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Chapter: Mathematics (maths) - Partial Differential Equations

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Formation of Partial Differential Equations

Partial differential equations can be obtained by the elimination of arbitrary constants or by the elimination of arbitrary functions.


Formation of Partial Differential Equations

 

Partial differential equations can be obtained by the elimination of arbitrary constants or by the elimination of arbitrary functions.

 

By the elimination of arbitrary constants

Let us consider the function

f( x, y, z, a, b  ) = 0  -------------  (1)

where a & b are arbitrary constants        

Differentiating equation (1) partially      w.r.t x & y, we get


Eliminating a and b from equations (1), (2) and (3), we get a partial differential equation of the first order of the form f (x,y,z, p, q) = 0

 

Example 1

Eliminate      the arbitrary constants a & b        from  z = ax + by + ab

Consider      z  = ax + by + ab     ____________ (1)

Differentiating        (1) partially w.r.t     x & y, we get


Using (2)      & (3)  in (1), we get

          z        = px +qy+ pq

which is the required partial differential equation.

Example 2

Form the partial differential equation by eliminating the arbitrary constants a and b from 

z = ( x2 +a2 ) ( y2 + b 2)

Given z = ( x2 +a2 ) ( y2 + b2)         ……..(1)

Differentiating        (1) partially w.r.t     x & y , we get

p        = 2x   (y2 + b2 )       

q        = 2y   (x  + a  )       

 

Substituting the values of p and q in (1), we get 

4xyz = pq

which is the required partial differential equation.

Example 3

 

Find the partial differential equation of the family of spheres of radius one whose centre lie in the xy - plane.

 

The equation of the sphere is given by

 

( x –a )2 + ( y- b) 2 +  z2  = 1    _____________ (1)

 

Differentiating (1) partially w.r.t x & y , we get

2        (x-a ) + 2 zp   =       0

2 ( y-b ) + 2 zq  =    0

 

From these equations we obtain

x-a = -zp _________ (2) 

y -b = -zq _________ (3)

Using (2) and (3) in (1),  we get

z2p2 + z2q2 + z 2    = 1

 

or  z2 ( p2  + q2  + 1) = 1

 

 

Example 4

Eliminate the arbitrary constants a, b & c  from 


and form the partial differential equation.

The given equation is


or  -zp + xzr + p2x = 0

 


By the elimination of arbitrary functions

     Let   u and v   be   any two   functions arbitrary function. This relation can be expressed as

u = f(v)  ______________ (1)

Differentiating  (1)  partially w.r.t    x &  y and eliminating    the arbitrary functions from these  relations, we get  a partial differential equation  of the first  order of the form  

f(x, y, z, p, q )  = 0. 

 

Example 5

 

Obtain the partial differential equation  by eliminating „f„from  z = ( x+y ) f ( x2 -  y2 )

 

Let us now consider the equation

 

z = (x+y ) f(x2- y2) _____________ (1) 

Differentiating (1) partially w.r.t x & y , we get

p  = ( x + y ) f ' ( x2 -  y2 ) . 2x  +  f ( x2 -  y2 )

 q  =  ( x + y ) f ' ( x2 -  y2 ) . (-2y)     + f ( x2 -  y2 )        


i.e, py - yf( x2 - y2 ) = -qx +xf ( x2 - y2 )

i.e, py +qx   = ( x+y ) f ( x2 -  y2 )

Therefore, we have by(1),  py +qx  = z

 

 

Example 6

 

Form the partial differential equation by eliminating the arbitrary function f

from

 

z = ey f (x + y)

 

Consider  z  = ey f ( x +y )  ___________  ( 1)

 

Differentiating  (1) partially  w .r. t  x & y, we get

 

p     = ey f ' (x  + y) 

 

q     = ey f '(x  + y) + f(x + y). ey

Hence, we have

 

q = p + z

 

Example 7



Exercises:

1. 1. Form the partial differential equation by eliminating the arbitrary constants „a‟ & „b‟ from the following equations.


 

2.                 Find the PDE of the family of spheres of radius 1 having their centres lie on the xy plane{Hint: (x –a)2 + (y –b)2 + z2 = 1} 

 

3.                 Find the PDE of all spheres whose centre lie on the (i) z axis (ii) x-axis 

 

4.                 Form the partial differential equations by eliminating the arbitrary functions in the following cases. 

 

(i)                z = f (x + y) 

(ii)             z = f (x2 –y2

(iii)           z = f (x2 + y2 + z2)

(iv)           f(xyz, x + y + z) = 0

(v)             F (xy + z2, x + y + z) = 0

(vi)           z = f (x + iy) +f (x –iy)

(vii)        z = f(x3 + 2y) +g(x3 –2y)

 

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