7. Non-Homogeneous Linear Equations

7th Topic Partial Differential Equations Non-homogeneous linear partial differential equations

Prepared by: Dr. Sunil NIT Hamirpur (HP) (Last updated on 03-09-2007) Non-homogeneous linear partial differential equations: Definition: 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 −1 n −1 n −1   ∂ n −1z ∂ ∂ ∂ z z z  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 = F(x, y ). + mn .............. +  m 0 ∂y   ∂x i.e. f (D, D ′)z = F(x, y ) 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. Since RHS is not zero, so this equation is called a non homogeneous linear partial differential equation of the nth order with constant coefficients. Some authors call an equation non-homogeneous if LHS of the equation is not of the same order. Then definition is like this:

In the equation f (D, D′)z = F( x, y) ,

(i)

2 Partial Differential Equations: Non-homogeneous linear partial differential equations Prepared by: Dr. Sunil, NIT Hamirpur (HP)

if the polynomial f (D, D′) in D, D ′ is not homogeneous, then (i) is called a non homogeneous linear partial differential equation . Complete solution:

As in the case of homogeneous linear partial differential equations, its complete solution is = C.F. + P.I. To find the particular integral (P.I.):

The method to find P.I. is exactly the same as those for homogeneous linear partial differential equations. To find complementary function (C.F.):

We resolve φ(D, D ′) into linear factors of the form D − mD′ − a . To find the solution of (D − mD ′ − a )z = 0 .

(ii)

This can also be written as p − mq = az . Lagrange’s auxiliary equations are

dx 1

=

dy

−m

From the first two members dy + mdx = 0 . From the first and last members

dz z

= adx .

=

dz az

.

∴ y + mx = b . ∴ log z = ax + log c ⇒ z = ceax .

∴ The complete solution of (ii) is z = eax f (y + mx ) . Hence, the C.F. of (i), i. e. the complete solution of

(D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0 is z = ea1x f 1(y + m1x ) + ea 2 x f 2 (y + m 2 x ) + ............ + ea n x f n (y + m n x ) . 3

Also, in the case of repeated factor, for example (D − mD′ − a ) z = 0 ax ax 2 ax We have z = e f 1 (y + mx ) + xe f 2 (y + mx ) + x e f 3 (y + mx ) .

Now let us solve some non-homogeneous linear partial differential equation:

Q.No.1.: Solve the following non-homogeneous linear partial differential equation:

D 2 + 2DD ′ + D′ 2 − 2D − 2D ′ z = sin (x + 2 y ) . 2 2 Sol.: The given equation is D + 2DD ′ + D′ − 2D − 2D ′ z = sin (x + 2 y ) .


This is a non-homogeneous linear partial differential equation with constant co-efficients. Step 1. To find the complementary function

Here f ( D, D ′) z = ( D + D′ − 0)(D + D ′ − 2) z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0

is z = ea1x f 1(y + m1x ) + ea 2 x f 2 (y + m 2 x ) + ............ + ea n x f n (y + m n x ) . Thus, C.F. = f 1 (y − x ) + e 2 x f 2 (y − x ) . Step 2. To find the particular integral

P.I. =

1 2

D + 2DD′ + D' −2D − 2D′

=−

=− = =

2

1 2(D + D ′) + 9

sin (x + 2 y ) = −

2(D + D′) − 9 4[− 1 + 2(− 2) − 4] − 81 1

117 1 39

sin (x + 2 y ) =

(

1

− 1 + 2(− 2) + (− 4) − 2D − 2D′

2(D + D′) − 9 2

4 D + 2DD′ + D′

2

)− 81

sin (x + 2 y )

sin (x + 2 y )

sin (x + 2 y )

[2{cos(x + 2 y ) + 2 cos(x + 2y )} − 9 sin (x + 2 y )]

[2 cos(x + 2 y ) − 3 sin (x + 2 y)] .

Step 3. To find the complete solution

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

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

1 39

[2 cos(x + 2 y) − 3 sin (x + 2 y )] .Ans.


(D 2 − D′2 + 3D′ − 3D)z = e x + 2 y + xy . Sol.: The given equation is D 2 − D′ 2 + 3D ′ − 3D z = e x + 2 y + xy .


Here f (D, D′)z = D 2 − D ′ 2 + 3D′ − 3D z = (D − D′)(D + D ′ − 3) z

= (D − D′ − 0 )[D + D′ − 3]z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0


is z = e

a1 x

f 1(y + m1x ) + e

a 2x

f 2 (y + m 2 x ) + ............ + e

anx

f n (y + m n x ) .

Thus, C.F. = f 1 (y + x ) + e3x f 2 (y − x ) . Step 2. To find the particular integral (i) P.I. corresponding to e x

=

1

(D − D′)(D + D′ − 3)

= −e x

1

(1 + D′ − 3)

= −e x + 2 y

1 D′

+2y

e x + 2y =

e 2 y = −e

.1 = − e x + 2 y

(D + D′ − 3)(1 − 2) 1

x

∫

1

D′ − 2

e x + 2y = −

e 2 y = −e x .e 2 y

.1dy = − yex

+y

1 D′ + 2 − 2

1

(D + D′ − 3)

e x + 2y

.1

.

(ii) P.I. corresponding to xy

1  D ′  = .xy = − 1 −  (D − D′)(D + D′ − 3) 3D  D  1

−1

 D D′  1 − −   3 3 

=−

1  D′  D D′ 2DD′  + ......... xy 1 + + ........1 + + + 3D  D 9  3 3 

=−

1  D D ′ D′ D′ 2DD′  ......... xy 1 + + + + + 3D  3 3 D 3 9 

−1

xy

1  y x 1 x 2  1  x y xy x 2x x  + + + + . =−  xy + + + x + +  = −  3  2 3 3 9 6  3D  3 3 D 3 9  2

2

3


Now, since the complete solution is z = C.F. + P.I. 3x

∴ z = f 1 (y + x ) + e f 2 (y − x ) − ye

x + 2y

1  x y 2

−

3  2

+

xy 3

+

x

2

2

+

2x 9

+

3 x 

6 

.


(D − 3D′ − 2)3 z = 6e 2 x sin (3x + y ) . 3

Sol.: The given equation is (D − 3D′ − 2) z = 6e 2x sin (3x + y ) .

This is a non-homogeneous linear partial differential equation with constant co-efficients. Step 1. To find the complementary function 3

Here f (D, D′)z = (D − 3D ′ − 2) z .

5 Partial Differential Equations: Non-homogeneous linear partial differential equations Prepared by: Dr. Sunil, NIT Hamirpur (HP) 3

Note: Since the solution of (D − mD ′ − a ) z = 0

is z = e

ax

f 1 (y + mx ) + xe

ax

2 ax

f 2 (y + mx ) + x e

f 3 (y + mx ) .

Thus, C. F. = e2 x f 1(y + 3x ) + xe 2 x f 2 (y + 3x ) + x 2e2 x f 3 (y + 3x ) . Step 2. To find the particular integral

P.I. =

1 3

(D − 3D′ − 2)

= 6e

6e 2 x sin (3x + y ) = 6e 2 x

1

2x

2x

3

(D − 3D′)

sin (3x + y ) = 6e .

x3 6

1 3

(D + 2 − 3D′ − 2)

sin (3x + y )

.sin (3x + y ) = x 3e 2x sin (3x + y ) .



∴ z = e 2x f 1(y + 3x ) + xe2 x f 2 (y + 3x ) + x 2e 2 x f 3 (y + 3x ) + x 3e2 x sin (3x + y) . Q.No.4.: Solve the following non-homogeneous linear partial differential equation:

(D3 − 3DD′ + D′ + 4)z = e 2x + y . Sol.: The given equation is D 3 − 3DD ′ + D′ + 4 z = e 2 x + y .


Here D 3 − 3DD′ + D ′ + 4 cannot be resolved into linear factors in D and D ′ . Let z = Aehx

+ ky

.

∴ D 3 − 3DD′ + D′ + 4 z = A h 3 − 3hk + k + 4 e hx + ky . Here D 3 − 3DD′ + D ′ + 4 z = 0 , iff h 3 − 3hk + k + 4 = 0 . Thus, C.F. =

∑

Ae hx + ky , where h 3 − 3hk + k + 4 = 0 .

Step 2. To find the particular integral

P.I. =

1 3

D − 3DD ′ + D′ + 4

e

2x + y

=

e2x + y 3

2 − 3(2)(1) + 1 + 4



=

1 7

e 2x

+y

.


1

∴ z = ∑ Aehx + ky + e2 x + y , where h 3 − 3hk + k + 4 = 0 . 7

Equation reducible to Partial Differential Equations with constant co-efficients:

An equation in which the co-efficient of derivative of any order say k is a multiple of the variable of the degree k, then it can be reduced to linear partial differential equation with constant co-efficients in the following way. Let x = eX , y = e Y so that X = log x , Y = log y . Then

∂z ∂z ∂X 1 ∂z dz dz ⇒x = = . = . dx dX ∂x ∂X ∂x x ∂X

∴x

∂  ∂  = D = . ∂x ∂ X  

k −1  k k −1 ∂  z z k −1 ∂ k ∂ z k −1 ∂  Now x x =x + (k − 1)x ∂x  ∂x k −1  ∂x k ∂x k −1

⇒ x k

−

∂ k z

 ∂  k −1 ∂ k 1z = x − k + 1 x ∂x k  ∂x ∂x k −1 

Putting k = 2, 3, ……….., we get x

2

x

3

∂ 2z ∂x 2 ∂ 3z ∂x

3

= (D − 1)x

∂z = (D − 1)Dz ∂x

= (D − 2)x

Similarly, y

dz dy

2

∂ 2z ∂x

2

= (D − 2)(D − 1)Dz etc.

= D′z , y

2

∂ 2z ∂y

2

= (D′ − 1)D′z , y

3

∂3z ∂y

3

= (D′ − 2)(D′ − 1)D′z etc.

∂ 2z = DD′z........ and xy ∂x∂y Substituting in the given equation it reduces to ψ (D − D′)z = V , which is a linear partial differential equation with constant co-efficients. Q.No.5.: Solve the following partial differential equations: 2 ∂ 2z ∂z 2 ∂ z − + + = x 3y4 . x 4 xy 4 y 6 y 2 2 ∂x∂y ∂y ∂x ∂y 2

∂ 2z

2 ∂ 2z ∂z 2 ∂ z − + + = x3y4 . 4 xy 4 y 6 y Sol.: The given equation is x 2 2 ∂x∂y ∂y ∂x ∂y 2

∂ 2z


This equation is not a linear partial differential equation with constant co-efficients, but it can be reduced to linear partial differential equation with constant co-efficients in the following way. Substituting x = e X , y = eY and denoting D =

∂ ∂ , D′ = . ∂Y ∂X

Given equation reduces to [D(D − 1) − 4DD′ + 4(D′ − 1)D ′ + 6D ′]z = e 3X + 4Y

⇒ (D − 2D′)(D − 2D′ − 1)z = e 3X + 4Y . Now this is a non-homogeneous linear partial differential equation with constant coefficients. Step 1. To find the complementary function

Here f (D, D′)z = (D − 2D′)(D − 2D′ − 1)z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0

is z = ea1x f 1(y + m1x ) + ea 2 x f 2 (y + m 2 x ) + ............ + ea n x f n (y + m n x ) . Thus, C.F. = f 1 (Y + 2X ) + e X f 2 (Y + 2X )

= f 1 (log y + 2 log x ) + e log x f 2 (log y + 2 log x )

( ) ( ) = g1 (yx 2 ) + xg 2 (yx 2 ). = f 1 log yx 2 + xf 2 log yx 2


P.I. =

1

(D − 2D′)(D − 2D′ − 1)

e

3X + 4 Y

=

e3X + 4Y

[3 − 2(4)][3 − 2(4) − 1]

=

1 30

x 3y4 .



( )

( )

∴ z = g1 yx 2 + xg 2 yx 2 +

1 30

x 3y4 .

Q.No.6.: Solve the following partial differential equations:

D 2 − DD ′ + D′ − 1 z = cos(x + 2 y ) . Sol.: The given equation is D 2 − DD′ + D ′ − 1 z = cos(x + 2y ) .

This is a non-homogeneous linear partial differential equation with constant co-efficients.


Step 1. To find the complementary function

Here f (D, D′)z = D 2 − DD′ + D′ − 1 z = (D + 1)(D − 1)z − D′(D − 1)z

= (D − 1)(D − D′ + 1) = (D − 0D′ − 1)[D − D′ − (− 1)] . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0

is z = e

a1 x

f 1(y + m1x ) + e

Thus, C.F. = e x f 1(y + 0x ) + e

−x

a 2x

f 2 (y + m 2 x ) + ............ + e

f 2 (y + x ) ⇒ e x f 1(y ) + e

−x

anx

f n (y + m n x ) .

f 2 (y + x ) .


P.I. =

=

=

1 2

D − DD′ + D ′ − 1

cos(x + 2 y )

1

− 12 − (− 1.2) + D′ − 1 1 D′

cos(x + 2 y ) =

cos(x + 2y )

∫

[putting D 2 = −12 and DD′ = −1.2 ] 1

cos(x + 2 ydy) =

2

x constant

sin (x + 2 y ) .


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

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

∂ 2z

∂ 2 z ∂z Q.No.7.: Solve the following partial differential equations: + + − z = e− x . 2 ∂x∂y ∂y ∂x ∂ 2z

∂ 2z ∂z −x z e . Sol.: The given equation is + + − = ∂x 2 ∂x∂y ∂y

(

)

This equation in symbolic form can be written as D 2 + DD′ + D ′ − 1 z = e − x . This is a non-homogeneous linear partial differential equation with constant co-efficients. Step 1. To find the complementary function

(

)

Here f (D, D′)z = D 2 + DD′ + D ′ − 1 z = (D + 1)(D + D′ − 1)z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0

is z = e

a1 x

f 1(y + m1x ) + e

a 2x

f 2 (y + m 2 x ) + ............ + e

Thus, C.F. = e − x f 1 (y ) + e x f 2 (y − x ) .

anx

f n (y + m n x ) .



P.I. =

=

1 2

D + DD′ + D ′ − 1 1

1

D + 1 (− 1 + 0 − 1)

e− x =

1

(D + 1)(D + D′ − 1)

e− x = −

1

1

2 D +1

e− x = −

e− x

1 1 −x 1 = − xe − x . x e 2 1 2



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

1 2

xe − x . Ans.


(D − D′ − 1)(D − D′ − 2)z = e 2x − y . Sol.: The given equation is (D − D′ − 1)(D − D′ − 2 )z = e 2 x − y .


Here f (D, D′)z = (D − D′ − 1)(D − D ′ − 2 )z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0

is z = ea1x f 1(y + m1x ) + ea 2 x f 2 (y + m 2 x ) + ............ + ea n x f n (y + m n x ) . Thus, C.F. = e x f 1 (y + x ) + e 2 x f 2 (y + x ) . Step 2. To find the particular integral

P.I. =

1

(D − D′ − 1)(D − D′ − 2)

e 2x − y =

1

(2 + 1 − 1)(2 + 1 − 2)

e 2x − y =

1 2x − y e . 2



∴ z = e x f 1 (y + x ) + e 2x f 2 (y + x ) +

1 2x − y . Ans. e 2


(D + D′ − 1)(D + 2D′ − 3)z = 4 + 3x + 6 y . Sol.: The given equation is (D + D′ − 1)(D + 2D′ − 3)z = 4 + 3x + 6y .



Here f (D, D′)z = (D + D ′ − 1)(D + 2D′ − 3)z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0

is z = ea1x f 1(y + m1x ) + ea 2 x f 2 (y + m 2 x ) + ............ + ea n x f n (y + m n x ) . Thus, C.F. = e x f 1 (y − x ) + e 3x f 2 (y − 2x ). Step 2. To find the particular integral

P.I. =

1

(D + D′ − 1)(D + 2D′ − 3) −1 

(4 + 3x + 6 y ) −1

 = (1 − D − D′) 1 − − D′  (4 + 3x + 6 y ) 3  3 3  1

D

 

1

= [1 + D + D′ + .....]1 + 3

= =

D 3

2

+

2 3

 

D ′ + ....... (4 + 3x + 6 y)

1 D 2  ′ ′ + + + + + 1 D D D ..... [4 + 3x + 6 y] 3  3 3 1

1 2  1 4 3 x 6 y . 3 . 6 3 6 + + + + + +  = 3 [3x + 6 y + 18] = x + 2 y + 6 . 3  3 3



∴ z = e x f 1 (y − x ) + e 3x f 2 (y − 2x ) + x + 2 y + 6 . Ans. Q.No.10.: Solve the following partial differential equations:

∂ 2z

∂ 2 z ∂z − + = x2 + y2 . ∂x 2 ∂x∂y ∂x Sol.: The given equation is

∂ 2z

∂ 2 z ∂z − + = x 2 + y2 . 2 ∂x∂y ∂x ∂x

This equation in symbolic form can be written as D 2 − DD′ + D z = x 2 + y 2 . This is a non-homogeneous linear partial differential equation with constant co-efficients. Step 1. To find the complementary function

Here f (D, D′)z = D 2 − DD′ + D z = D(D − D ′ + 1)z = (D − 0D′ − 0)(D − D ′ + 1)z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0


is z = e

a1 x

f 1(y + m1x ) + e

a 2x

f 2 (y + m 2 x ) + ............ + e

anx

f n (y + m n x ) .

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

P.I. =

1 −1 ( x 2 + y 2 ) = (1 + D − D′) (x 2 + y 2 ) D(D − D ′ + 1) D 1

=

2 2 2 [ 1 − (D − D′) + (D − D′) − ........](x + y ) D

=

[ 1 − D + D ′ + D 2 + D ′ 2 − 2DD′ − ..........](x 2 + y 2 ) D

= =

1 1

1 D

(x

2

)

2

+ y − 2x + 2 y + 2 + 2 =

x

3

3

2

+ xy − 2

x

2

2

+ 2 xy + 2 x + 2x

1 3 x − x 2 + xy 2 + 2 xy + 4 x . 3


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

∴ z = f 1 (y ) + e − x f 2 (y + x) + x 3 − x 2 + xy 2 + 2xy + 4x . Ans. 3


(2DD′ + D′2 − 3D′)z = 3 cos(3x − 2y) . 2 Sol.: The given equation is 2DD ′ + D′ − 3D′ z = 3 cos(3x − 2 y ) .


Here f (D, D′)z = D ′[2D + D ′ − 3]z = (D ′ − 0)[2D + D′ − 3]z . Note: Since the solution of (D − m1D′ − a1 )(D − m 2 D′ − a 2 ).............(D − m n D′ − a n )z = 0

is z = ea1x f 1(y + m1x ) + ea 2 x f 2 (y + m 2 x ) + ............ + ea n x f n (y + m n x ) . Thus, C.F. = f 1 (x ) + e 3x f 2 (2 y − x ) . Step 2. To find the particular integral

P.I. =

1 2

2DD ′ + D′ − 3D ′

3 cos(3x − 2 y ) =

3 2(6) − 4 − 3D′

cos(3x − 2 y )


= =

3 8 − 3D′ 3 100

cos(3x − 2 y ) =

3(8 + 3D ′) 64 − 9D ′ 2

(8 + 3D′) cos(3x − 2 y ) =

3 50

cos(3x − 2 y)

[4 cos(3x − 2 y ) + 3 sin(3x − 2 y )]



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

3 50

[4 cos(3x − 2 y ) + 3 sin (3x − 2 y )] . Ans.

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

DD ′(D + 2D′ + 1)z = 0 . Ans.: z = f 1 (y ) + f 2 (− x ) + e − f 3 (y − 2x ) . x

Q.No.2.: Solve the following partial differential equation:

r + 2s + t + 2p + 2q = 0 x x Ans.: z = e − f 1 (y − x ) + xe − f 2 (y − x ) .


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

3x

f 2 (y + 2x ) −

1 x + 2y . e 6


(D − D′ − 1)(D − D′ − 2)z = e x + 2 y + x . x

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

2x

f 2 (y + x ) +

1 2x − y 1 + (2x + 3) . e 2 4


(D + D′ − 1)(D + 2D′ − 3)z = 4 + 3x + 6 y .


Ans.: z = e x f 1 (y − x ) + e 3x f 2 (y − 2x ) + x + 2 y + 6 . Q.No.6.: Solve the following partial differential equation:

(D 2 + DD′ + D′ − 1)z = cos(x + 2y) + e y . x

Ans.: z = e f 1 (y ) + e

−x

f 2 (y + x ) +

1 2

sin (x + 2 y ) − xe y .


D 2 − D′ z = 2 y − x 2 . Ans.: z =

∑

Ae

hx + h 2 y

2

−y −

x4 12

.


x 2 r + 2xys + y 2 t = 0 .

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

Ans.: z = f 1 


x 2 r − 3xys + 2 y 2 t + px + 2qy = x + 2 y . 2 Ans.: z = f 1 (xy ) + f 2 x y + x + y .

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7. Non-Homogeneous Linear Equations

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