SEMESTER_3_M3 important 2_ marks

PSG Institute of Technology and Applied Research, Coimbatore - 641 062

Department of Mathematics
MA6351 TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS
Two Marks with Answers

UNIT-I
PARTIAL DIFFERENTIAL EQUATIONS

Q1. Form the PDE by eliminating a and b from z = (x2+a2)(y2+b2).

A: Given z = (x2+a2)(y2+b2) (1)
Partially differentiating (1) with respect to x and y we get,
p  2x ( y 2  b 2 )
p
 ( y2  b2 ) (2)
2x
q  2 y (x 2  a 2 )
q
 (x 2  a 2 ) (3)
2y
Substituting (2) and (3) in (1), we get
4 xyz  pq


Q2. Obtain the partial differential equation by eliminating the arbitrary constants a
and b from (x-a)2 + (y-b)2 + z2 = 1.

A: Given (x-a)2 + (y-b)2 + z2 = 1 (1)

Partially differentiating (1) with respect to x and y we get,

z
2(x – a) + 2z 0
x
z
2(y – b) + 2z 0
y
(x – a) = – zp
(y – b) = – zq
using these in (1), we get z2p2+z2q2 +z2 = 1



1

,Q3. Form the partial differential equation by eliminating the arbitrary constants a
and b from z = axn + byn

A: z = axn + byn (1)
Partially differentiating with respect to x and y we get,
z
 nax n 1
x
z
 nby n 1
y
p
a 
nx n 1
q
b 
ny n 1
Using these values in (1) we get nz = px + qy.

Q4. Find the partial differential equation of all spheres of radius ‘c’ units having
their centre’s on xoz plane.

A: Let the centre be (a , 0 , b) The equation of the sphere is
(x – a)2 + (y – 0)2 + (z – b)2 = c2 (1)
Even though there are 3 constants, only a and b are arbitrary constants. Hence we
have to eliminate only a and b]
Partially differentiating (1) with respect to x and y we get
2(x – a) + 2(z – b)p = 0
2y + 2(z – b)q = 0
(x – a) = – (z – b)p
(z – b) = – (y/q)
p2 y2
(x – a)2 = (z – b)2p2 =
q2

p2 y2 y
using (x-a)2 = , (z-b) = – in (1) we get
q2 q

p2 y2 y2
+ y 2
+  c2
q2 q2
y2(p2+q2+1) = c2q2

2

, Q5. Find the Partial differential equation of all spheres whose centres lie on the z
axis.

A: Let the centre and radius of the sphere be (o ,o ,a) and b
 Equation of the sphere is x2 + y2 + (z – a)2 = b2 (1)
Partially differentiating with respect to x and y we get
x
2x + 2(z – a)p = 0 , (z – a) = – (2)
p
y
2y + 2(z – a)q = 0, (z – a) = – (3)
q
x y
From (2) and (3) we get = (or) py = qx
p q


Q6. Eliminate the function ' f ' from z = y f (y/x)

A: Given z = y f (y/x) (1)
Differentiating (1) partially with respect x and y, we get
z y  y
p =  yf '   
x x  x2 

y2  y 
p=  f '  (2)
x2  x 
z  y  y  1 
q =  f    yf '    (3)
y x  x  x 
Using (1) and (2) in (3), we get

z y   px 2 
q = +  
y x  y 2 
q y = z – px (or) z = px +qy

Q7. Find the partial differential equation by eliminating ' f ' from
1 
z = y2+2 f   log y 
x 

1 
A: Given z = y2+2 f   log y  (1)
x 

3