Page 126 - Engineering Mathematics Workbook_Final

Differential Equations & Partial Differential Equations

CLASSIFICATION OF P.D.E  ( n t cosn )
2
−
u
, x
(b) ( ) t =  A e x with
n
252. Consider the following partial n= 0
2
differential equation: A =    f x
( )cosnx dx
n
 2   2   2  0
0
3 + B + 3 + 4 = (c)
 x 2  x y  y 2

   −   2n+ 1   2        

For this equation to be classified as u ( ) t =  A e    2         sin      2n + 1     x  
, x
n
2
2
parabolic, the value of B must be with n= 0   
_____ [GATE 2017] 2   2n + 1 

A = f x    x dx
( )sin

 
253. The type of partial differential n  0  2 
equation
 2 P +  2 P + 3  2 P + 2  P −  P = 0 (d)

 x 2  y 2    x  y u ( ) t = , x  A n exp − ( n t ) sin nx
2
x y
( )
is n= 1
2 
nx
(a) elliptic (b) parabolic with A =   0 f ( )sinx ( )dx
n
(c) hyperbolic (d) none of these
255. If u = (x, t) is such that
[GATE-2016-CE-SET 1]  2 u = 4  2 u , 0  x  , t  ,

0
t  2  x 2
254. The solution of the initial boundary
 u  2 u u ( ) u  ) 0
( ,t =
0,t =


value problem = 0 x 
t   x 2
, t  0 with boundary and initial u x ) 0;  u ( ,0 = x ) sin x then
( ,0 =
conditions t 
 u ( ) 0 u  ), 0      
0,t = =
,
 x ( ,t t  and u   3 6     is ________
( ,0 =
u x ) f x  
( ), 0 x  is ____
3 3
(a) (b)
(a) 4 8
  −  − 2n+  1  2  t       

)
u ( , x t =  A e    2         cos   2n + 1 x     (c) 3 (d) 3
n
n= 0  2  4 8
with
2    2n + 1  
A = f ( )cosx   x dx

 
n
 0  2 

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