6.Homogeneous Linear Equations With Constant Coefficients

6th Topic Partial Differential Equations Homogeneous linear equations with constant co-efficients (Complementary function and Particular Integral)

Prepared by: Dr. Sunil NIT Hamirpur (HP) (Last updated on 03-09-2007)

Homogeneous linear equations with constant co-efficients:

Consider a partial differential equation of the form

 ∂ n z ∂nz ∂nz ∂ n z  a0 + a1 + a2 + .......... + a n + n −1 n −2 2 n   ∂x n ∂x ∂y ∂x ∂y ∂y    ∂ n −1z ∂ n −1z ∂ n −1z ∂ n −1z   b0 + b1 + b2 + .......... + b n −1 + n−2 n −3 2 n −1   ∂x n −1 ∂ ∂ ∂ ∂ ∂ x y x y y    ∂z ∂z   + n 0 z = 0. + mn .............. +  m 0 ∂ ∂ x y   i.e. φ(D, D′) = 0 Here a 0 , a1 , a 2 ,............., a n , b 0 , b1 , b 2 ,............., b n −1 , m 0 , m1 , n 0 are all constants. In this equation the dependent variable z and its derivatives are linear, since each term in the LHS contains z or its derivatives. This equation is called a homogeneous linear partial differential equation of the nth order with constant co-efficients . Some authors call an equation homogeneous if each term in equation is of the same order. Then definition is like this:

An equation of the form

2 Partial Differential Equations: Homogeneous linear equations with constant co-efficients Prepared by: Dr. Sunil, NIT Hamirpur (HP)

a0

∂nz ∂x

n

+ a1

∂nz ∂x

n −1

∂y

+ a2

∂nz ∂x

n −2

∂y

2

+ .......... + a n

∂nz ∂y

n

= F( x, y) ,

(i)

where a 0 , a1 , a 2 ,............., a n are constants, is called a homogeneous linear partial differential equation of the nth order with constant co-efficients .

Here, all the partial derivatives are of the nth order. Writing D for

∂ ∂ and D′ for , (i) can be written as ∂x ∂y

D n + a1D n −1D′ + a 2 D n − 2 D′ 2 + .......... + a n D ′ n z = F(x , y) ⇒ φ(D, D ′)z = F( x, y) .

(ii)

As in the case of ordinary linear differential equations with constant co-efficients the complete solution of (ii) contains of two parts: (i)

The complimentary function (C.F.), which is the complete solution of the equation

φ(D, D′)z = 0 . It must contain n arbitrary functions where n is the order of the differential equation. (ii)

The particular integral (P.I.), which is a particular solution (free from arbitrary constants) of φ(D, D′)z = F( x, y) . The complete solution of (ii) is z = C.F. + P.I.

Rules for finding Complementary function (C.F.):

We explain the procedure for finding the complimentary function of a differential equation of the second order. It can be easily extended to differential equation of the higher order.

∂ 2z

∂ 2z ∂2z + a1 + a2 2 = 0 Consider the equation ∂x∂y ∂x 2 ∂y which in symbolic form is D 2 + a1DD′ + a 2 D ′ 2 z = 0 . Its auxiliary equation (A.E.) is m 2 + a 1m + a 2 = 0 . Here we put m =

D D′

(i) (ii)

. Let its roots be m1, m 2 .

Case I.: When the A.E. has distinct roots i. e. m1 ≠ m 2 then (ii) can be written as

(D − m1D′)(D − m 2 D′)z = 0 . Now the solution of (D − m 2D')z = 0 will be the solution of (iii)

(iii)


But (D − m 2 D′)z = 0 ⇒ p − m 2q = 0 , which is of Langrange’s form and the auxiliary equations are

dx 1

=

dy

− m2

=

dz 0

.

The first two members give dy + m 2dx = 0 ⇒ y + m 2 x = a Also dz = 0 ⇒ z = b

∴ z = f 2 (y + m 2 x ) is the solution of (D − m 2 D′)z = 0 Similarly, (iii) will also satisfied by the solution of

(D − m1D′)z = 0 i. e. by z = f 1 (y + m1x ) Hence the complete solution of (ii) is f 1(y + m1x ) + f 2 (y + m 2 x ) . Case II.: When the A. E. has equal roots, each = m, then (ii) can be written as

(D − mD′)(D − mD′)z = 0 .

(iv)

Let (D − mD′)z = u , then (iv) becomes (D − mD ′)u = 0 Its solution is u = f ( y + mx ) , as proved in case I.

∴ (D − mD′)z = u takes the form (D − mD′)z = f ( y + mx)

⇒ p − mq = f ( y + mx ) The auxiliary equations are

Now Also

dx 1 dx 1

= =

dy

−m dz f (a )

[Langrange’ form] dx 1

=

dy

−m

=

dz f ( y + mx )

.

gives dy + mdx = 0 ⇒ y + mx = a . gives dz = f (a )dx ⇒ z = f (a ) x + b i. e. z − xf ( y + mx ) = b .

∴ The complete solution of (ii) is z − xφ(y + mx ) = φ(y + mx ) ⇒ z = f ( y + mx) + xf ( y + mx ) .

(

)

2 2 Note1. We know that for the differential equation D + a1DD ′ + a 2 D′ z = 0 ,

the A.E. is m 2 + a1m + a 2 = 0 .

(i)

Now the roots of (i), considered as a quadratic in D / D′ , are the same as those of m 2 + a1m + a 2 .

∴ This quadratic in m can also be called the A.E. Hence, we shall write the A.E. in terms of m.

(ii)


Thus, the A.E. is obtained by putting D = m and D ′ = 1 in φ(D, D′) = 0 . Hence, the A.E. of differential equation φ(D, D′)z = F( x, y) is φ(m, 1 ) = 0 . Note 2. Generalizing the results of case I and case II, we have

(i)

if the roots of A.E. are m1, m 2 , m3 ,.............. (all are distinct roots), then C.F. = f 1(y + m1x ) + f 2 (y + m 2 x ) + f 3 (y + m3x ) + .................

(ii)

if the roots of A.E. are m1, m 2 , m3 ,.............. (two equal roots), then C.F. = f 1(y + m1x ) + xf 2 (y + m1x ) + f 3 (y + m 2 x ) + .................

(iii)

if the roots of A.E. are m1, m 2 , m3 ,.............. (three equal roots), then C.F. = f 1(y + m1x ) + xf 2 (y + m1x ) + x 2f 3 (y + m1x ) + .................

∂ 2z

∂ 2z ∂ 2z + 2 2 =0 . Q.No.1.: Solve the following partial differential equation: 2 2 + 5 ∂x∂y ∂x ∂y Sol.: The given equation in symbolic form is 2D 2 + 5DD′ + 2D ′ 2 z = 0 .

Its auxiliary equation is 2m 2 + 5m + 2 = 0 ⇒ m = −2, −

 

Here, the complete solution is f 1 (y − 2x ) + f 2  y −

1 2

.

1  x . 2 

This complete solution can also be written as φ1 (y − 2 x ) + φ 2 (2 y − x ) .

∂ 2z

∂2z ∂ 2z − −6 2 =0 . Q.No.2.: Solve the following partial differential equation: ∂x 2 ∂x∂y ∂y

(

)

Sol.: The given equation in symbolic form is D 2 − DD ′ − 6D ′ 2 z = 0

Its auxiliary equation is m 2 − m − 6 = 0 ⇒ (m − 3)(m + 2) = 0 ⇒ m = 3, − 2 . Hence, its complete solution is z = f 1(y + 3x ) + f 2 (y − 2x ) . 2

Q.No.3.: Solve the following partial differential equation: (D + 2D′)(D − 3D′) z = 0 . 2

Sol.: The given equation is (D + 2D')(D − 3D') z = 0 . 2

Its auxiliary equation is (m + 2)(m − 3) = 0 ⇒ m = −2, 3, 3 . Hence, the complete solution is z = f 1(y − 2x ) + f 2 (y + 3x ) + xf 3 (y + 3x ) . Q.No.4.: Solve the following partial differential equation: 4r − 12s + 9 t = 0 .


Sol.: The given equation is 4r − 12s + 9 t = 0 .

Its symbolic form is 4D 2 − 12DD′ + 9D′ 2 z = 0 .

∂ 2z

∂ 2z ∂ 2z 2 = = = = = Since r = D z , s DD ' z , t D ' z. ∂x∂y ∂x 2 ∂y 2 2

2

Its auxiliary equation is 4m 2 − 12m + 9 = 0 ⇒ (2m − 3) = 0 ⇒ m =

 

Hence, the complete solution is z = φ1 y +

3 3 , . 2 2

3   3  x  + xφ2  y + x  , 2   2 

This complete solution can also be written as z = f 1(2 y + 3x ) + xf 2 (2 y + 3x ) . Q.No.5.: Solve the following partial differential equation: r + 6s + 9t = 0 . Sol.: The given equation is r + 6s + 9t = 0 .

Its symbolic form is D 2 + 6DD′ + 9D′ 2 z = 0 .

∂ 2z ∂2z 2 ′ ′ = Since r = D z , s D D z , = = = = t D z. 2 ∂x∂y ∂x 2 ∂y ∂ 2z

2

Its auxiliary equation is m 2 + 6m + 9 = 0 ⇒ m = −3, − 3 . Hence, the complete solution is z = f 1(y − 3x ) + xf 2 (y − 3x ) . Rules for finding the P.I.: I. Short methods to find P.I.: (i). When F(x, y ) = eax

P.I. =

1

φ(D, D′)

+ by

e ax + by =

. 1

φ(a , b )

e ax + by (i. e. put D = a and D′ = b ) provided φ(a , b ) ≠ 0

If φ(a , b ) = 0 , then it is called a case of failure. (ii). When F(x , y ) = sin(ax + by ) .

P. I. =

(

1

φ D 2 , DD′, D′ 2

)

sin (ax + by ) =

(

1

φ − a 2 ,−ab,− b 2

)

sin (ax + by ) ,

(

)

2 2 (i. e. put D 2 = −a 2 , DD ′ = −ab , D′ = −b ) provided φ − a 2 ,−ab,−b 2 ≠ 0 .

(

)

If φ − a 2 ,−ab,− b2 = 0 , then it is a case of failure. Note: A similar rule holds when F(x , y ) = cos(ax + by ) .


(iii). When F(x , y ) = x p yq , where p, q are positive integers.

P. I. =

1

φ(D, D′)

x p y q = [φ(D, D ′)]−1 x p y q . −1

If p < q , expand [φ(D, D′)]

D

in powers of

D′ D′

If q < p , expand [φ(D, D′)]−1 in powers of Also, we have

1 D

∫

F( x , y) =

D

F( x , y)dx and

y constant

1 D′

∫

F( x, y) =

F( x, y)dy .

x constant

II. General method to find the P.I.:

F(x, y) is not always of the forms given above. Moreover, there are cases of failure in the short methods. The general method is applicable to all cases where F(x, y) is not of the forms given above or the short methods fail. Now, φ(D, D′) can be factorized, in general, into n linear factors.

∴ P.I. =

1

φ(D, D′)

=

1

F( x, y) =

.

1

(D − m1D′)(D − m 2 D′).........(D − m n D ′)

1

D − m1D′ D − m 2 D′

Let us evaluate

1 D − mD′

...........

1 D − m n D′

F( x , y)

F( x , y) .

Consider the equation, (D − mD′)z = F( x , y) ⇒ p − mq = F( x, y) . The auxiliary equations are

dx

=

1

dy

−m

=

dz F( x , y)

.

From the first two members dy + mdx = 0 ⇒ y + mx = c . From the first and last members, we have dz = F(x, y)dx = F(x , c − mx )dx

∴ z = ∫ F( x , c − mx )dx ⇒

1 D − mD′

F( x , y) =

∫

F( x , c − mx )dx ,

where c is replaced by y + mx after integration. By repeated application of above rules, the P. I. can be evaluated. Working procedure to solve the equation:

F( x , y)


∂n z ∂x n

+ k 1

∂nz

n

+ .......... + k

∂x n −1∂y

∂nz ∂y n

= F(x, y)

Its symbolic form is D n + k 1D n −1D′ + ......... + k n D′ n z = F( x, y) or briefly f ( D, D′) z = F( x , y) . Step1. To find the C.F.:

(i) Write the auxiliary equation (A.E.) i. e. m n + k 1m n

−1

+ .............. + k n = 0 and solve it for m.

(ii) Write the C.F. as follows

Roots of A. E.

C.F

1. m1, m 2 , m3 .......... (distinct roots)

f 1(y + m1x ) + f 2 (y + m 2 x ) + f 3 (y + m3x ) + ...

2. m1, m 2 , m3 .......... (two equal roots)

f 1(y + m1x ) + xf 2 (y + m1x ) + f 3 (y + m3x ) + .....

3. m1, m 2 , m3 .......... (three equal roots)

f 1(y + m1x ) + xf 2 (y + m1x ) + x 2f 3 (y + m1x ) + .....

Step2. To find the P.I.:

From the symbolic form, P. I. = (i) When F(x, y ) = eax

+ by

1 f (D, D′)

, P. I. =

F( x, y) .

1 f (D, D′)

e ax + by [put D = a and D′ = b ]

(ii) When F(x, y ) = sin (mx + ny ) or cos(mx + ny ) P. I. =

(

1 2

f D , DD′, D′

2

)

sin or cos(mx + ny ) . [put D 2 = −m 2 , DD′ = −mn , D'2 = −n 2 ]

(iii) When F( x, y) = x m y n , P.I. = −1

Expand [f (D, D ′)]

1 f (D, D′)

x m y n = [f (D, D′)]−1 x m y n .

in ascending powers of D or D ′ and operate on x m yn term by term.

(iv) When F(x, y) is any function of x and y, P. I. =

1 f (D, D ′)

F( x, y) .


1

Resolve

f (D, D ′)

into partial fractions considering f (D, D′) as a function of D alone and

operate each partial fraction on F(x, y) remembering that 1 D − mD′

F( x , y) =

∫

F( x , c − mx )dx ,

where c is replaced by y + mx after integration.

Now let us solve some homogeneous linear partial differential equation with constant co-efficients.

∂ 2z

∂ 2z Q.No.1.: Solve the following partial differential equation: − = cos x cos 2 y . 2 x y ∂ ∂ ∂x

(

)

Sol.: The given equation in symbolic form is D 2 − DD′ z = cos x cos 2 y . Step 1. To find the complementary function

Its auxiliary equation (A.E.) is m 2 − m = 0 ⇒ m(m − 1) = 0 ⇒ m = 0, 1 . Thus, C.F. = f 1 (y ) + f 2 (y + x ) . Step 2. To find the particular integral

1

Now P.I. =

D 2 − DD′

1 

= .

cos x cos 2 y =

1

2  D 2 − DD′

=

1 2

1

2 D 2 − DD′

. cos(x + 2 y ) +

cos(x + 2 y ) −

1 6

1

.

1 2

D − DD′

[cos(x + 2 y ) + cos(x − 2y )] 

cos(x − 2 y )



cos(x − 2 y ) .

Step 3. To find the complete solution

Now, since the complete solution is z = C.F. + P.I. 1

1

2

6

∴ z = f 1 (y ) + f 2 (y + x ) + cos(x + 2 y ) − cos(x − 2 y ) . Ans. ∂ 2z

∂ 2z Q.No.2.: Solve the following partial differential equation: − = sin x cos 2 y . ∂x 2 ∂x∂y Sol.: The given equation in symbolic form is D 2 − DD ′ z = sin x cos 2y . Step 1. To find the complementary function


Its auxiliary equation (A.E.) is m 2 − m = 0 ⇒ m(m − 1) = 0 ⇒ m = 0, 1 . Thus, C.F. = f 1 (y ) + f 2 (y + x ) . Step 2. To find the particular integral

Now P.I. =

1 D 2 − DD' 1

sin x cos 2 y =

1

= .

2 D 2 − DD'

1  ( ) 2 sin x cos 2 y   D 2 − DD'  2 1

[sin (x + 2 y ) + sin (x − 2 y )]

1  1 1  .sin (x + 2 y ) + sin (x − 2 y ) 2 2 2  D − DD' D − DD' 

= . =

1 2

sin (x + 2 y ) −

1 6

sin (x − 2 y ) .



1

2

6

∴ z = f 1 (y ) + f 2 (y + x ) + sin (x + 2 y ) − sin (x − 2 y ) . Ans. ∂ 3z

Q.No.3.: Solve the following partial differential equation:

(

∂x

3

−2

∂ 3z 2

∂x ∂y

= 2e2 x + 3x 2y .

)

Sol.: The given equation in symbolic form is D3 − 2D 2D' z = 2e2 x + 3x 2 y . Step 1. To find the complementary function 3 2 Its auxiliary equation (A.E.) is m − 2m = 0 ⇒ m = 0, 0, 2 .

Thus, C.F. = f 1 (y ) + xf 2 (y) + f 3 (y + 2x ) . Step 2. To find the particular integral

Now P.I. =

=2

=

1 4

1 3

D − 2D D '

1 23 − 2.22 (0) e2x +

2

(2e2x + 3x 2 y) = 2

1 3

2

D − 2 D D'

e2 x + 3

1

 

D3 1 −

2D' 

x 2y



D 

−1

e

2x

 3  2D'  1 2x 3  2 D ' 4 D '2 2  + 3 1 − + 2 + .....  x 2 y  x y = e + 3 1 +  D  4 D D  D  D 

3  2 2 2  1 2x 3  2 2 3   1 x y x . 1 e x y x  Q f ( x ) = + = + +    D 3   D D3  D3   4

∫

 

f ( x )dx 


=

=

1 4

e

2x

e2x 4

+ 3y

+

x5y 20

x5 3.4.5

+

x6 60

+ 2.

x6

 1 Q D3 f ( x ) = ∫ 

4.5.6

[∫ (∫

 

) ]

f ( x )dx dx dx 

.


Now, since the complete solution is z = C.F. + P.I.

∴ f 1(y ) + xf 2 (y ) + f 3 (y + 2 x ) +

( 15e 2 x + 3x 5 y + x 6 ) . Ans. 60 1


(D3 − 7DD'2 −6D'3 ) = sin(x + 2y) + e2x + y . + Sol.: The given equation is (D3 − 7 DD'2 −6D'3 ) = sin (x + 2 y ) + e2 x y . Step 1. To find the complementary function

The auxiliary equation (A.E.) is m3 − 7m − 6 = 0 ⇒ (m + 1)(m − 3)(m + 2) = 0

⇒ m = −1, 3, − 2 . Thus, C.F. = f 1 (y − x ) + f 2 (y + 3x ) + f 3 (y − 2x ) . Step 2. To find the particular integral

Now P.I. =

= =

=

= =

1 3

2

3

D − 7 DD' −6D' 1 3

2

3

D − 7DD' −6D'

1

( )

( )

2

2

[sin(x + 2 y) + e2x + y ]

sin (x + 2 y ) +

( ) 2

D D − 7D D' − 6D' D'

( ) − 7D(− 2 )− 6D' (− 2 ) 2

D −1

1 27 D + 24D'

( ) 2

2

sin (x + 2 y ) −

D

27 − 1 + 24(− 1.2)

3

2

1 12

e2x + y =

1 12

e2 x + y

1 2

3

1 12

2

27 D + 24DD' +y



=−

e2 x

+y

e 2x + y

D

e2 x

3

2 − 7(2)(1) − 6.1

sin (x + 2y ) −

sin (x + 2 y ) −

3

D − 7 DD' −6D'

sin (x + 2 y ) +

1

2

1

1 75

sin (x + 2 y ) −

cos(x + 2 y ) −

1 12

1 12

e2x + y

e2 x + y


∴ f 1 (y − x ) + f 2 (y + 3x ) + f 3 (y − 2 x ) −

1 75

cos(x + 2 y ) −

1 12

e2 x + y .

∂ 2z

∂ 2z ∂ 2z Q.No.5.: Solve the following partial differential equation: +3 + 2 2 = x +y . ∂x∂y ∂x 2 ∂y Sol.: The given equation in symbolic form is D 2 + 3DD'+2D' 2 z = x + y . Step 1. To find the complementary function

Its auxiliary equation (A.E.) is m 2 + 3m + 2 = 0 ⇒ m = −1, − 2 . Thus, C.F. = f 1 (y − x ) + f 2 (y − 2x ) . Step 2. To find the particular integral

Now P.I. =

1 2

D + 3DD'+2D'

(x + y ) = 2

1

 D' D' 2  2 D 1+ 3 + 2  D D 2  

(x + y )

−1

2 1   3D' 2D'   1  3D'   1 +  (x + y ) = = + 1− + .........(x + y )  2 2 2  D D  D   D   D 

=

=

1 

3

2

D

x + y −

D  1 D

(yx − x ) = 2

 

D

yx 2

x3

(1) =

2

−

1

3

[x + y − 3x ] = 2

1 D

2

(y − 2 x )

.



∴ f 1 (y − x ) + f 2 (y − 2x ) +

yx 2 2

−

x3 3

.

Q.No.6.: Solve the following partial differential equation: r − 4s + 4t = e

2x + y

∂ 2z

∂ 2z ∂ 2z 2x + y Sol.: The equation can be written as − + = 4 4 e . ∂x∂y ∂x 2 ∂y 2 The given equation in symbolic form is D 2 − 4DD'+4D' 2 z = e 2 x + y . Step 1. To find the complementary function 2

Its auxiliary equation (A.E.) is m 2 − 4m + 4 = 0 ⇒ (m − 2) = 0 ⇒ m = 2, 2 .

.


Thus, C.F. = f 1 (y + 2x ) + xf 2 (y + 2x ) . Step 2. To find the particular integral

Now P.I. =

1 2

(D − 2D')

+y

e2 x

, putting D = 2 and D' = 1 , the denominator vanishes.

∴ It is a case of failure. Since

1 D − mD'

F( x, y) =

∫

F(x, c − mx )dx , where c is to be replaced by (y + mx ) after

integration.

∴ P.I. = =

1

.

1

D − 2D' D − 2D' 1 D − 2D '

xe y

+ 2x

e 2x + y =

1

D − 2D ' ∫

e 2 x + (c−2 x ) dx =

= ∫ xe(c − 2x )+ 2 x dx = ec ∫ xdx =

1 2

1

D − 2 D' ∫

x 2 .ec =

1 2

x 2e y

e c dx + 2x

.



∴ f 1 (y + 2x ) + xf 2 (y + 2x ) +

1 2

x 2e y + 2 x .


(2D2 − 5DD'+2D'2 )z = 5sin(2x + y) . Sol.: The given equation is (2D 2 − 5DD'+2D'2 )z = 5 sin (2 x + y ) . Step 1. To find the complementary function

Its auxiliary equation (A.E.) is 2m 2 − 5m + 2 = 0 ⇒ (2m − 1)(m − 2) = 0 ⇒ m = Thus, C.F. = f 1 (2 y + x ) + f 2 (y + 2x ) . Step 2. To find the particular integral

Now P.I. = 5.

1 2D 2 − 5DD'+2D'2

sin (2x + y ) .

Putting D 2 = −2 2 , DD' = −2.1 , D'2 = −12 . Denominator = 2(− 4) − 5(− 2) + 2(− 1) = 0 .

∴ It is case of failure. Now P.I. = 5.

1

(2D − D')(D − 2D')

sin (2x + y ) = 5.

1 2D − D'

∫

sin[2x + (c − 2x )]dx

1 2

, 2.


5

1 1

= . 2

D−

5

=

5 2

D−

∫

2

D'

 

3

3

2

1 1

.

D− 5

2∫

2

x sin c D'

 

x sin  c −

 1  x  + 2x  dx 2  

  3   3   − cos c + x  − cos c + x    5  3   2  − 1.  2  dx  x sin c + x dx =  x. ∫ 3 3 2   2    2 2  

5

−5

5

x sin (y + 2x ) =

= − x cos c +

=

sin cdx =

D'

1 1

= . 2

2

∫

 

3  10  3  x  + sin  c + x  2  9  2 

x cos  y +

 1  3  1  3  10 x  + x  + sin  y + x  + x  2  2  9  2  2 

5

10

3

9

= − x cos(y + 2x ) +

sin (y + 2 x ) .



10

3

9

∴ z = f 1 (2 y + x ) + f 2 (y + 2x ) − x cos(y + 2x ) + Let f 2 (y + 2x ) +

10 9

sin (y + 2x )

sin (y + 2x ) = f 2 (y + 2 x ) + φ( y + 2x ) = F2 ( y + 2x ) . 5

∴ z = f 1 (2 y + x ) + F2 (y + 2x ) − x cos(y + 2x ) . 3

∂ 2z

∂ 2z ∂ 2z + − 6 2 = y cosx . Q.No.8.: Solve the following partial differential equation: 2 ∂ ∂ x y ∂x ∂y

(

)

Sol.: The given equation in symbolic form is D 2 + DD'−6D' 2 z = y cos x . Step 1. To find the complementary function

Its auxiliary equation (A.E.) is m 2 + m − 6 = 0 ⇒ m = −3, 2 . Thus, C.F. = f 1 (y − 3x ) + f 2 (y + 2x ) .


Step 2. To find the particular integral

1

Now P.I. =

2

D + DD'−6D' 1

=

2

1

.

D + 3D' D − 2D'

1

y cos x =

y cos x =

(D + 3D')(D − 2D') 1 D + 3D'

∫

y cos x

(c − 2x )cos xdx ,

where c is to replaced by y + mx = y + 2x after integration

= =

[(c − 2x )sin x − ∫ D + 3D′ 1

1 D + 3D'

]

1

− 2 sin xdx =

D + 3D′

[(y + 2x − 2 x )sin x − 2 cos x ]

(y sin x − 2 cos x ) = ∫ [(c + 3x )sin x − 2 cos x ]dx

where c is to replaced by y + mx = y − 3x after integration

= (c + 3x )(− cos x ) − ∫ 3(− cos x )dx − 2 sin x = (y − 3x + 3x )(− cos x ) + 3 sin x − 2 sin x = − y cos x + sin x . Step 3. To find the complete solution


∴ z = f 1(y − 3x ) + f 2 (y + 2x ) − y cos x + sin x . Ans. Q.No.9.: Solve the following partial differential equation:

∂ 3z ∂x

3

−3

∂ 3z 2

∂ x∂y

+4

∂ 3z ∂y

3

= ex + 2y .

(

3

2

3

)

Sol.: The given equation in symbolic form is D − 3D D'+4D' z = e

x +2y

Step 1. To find the complementary function

(

)

Its auxiliary equation (A.E.) is m3 − 3m 2 + 4 = 0 ⇒ (m + 1) m 2 − 4m + 4 = 0

⇒ (m + 1)(m − 2)2 = 0 ⇒ m = −1, 2, 2 . Thus, C.F. = f 1 (y − x ) + f 2 (y + 2 x ) + xf 3 (y + 2x ) . Step 2. To find the particular integral

Now P.I. =

1

x e D3 − 3D 2D'+4D'3

+ 2y

=

1

x e 13 − 3.12.2 + 4.23

+2y

=

1 27

ex

+2y

.

(putting D = 1, D' = 2 ) Step 3. To find the complete solution



∴ f 1(y − x ) + f 2 (y + 2x ) + xf 3 (y + 2x ) +

1 27

ex

+2y

. Ans.


∂ 3z ∂x 3

−4

∂ 3z ∂x 2∂y

+4

∂ 3z

= 0.

∂x∂y 2

(

)

Sol.: The given equation in symbolic form is D 3 − 4D 2 D ′ + 4DD′ 2 z = 0 . Step 1. To find the complementary function

Here auxiliary equation (A.E.) is m 3 − 4m 2 + 4m = 0 , where m =

D D′

.

⇒ m(m − 2)2 = 0 ∴ m = 0, 2, 2 . Step 2. To find the particular integral Now P.I. = 0 Step 3. To find the complete solution


∴ z = f 1 ( y) + f 2 ( y + 2x ) + xf 3 ( y + 2x ) . Ans. 3x −2 y

Q.No.11.: Solve the following partial differential equation: 4r + 12s + 9t = e

∂ 2z

∂ 2z ∂ 2z Sol.: The given equation can be written as 4 + 12 +9 = e 3x − 2 y . ∂x∂y ∂x 2 ∂y 2

(

)

The given equation in symbolic form is 4D 2 + 12DD′ + 9D ′ 2 z = e 3x − 2 y . Step 1. To find the complementary function

Here auxiliary equation (A.E.) is 4m 2 + 12m + 9 = 0 , where m = 3

3

2

2

⇒ (2m + 3)2 = 0 ⇒ m = − ,− .  

Thus, C.F. = f 1  y −

3   3  x  + xf 2  y − x  . 2   2 

Step 2. To find the particular integral

Now P.I. =

1 4D 2 + 12DD′ + 9D′ 2

Here the usual rule fails

e

3x − 2 y

.

D D′

.

.


Q

2

2

4D + 12DD′ + 9D′ = 0 for D = 3, D ′ = −2

=

=

=

1

(2D + 3D′)2 1

(2D + 3D′) ∫ 1

(2D + 3D′) ∫

= ∫ xe =

1

e 3x − 2 y =

e

(2D + 3D ) (2D + 3D′)

 3  3x − 2 c + x   2 

e − 2c dx =

 3  3 x − 2 c + x   2 

dx =

1

.

∫

e 3x − 2 y

3 3   − = = + y x 0 , y c x  2 2 

dx

1

(2D + 3D′) xe −2c dx =

xe − 2c =

1

(2D + 3D′)

xe

 

3 

− 2y − x 2



1 2 − 2c 1 2 − (2 y − 3x ) = x e x e 2 2

1 2 3x − 2 y x e . 2



 

∴ z = f 1  y −

3   3  1 2 3x 2 y x  + xf 2  y − x  + x e − 2   2  2 1

⇒ z = f 1 (2 y − 3x ) + xf 2 (2 y − 3x ) + x 2 e 3x − 2 y . Ans. 2

∂2z

∂ 2 z ∂ 2z −2 + = sin x . Q.No.12.: Solve the following partial differential equation: ∂x∂y ∂y 2 ∂x 2

(

)

Sol.: The given equation in symbolic form is D 2 − 2DD′ + D ′ 2 z = sin x . Step 1. To find the complementary function 2 Here auxiliary equation (A.E.) is m − 2m + 1 = 0 ⇒ m = 1, 1 , where m =

Thus, C.F. = f 1 (y + x ) + xf 2 (y + x ) . Step 2. To find the particular integral

Now P.I. =

1 D 2 − 2DD′ + D ′ 2

sin x =

1

− 12 − 0 + 0



sin x = − sin x .

D D′

.


∴ z = f 1 (y + x ) + xf 2 (y + x ) − sin x . Ans. Q.No.13.: Solve the following partial differential equation:

∂2y ∂t

2

−a

2

∂2 y ∂x

2

= E sin pt .

Sol.: The given equation in symbolic form is D 2 − a 2 D ′ 2 y = E sin pt .

Here D =

D ∂ ∂ , D′ = , m= . ∂t ∂x D′

Step 1. To find the complementary function

Here auxiliary equation (A.E.) is m 2 − a 2 = 0 ⇒ m = +a , − a . Thus, C.F. = f 1 (x − at ) + f 2 (x + at ) . Step 2. To find the particular integral

Now P.I. =

1 2

2

D − a D′ 2

E sin pt =

1 2

2

− p − a .0

E sin pt = −

E p

2

sin pt .


Now, since the complete solution is y = C.F. + P.I.

∴ y = f 1 (x − at ) + f 2 (x + at ) −

E p

2

sin pt . Ans.


∂2z ∂x 2

−

∂ 2z ∂y 2

= cos 2x cos 3y .

Sol.: The given equation in symbolic form is D 2 − D′ 2 z = cos 2x cos 3y . Step 1. To find the complementary function 2 Here auxiliary equation (A.E.) is m − 1 = 0 ⇒ m = −1, 1 .

Thus, C.F. = f 1 (y + x ) + f 2 (y − x ) Step 2. To find the particular integral

Now P.I. =

=

1 D 2 − D′ 2 1

1

2  − 4 + 9

1

cos 2x cos 3y = cos(2x + 3y ) +

= cos 2x cos 3y . 5

1

1

2 D 2 − D′ 2 1

−4+9

[cos(2x + 3y ) + cos(2 x − 3y )]  

cos(2 x − 3y )



Now, since the complete solution is z = C.F. + P.I. 1!

∴ z = f 1 (y + x ) + f 2 (y − x ) + cos 2x cos 3y . Ans. 5

(

)

Q.No.15.: Solve the following partial differential equation: D 2 + 3DD'+2D'2 z = 24xy .

(

)

Sol.: The given equation is D 2 + 3DD'+2D'2 z = 24xy . Step 1. To find the complementary function

(

)

Here auxiliary equation (A.E.) is m 2 + 3m + 2 = 0

⇒ (m + 1)(m + 2) = 0 ⇒ m = −1, − 2 . Thus, C. F. = f 1 (y − x ) + f 2 (y − 2x ) Step 2. To find the particular integral 2 1  3D′ 2D′  Now P.I. = 24xy = 24. + 1 +  D D 2 + 3DD ′ + 2D′ 2 D 2  D 2 

1

=

−1

xy

 24  3D′ 2D′ 2 − − − ......... xy 1 2 2 D D  D 

3x 2  x3 3 x4  24  = xy − D .x.1 = 2  xy − 2  = 24.y 2.3 − 24. 2 . 3.4 D  D2   24 

3

= 4x 3 y − 3x 4 . Step 3. To find the complete solution


∴ z = f 1 (y − x ) + f 2 (y − 2x ) + 4x 3 y − 3x 4 . Ans. Q.No.16.: Solve the following partial differential equation:

∂ 2z

∂ 2z ∂ 2 z 2 2 + + = + + 2 x xy y . ∂x∂y ∂y 2 ∂x 2 Sol.: The given equation in symbolic form is D 2 + 2DD′ + D′ 2 z = x 2 + xy + y 2 . Step 1. To find the complementary function

(

)

2 Here auxiliary equation (A.E.) is m 2 + 2m + 1 = 0 ⇒ (m + 1) = 0 ⇒ m = −1, − 1 .


Thus, C.F. = f 1 (y − x ) + xf 2 (y − x ) . Step 2. To find the particular integral 2 1  D′ D ′  2 2 Now P.I. = x + xy + y = 1 + 2 + 2  2 2 D D 2  D  D + 2DD′ + D′

(

1

)

2   2   2  ′ ′ ′ ′ 1   D D D D  + 2 +  − ..... = + 1− 2 2  2 2     D D   D D D    

=

D

=

2

[3x

2

− 3xy + y

2

4

3

] = 3 4.3 − 3y 2.3 + x

x

(x 2 + xy + y 2 )

2 2

y x 2

[ x 4 + 2x 2 y 2 − 2x 3 y]. 4

1



 1 x ∴ z = f 1 (y − x ) + xf 2 (y − x ) + 3 + 2 x 2 y 2 − 2 x 3 y . 4  3  4


(D3 + D2D'−DD'2 −D'3 )z = ex cos 2y . Sol.: The given equation is (D3 + D 2D'−DD'2 −D'3 )z = e x cos 2 y . Step 1. To find the complementary function

(

)

Here auxiliary equation (A.E.) is m 3 + m 2 − m − 1 = 0

⇒ (m + 1)(m + 1)(m − 1) = 0 ⇒ m = −1, − 1, 1 . Thus, C.F. = f 1 (y − x ) + xf 2 (y − x ) + f 3 (y + x ) . Step 2. To find the particular integral

Now P.I. =

1 D 3 + D 2 D′ − DD′ 2 − D′ 3

Take as particular integral

e x cos 2 y .

(x 2 + xy + y 2 )

−1

  x 2  2x 2 1  2 2   x + xy + y − 2 = + 2 yx  − + 4x 2    2  D 2   2  1

−1


z = Ae x cos 2 y + Be x sin 2 y . Then D 3 z = Ae x cos 2 y + Be x sin 2y DD′ 2 z = −4Ae x cos 2 y − 4Be x sin 2 y D 2 D ′z = −2Ae x sin 2 y + 2Be x cos 2 y D′ 3 z = 8Ae x sin 2 y − 8Be x cos 2 y . Substituting in the given equation, we get

(5A + 10B)e x cos 2 y + (5B − 10A )e x sin 2 y = e x cos 2 y . So that 5A + 10B = 1 and 5B − 10A = 0

∴A =

1 25

and B =

∴ P.I. is z =

1 25

2 25

x

.

e cos 2 y +

2 25

x

e sin 2 y .



∴ z = f 1 (y − x ) + xf 2 (y − x ) + f 3 (y + x ) +

e

x

25

[cos 2 y + 2 sin 2 y]. Ans.


∂2z

∂2z ∂ 2z + −6 = cos(2x + y ) . 2 2 x y ∂ ∂ ∂x ∂y

(

)

Sol.: The given equation in symbolic form is D 2 + DD′ − 6D′ 2 z = cos(2x + y ) . Step 1. To find the complementary function 2 Here auxiliary equation (A.E.) is m + m − 6 = 0 ⇒ m = −3, 2 .

Thus, C.F. = f 1 (y − 3x ) + f 2 (y + 2x ) . Step 2. To find the particular integral 2

Since D 2 + DD ′ − 6D′ 2 = −2 2 − (2)(1) − 6(− 1) = 0 .

∴ It is a case of failure and we have to apply the general method. P.I. =

1 D 2 + DD′ − 6D′ 2

cos(2 x + y ) =

1

(D + 3D′)(D − 2D′)

cos(2x + y )


= =

1

D + 3D′ ∫ 1 D + 3D′

(

)

cos 2x + c − 2x dx =

x cos c =

1

1 D + 3D ′

∫

cos cdx

x cos(y + 2x )

D + 3D′

= ∫ x cos(c + 3x + 2x )dx = ∫ x cos(5x + c )dx = = =

x sin (5x + c) 5 x 5

+

cos(5x + c)

(

)

sin 5x + y − 3x +

x 5

sin (2 x + y ) +

[Integrating by parts]

25

1 25

1 25

(

cos 5x + y − 3x

)

cos(2x + y ) .



∴ z = f 1 (y − 3x ) + f 2 (y + 2x ) +

x 5

sin (2x + y ) +

1 25

cos(2x + y ) . Ans.


∂ 3z ∂x 2 ∂y

−5

∂ 3z ∂x∂y 2

+6

∂ 3z ∂y 3

= ex .

(

)3

Sol.: The given equation in symbolic form is D 2 D′ − 5DD′ 2 + 6D′ z = e x . Step 1. To find the complementary function 3 2 Here auxiliary equation (A.E.) is 6m − 5m + m = 0 ⇒ m(3m − 1)(2m − 1) = 0

⇒ m = 0,

1 1 , . 3 2

Thus, C.F. = f 1 (x + m1 y ) + f 2 (x + m 2 y ) + f 3 (x + m 3 y )

 

= f 1 (x + 0) + f 2  x +

1   1  y  + f 3  x + y  2   3 

= f 1 (x ) + f 2 (y + 2x ) + f 3 (y + 3x ) . Step 2. To find the particular integral

Now P.I. =

1 D 2 D′ − 5DD′ 2 + 6D ′3

e

x

=

1 D′(3D′ − D )(2D′ − D )

e

x


=

1

1

D ′ (0 − 1)(0 − 1)

ex =

1 D′

ex =

∫

e x dy = ye x .



∴ z = f 1 (x ) + f 2 (y + 2x ) + f 3 (y + 3x ) + ye x . Ans. Q.No.20.: Solve the following partial differential equation:

∂ 2z

∂ 2z ∂ 2z +2 + = x2y 2 2 ∂x∂y ∂y ∂x Sol.: The given equation in symbolic form is D 2 + 2DD′ + D′ 2 z = x 2y . Step 1. To find the complementary function 2

Here auxiliary equation (A.E.) is m + 2m + 1 = 0 .

⇒ (m + 1) 2 = 0 ⇒ m = −1, − 1 . Thus, C.F. = f 1 (y − x ) + xf 2 (y − x ) . Step 2. To find the particular integral 2 1  2D ′ D′  2 Now P.I. = x y= + 1 +  2 2 2 D D + 2DD′ + D′ D  D 2 

1

−1

x2y

2    2   2  ′ ′ ′ ′ 1   2D D   2D D  1− = + + + − .......... x 2 y 2 2 2  D  D  D   D   D   

1  2 1 1 2 x y 2 x = − =  2  2 D D  D

=

x4y 12

−

2 60

5

x =

x4 60

4

5

2 3 x y 2 x  2 − x y x = −  3.4 3 4.5 3 

(5y − 2x ) .



∴ z = f 1 (y − x ) + xf 2 (y − x ) +

x

4

60

(5y − 2x ) . Ans.

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********************************

Home Assignments Q.No.1.: Solve the following partial differential equation:

∂ 3z

−4

∂x 3

∂ 3z ∂x 2∂y

+5

∂ 3z ∂x∂y 2

−2

∂ 3z ∂y3

= e2x + y .

Ans.: z = f 1 (y + x ) + xf 2 (y + x ) + f 3 (y + 2x ) − e

2x + y

.


∂ 3z ∂x 3

−4

∂ 3z ∂x 2∂y

+4

∂ 3z ∂x∂y 2

= 2 sin (3x + 2y ) .

Ans.: Q.No.3.: Solve the following partial differential equation:

∂ 2z

∂ 2z ∂ 2z −4 + 3 2 = x + 3y . 2 x y ∂ ∂ ∂x ∂y Ans.: z = f 1 (y + x ) + f 2 (y + 3x ) +

1 6

(3c − 2x )3 / 2 −

1 24

(3c − 8x )3 / 2


D 2 − DD'−2D'2 z = (y − 1)e x . Ans.: z = f 1 (y − x ) + f 2 (y + 2x ) + ye x Q.No.5.: Solve the following partial differential equation:

(D2 + 2DD'+D'2 )z = 2 cos y − x sin y . Ans.: z = f 1 (y − x ) + xf 2 (y − x ) + x sin y Q.No.6.: Solve the following partial differential equation:

∂ 3z ∂x

3

−7

∂ 3z ∂x∂y

2

+6

∂ 3z ∂y

3

= 0.

Ans.: z = f 1 (y + x ) + f 2 (y + 2x ) + f 3 (y − 3x ) . Q.No.7.: Solve the following partial differential equation:


25r − 40s + 16t = 0 . Ans.: z = f 1 (5x + 4 y ) + xf 2 (5y + 4x ) . Q.No.8.: Solve the following partial differential equation:

∂2z ∂x 2

−7

∂2z ∂ 2z + 12 = ex − y . 2 ∂x∂y ∂y

Ans.: z = f 1 (y + 3x ) + f 2 (y + 4x ) +

1 20

e

x−y

.


D 2 + 2DD′ + D′ 2 z = e 2 x + 3y . Ans.: z = f 1 (y − x ) + xf 2 (y − x ) +

1 25

e 2 x + 3y .


∂ 2z

∂ 2z ∂2z + −6 = cos(3x + y ) . 2 2 ∂ ∂ x y ∂x ∂y Ans.: z = f 1 (y − 3x ) + f 2 (y + 2x ) −

1 6

cos(3x + y) .


D 3 − 4D 2 D ′ + 4DD′ 2 z = 6 sin (3x + 2y ) . Ans.: z = f 1 (y ) + f 2 (y + 2 x ) + xf 3 (y + 2 x ) + 2 cos(3x + 2 y ) . Q.No.12.: Solve the following partial differential equation:

∂ 2z

∂2z ∂ 2z −3 +2 = e 2 x + 3y + sin (x − 2 y ) . ∂x∂y ∂x 2 ∂y 2 Ans.: z = f 1 (y + x ) + f 2 (y + 2x ) +

1 2 x + 3y 1 − sin (x − 2 y ) . e 4 15


∂ 2z

∂2z ∂ 2z +3 +2 = 12xy . 2 2 ∂ ∂ x y ∂x ∂y 3

Ans.: z = f 1 (y − x ) + f 2 (y − 2x ) + 2x y −

3 4 x . 2



D 2 − 6DD′ + 9D′ 2 z = 6x + 2 y . 2 Ans.: z = f 1 (y + 3x ) + xf 2 (y + 3x ) + x (3x + y ) .


D 2 − 6DD′ + 9D′ 2 z = 12x 2 + 36xy . Ans.: z = f 1 (y + 3x) + xf 2 (y + 3x ) + 10x 4 + 6x 3 y . Q.No.16.: Solve the following partial differential equation:

(D 2 − 2DD′ + D′ 2 )z = e x + y . Ans.: z = f 1 (y + x ) + xf 2 (y + x ) +

1 2 x+y . x e 2


r + 2s + t = 2(y − x ) + sin(x − y) . 2

3

Ans.: z = f 1 (y − x ) + xf 2 (y − x ) + x y − x +

1 2 x sin (x − y ) 2


r + s − 6t = cos(2 x + y ) . Ans.: z = f 1 (y − 3x ) + f 2 (y + 2x ) +

1 5

x sin (2x + y ) .


∂ 3z ∂x

3

−4

∂ 3z 2

∂x ∂y

+4

∂ 3z ∂x∂y

2

= 4 sin (2x + y ) . 2

Ans.: z = f 1 (y ) + f 2 (y + 2x ) + xf 3 ( y + 2x ) − x cos(2x + y) . Q.No.20.: Solve the following partial differential equation:

4r − 4s + t = 16 log( x + 2 y) . Ans.: z = f 1 (2 y + x ) + xf 2 (2 y + x ) + 2x 2 log(x + 2 y ) .

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6.Homogeneous Linear Equations With Constant Coefficients

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