Page 130 - Engineering Mathematics Workbook_Final

Differential Equations & Partial Differential Equations

x ay (a) The value of u(2, 2) = -1
(a) z = + + b
a x
(b) The value of u(2, 2) = 1

x ay
(b) z = + + b   1 1   1
b x (c) The value of u  ,    =
 2 2  2

2
b
(c) z = ( 4 ax + ) y +  1 1  1
(d) The value of u   ,     =
2
−
(d) (z b ) = ( 4 ax + ) y  2 2  2
Partial Differential Equations
275. For an arbitrary continuously
differentiable function f, which of the 277. The one dimensional heat conduction
following is a general solution of partial differential equation
)
−
( z px qy = y − x 2
2
2
 T =  T , is
2
2
2
(a) x + y + z = f xy t   x 2
( )
(a) parabolic (b) hyperbolic
(b) (x + ) y 2 + z = f xy
2
( )
(c) elliptic (d) mixed
2
2
2
(c) x + y + z = ( f y x− )
[GATE – 1996]
(d) x + y + z = f ( ( x + ) y 2 + z 2 ) 278. The one dimensional heat conduction
2
2
2
partial differential equation

, x
276. Let ( ) t be the solution of initial  T =  2 T is
t   x 2
boundary value problem
 2 u  2 u (a) parabolic (b) hyperbolic

0
, t
= , 0 x    ;
t  2  x 2 (c) elliptic (d) mixed
  x  [GATE – 1996 (ME)]
u ( ,0x ) cos=       , 0 x ;
 2  279. The number of boundary conditions
required to solve the differential
 u
x
( ,0 =  2   2 
) 0, 0 x ,
t  equation + is
 x 2  y 2
 u ( ) 0t = , t  0.
0,
 x (a) 2 (b) 0

(c) 4 (d) 1

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