Chapter: Mathematics (maths) - Partial Differential Equations

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Non Homogeneous Linear Equations

The methods for finding the Particular Integrals are the same as those for homogeneous linear equations.

Non –Homogeneous Linear Equations

 

Let us consider the partial differential equation

f (D,D') z = F (x,y)------- (1)

If f (D,D') is not homogeneous, then (1) is a non–homogeneous linear partial differential equation. Here also, the complete solution = C.F + P.I.

 

 

The methods for finding the Particular Integrals are the same as those for homogeneous linear equations.

 

 

But for finding the C.F, we have to factorize f (D,D') into factors of the form D –mD' –c.

 

 

Consider now the equation

 

(D –mD' –c) z = 0 -----------  (2).

 

This equation can be expressed as

 

 

p –mq = cz ---------(3),

 

 

which is in Lagrangian form.

 

 

The subsidiary equations are


 

The solutions of (4) are y + mx = a and z = becx.

 

 

Taking b =  f (a), we get z = ecx f (y+mx) as the solution of (2).

 

Note:

 

1.  If   (D-m1D' –C1) (D –m2D'-C2)   …… –m(Dn'-Cn) z = 0 is the partial

differential equation, then its complete solution is

 

z = ec1x f1(y +m1x) + ec2x f2(y+m2x) + . . . . . + ecnx fn(y+mnx)

 

2.  In the case of repeated factors, the equation (D-mD' –C)nz = 0 has a complete

 

solution z = ecx f1(y +mx) + x ecx f2(y+mx) + . . . . . +x n-1 ecx fn(y+mx).

 

 

Example 31

 

Solve (D-D'-1) (D-D' –2)z = e 2x –y

 

Here  m1 = 1, m2 = 1, c1 = 1, c2 = 2.

 

Therefore, the C.F is ex f1 (y+x) + e2x f2 (y+x).


 

Example 32

 

Solve (D2 –DD' + D' –1) z = cos (x + 2y)

 

The given equation can be rewritten as

 

(D-D'+1) (D-1) z = cos (x + 2y) 

Here m1 = 1, m2 = 0, c1 = -1, c2 = 1. 

Therefore, the C.F = e–x f1(y+x) + ex f2 (y)



 

 

Example 33

 

Solve [(D + D'–1) (D + 2D' –3)] z = ex+2y + 4 + 3x +6y 

Here m1 = –1, m2 = –2 , c1 = 1, c2 = 3.

Hence the C.F is z = ex f1(y –x) + e3x f2(y –2x).


 

 


It is the complete solution. 


 

 

Exercises

 

(a) Solve the following homogeneous Equations.


 

6. (D2 + 4DD' –5D'2) z = 3e2x-y + sin (x –2y) 

 

7. (D2 –DD' –30D'2) z  = xy + e6x+y

 

8. (D2 –4D' 2) z = cos2x. cos3y 

9. (D2 –DD'  –2D'2) z = (y –1)ex

 

10.4r + 12s + 9t =  e3x –2y

 

(b)Solve the following non –homogeneous equations. 

 

1.     (2DD' + D' 2 –3D') z = 3 cos(3x –2y) 

 

2.     (D2 + DD' + D' –1) z = e-x

 

3.     r –s + p = x2 + y2

 

4.     (D2 –2DD' + D'2 –3D + 3D' + 2)z = (e3x + 2e-2y)2


(D2 –D'2 –3D + 3D') z = xy + 7.

 

 

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