  B  e3x

Case (iii) complex roots If the auxiliary equation

am2  bm  c  0

has complex roots of the form

m    j where j   1

the solution is

y  ex  A cos x  B sin x  d2y

Example Solve the equation

2

dy  4y  0 dx

dx 2 m2  2m  4  0

The auxiliary equation is

2  12 2 22 j 3   1 j 3 2 y  e x  A cos x 3  B sin x 3  .   m

which gives



The general solution is







Exercise 1 Solve the equations d 2 y dy 1.  0 dx 2 dx 4. 7. 10.

d2y 2

dx d2y dx 2 d2y dx 2



dy  2y  0 dx

y0 3

dy  10 y  0 dx

d2y

dy 13.  y0 dx 2 dx 15.

d2y

2

dy  4y  0 dx

2. 2

d2y



dy 0 dx

dx 2 d2y dy 5. 2  y  0 2 dx dx d2y dy 8. 3 2  y 0 dx dx 2 11.

d2y dx 2

14.

 9y  0

d2y dx

16. 3

2

 10

d2y

4

3. 6.

d2y dx 2 d2y 2

 4y  0 5

dy  6y  0 dx

dx d2y dy 9. 4 4  y 0 dx dx 2 d2y 12.  9y  0 dx 2

dy  25 y  0 dx dy y0 dx

dx 2 dx 2 d2y dy d2y dy 17. a 2  2a  y  0 18.  6  13 y  0 dx dx dx 2 dx 2 2 2 dx d x dx k  u, when t  0. 19. Solve the equation k  x  0 , given x  0, 2 dt dt 2 dt ds d 2s ds  u, when t  0. 20. Solve the equation  4  13s  0 , given s  a, 2 dt dt dt 2

Type 3 Linear equations with constant coefficients of the form a

d2y dx

2

b

dy  cy  f x  dx

This is an extension of type 2.

We can get the solution of in the form

d2y

dy  cy  0 dx dx y  Au1  Bu2 where u1 and u2 are functions of x .

a

2

b

If we can find a particular function y  g ( x) which satisfies the equation a

d2y dx

2

b

dy  cy  f x  dx

then we can write the general solution as y  Au1  Bu2  g( x) The main problem is finding the function y  g ( x) !

y  Au1  Bu2 is often called the and y  g ( x) the

complementary function

particular integral.

It quite logical to assume that the function y  g ( x) will be the same sort of function as f ( x ) . For example if f ( x ) is of the form e x then g ( x ) must be of the form ke x or kxe x ... as there are no combinations of other functions which will give the correct f ( x ) . There are a number of different possibilities three of which are discussed here.

3

case (i) f ( x) of the form ke mx Examples d2y

dy d 2 y dy 3x Solve the equations (a)   2y  e (b)   2 y  e2 x 2 2 dx dx dx dx d 2 y dy In both the complementary function is the solution of   2y  0 dx 2 dx

This has auxiliary equation which gives

m2  m  2  0 m1  2 , m2  1.

Hence complementary function is

y  Ae 2 x  Be  x .

(a)

d2y



2

dy  2 y  e3 x has Complementary function y  Ae 2 x  Be  x dx

dx To find a particular integral try

y  ke3x

y  ke3 x 

So, for y  ke3x to satisfy

then

d2y dx 2



dy d2y  3ke3x   9ke3 x 2 dx dx

dy  2 y  e3 x dx

9ke3 x  3ke3 x  2ke3 x  e3 x 4ke3 x  e3 x  k  1

4

Hence the general solution is

(b)

d2y 2



y  Ae 2 x  Be  x  1 e3x 4

dy  2 y  e2 x has Complementary Function y  Ae 2 x  Be  x dx

dx There is a term Ae2x in the complementary function so the particular integral cannot

be of the form y  ke2x

so we try

y  kxe2 x

dy d2y  ke2 x  2kxe2 x   2ke2 x  2ke2 x  4kxe 2 x  4ke2k  4kxe 2 x 2 dx dx d 2 y dy 2x So, for y  kxe to satisfy   2 y  e2 x 2 dx dx 2x e  4k  4kx    k  2kx   2kx   e 2 x

y  kxe2 x



e 2 x  4k  4kx  k  2kx  2kx   e 2 x 3ke2 x  e k  13

Hence the general solution is

y  Ae 2 x  Be  x  1 xe 2 x 3

4

case (ii) f ( x) is a polynomial of degree n A particular integral can be found by substituting y  a0 x n  a1x n 1  a2 x n 2  a3 x n 3  .........  an

The highest power in the particular function cannot be higher than the highest power in the function f ( x ) .

Example Solve the equation

d 2s 2

 4s  t 2  2

dt The auxiliary equation is m  4  0 which gives m  2 j . Hence the complementary function is s  A cos 2t  B sin2t 2

To find a particular integral try s  a0t 2  a1t  a2 ds d 2s  2a0t  a1   2a0 dt dt 2 d 2s To satisfy the equation  4s  t 2  2 2 dt s  a0t 2  a1t  a2 





2a0  4 a0t 2  a1t  a2  t 2  2 4a0t 2  4at1  2a0  4a2  t 2  2

Equating coefficients of t 2 term t term

constant term

 a0 

4a0  1 4a1  0

1 4

 a1  0

 a2 

2a0  4a2  2

3 8





Particular Integral is s  1 t 2  3  1 2t 2  3 4

8

8





General solution s  A cos 2t  B sin2t  1 2t 2  3 8

5

case (iii) f ( x) is of the form k1 cos cx  k2 sin cx (where k1 or k2 may be zero). In general a particular integral can be found by trying

y  p cos cx +q sincx

Example Solve the equation

d 2r d 2

4

dr  5r  12 sin2 d

Auxiliary equation

m2  4m  5  0

gives complementary function

r  Ae 5  Be r  p cos 2 +q sin2 dr  2 p sin2 +2q cos 2 d

Try particular integral

d 2r d 2 d 2r

To satisfy the equation

d

2

 4 p cos 2 4q sin2

4

dr  5r  12 sin2 d

 4 p cos 2  4q sin2  4 2 p sin2  2q cos 2   5 p cos 2  q sin2   12 sin  4 p cos 2  8q cos 2  5 p cos 2  4q sin2  8 p sin2  5q sin2  12 sin

cos 2  9 p  8q   sin2  9q  8 p   12 sin Equating terms in

cos 2 :  9 p  8q  0 sin2 :  9q  8 p  12

Solving  9 p  8q  0   9 p  8q  0   81 p  72q  0      9q  8 p  12   8 p  9q  12   64 p  72q  96  Adding gives Substituting in gives General solution

 145 p  96  p   96

145

q   108 145

r  Ae 5  Be  96 cos 2  108 sin2 145

145

6

Further examples d 2r

(a) Solve the equation

d

2

4

dr  5r  7  34  15 2 d

From the previous example

r  Ae 5  Be

the complementary function is

r  a 2  b  c dr d 2r  2a  b;  2a d d 2

Try particular integral

d 2r

To satisfy

d

2

4



dr  5r  7  34  15 2 d



2a  42a  b   5 a 2  b  c  7  34  15 2

2a  4b  5c   8a  5b  5a 2  7  34  15 2 Equating

 2 terms  terms constant terms

 5a  15 8a  5b  34 2a + 4b - 5c = -7

Solving gives

a  3 , b  2 , c  1

General Solution

r  Ae 5  Be  3 2  2  1

(b) Solve the equation

d2y dx

2

 16 y  2 cos 4 x  20e 2 x (a more complex equation!)

Auxiliary Equation

m2  16  0  m  4 j

Complementary Function

y  A cos 4 x  B sin4 x

The Particular Integral must have terms to give both the 2cos 4 x term and the 20e2 x term when put into

d2y dx 2

 16 y .

We would normally go for y  p cos 4 x  q sin4 x  ke2 x but we already have a cos 4x term in the CF so we need to use

y  x p cos 4 x  q sin4 x   ke2 x dy  x 4 p sin 4 x  4q cos 4 x    p cos 4 x  q sin 4 x   2ke 2 x dx d2y

This gives dx 2

 x 16 p cos 4 x  16q sin 4 x    4 p sin 4 x  4q cos 4 x    4 p sin 4 x  4q cos 4 x   4ke 2 x  16 x p cos 4 x  q sin 4 x   8 p sin 4 x  q cos 4 x   4ke 2 x 7

To satisfy the differential equation

d2y dx

2

 16 y  2 cos 4 x  20e 2 x

 16 x p cos 4 x  q sin 4 x   8 p sin 4 x  q cos 4 x   4ke2 x  16 x p cos 4 x  q sin 4 x   16ke2 x  2 cos 4 x  20e 2 x  8 p sin 4 x  8q cos 4 x  20ke2 x  2 cos 4 x  20e 2 x sin4 x :  8 p  0  p  0 cos 4 x : 8q  2  q  1

Equating terms in

4

e 2x :

20k  20  k  1

Hence General Solution is y  A cos 4 x  B sin4 x  1 x sin4 x  e 2 x 4

Exercise 2 Solve the following 1. 3.

d2y

 9 y  2e 2 x

dx 2 d2y

 y  3 cos 2 x

dx 2

d y dx

4.

2

2

 y  x  3x

6. 3

d y

dy x 7. 2   3y  2 dx 2 dx

11.

d2y dx 2 d2y dx 2

13. 4 15.

dx 2

d2y

8.

 4 y  4e  2 x 2

d2y dx 2

10.

dy  5 y  80e 3 x dx

12.

 y  ex  x

dy 6  14 y  6 cos x  13 sin x dx dx 2

17. Solve the equation 2

dx 2 d2y dx 2 d2y dx 2

dt 2

dy  16 y  e  x dx

2

dy  y  100 sin3x dx

2

dy  5 y  15 dx

8

dy  25 y  50 dx

2

dy  2y  1  x  x2 dx

dx 2

dx 2

dy y2 dx

6

d2y

d2y

3



dy  4x  1 dx

 9 y  3 cos 2 x  sin2 x

 4  sint given that  

 6

,

d  0 when t  0 . dt

dy  2 y  4  x , find y in terms of x , given that, when x  0 , dx dx dy y  4,  2. dx d2y 19. Obtain the solutions of the equation  9 y  2 sin3x given that when x  0 , 2 dx dy y  0, 0. dx

18. If

d y

d 2

16.

2

d2y

14. 2

d2y

2

d2y dx

2

9.

d2y

dx

2

5. 4

2. 4

2

3

8

ANSWERS Exercise 1  2x

3. y  Ae 2 x  Be 2 x

1. y  Ae x  B

2. y  Ae

4. y  Ae 2 x  Be  x

5. y   Ax  B e x

7. y  A cos x  B sin x

8. y  Ae 3  Be  x

10. y  Ae5 x  Be 2 x

13. y  e

11. y  Ae3x  Be 3x

 

 

x

2

12. y  A cos 3x  B sin3x

x

16. y  Ae x  Be 3

15. y  e x A cos x 3  B sin x 3 x

18. y  e 3 x  A cos 2 x  B sin 2 x 

2 

20. s  e 2t a cos 3t  u  2a sin3t

17. y   Ax  B e a



 19. x  2u e 2 sin kt k

9. y   Ax  B e

14. y   Ax  B e 5 x

 3   3   A cos 2 x   B sin 2 x 

kt

6. y  Ae3x  Be 2 x

x

x  2



B

3



Exercise 2 1. y  Ae3x  Be 3x  2 e2 x

2. y  Ae

3. y  A cos x  B sin x  cos 2 x

4. y  Ae8 x  Be 2 x  1 e x

5

x

4

 Be x  2 9

5. y 

x Ae 2



x Be 2

7. y  Ae x  Be

 3x 2

6. y   Ax  B e  x  23 cos 3x  4 sin3x 

 x 2  3x  8



 1 9 x3  9 x 2  42 x  26 54

8. y  e  x  A cos 2 x  B sin2 x   3 10. y  e 4 x  A cos 3x  B sin3x   2





9. y  Ae 2 x  Be 2 x  xe 2 x 11. y  e x  A cos 2 x  B sin2 x   10e 3 x



12. y  e x  A cos x  B sin x   1 x 2  x  1

2 

2

2 

13. y  A cos x  B sin x  x  1 e x

15. y  e

3x

5

x

14. y  A  Be 2  2 x 2  9 x

A cosx 5   B sinx 5  sin x

16. y  A cos 3x  B sin3x  1 3 cos 2 x  sin2 x  5

17.   1  cos 2t  sin2t  2 sint  19. y  1 (e 3x  e 3x  2 sin3x) 6 18





18. y  1 4e x  e 2 x  2 x  11 4

9