NOUNSPOTEDU€¦ · Web viewPartial Differential Credit Unit: 3 Time Allowed: 3 Hours Total: 70...
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NATIONAL OPEN UNIVERSITY OF NIGERIAPlot 91, Cadastral Zone, Nnamdi Azikwe Express Way, Jabi-Abuja
FACULTY OF SCIENCEOctober\November Examination 2016
Course Code: MTH422 Course Title: Partial Differential Equations Credit Unit: 3Time Allowed: 3 HoursTotal: 70 MarksInstruction: Answer Any 4 Questions
1a. Solve the Cauchy Problem: Let 2 zx−3 z y+( x+ y ) z=0 that z (x ,0 )=x2
using the Lagrange methods. (6marks)
1b. Find the general solution of
(Zx i Zy i −1 ) ( A ,B ,C )
By method of Lagrange multiplier (8marks)
2a. Solve the boundary value problem:
ut−2ktu xx=0 . , 0<x<π , . t>0
u (0 , t )=u (π , t )=0 , t≥0 .
u ( x ,0 )=2 sin2 x−5 sin 3 x . 0<x<π , . (14marks)
3a. A Dirichlet problem in a circular region is given as follows:
∇2u=0 . (r , θ )∈D⊂ℜ2
u (a ,θ )=b0 cos2θ .where D is the circular region with centre at the origin and radius a. Here b is an arbitrary constant.(i) What other conditions we need for the existence of the solution? (6marks)(ii) Find the solution of this boundary value problem. (8marks)
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4a. Solve the vibration of an elastic string governed by the one-dimensional wave equation
∂2u∂ t 2
=c2 ∂2u∂ x2
where u(x, y) is the deflection of the string. Since the string is fixed at the ends x
= 0 and x = l , we have the two boundary conditions thus:
u(0 , t ) = 0 , u( l , t )= 0 for all t
The form of the motion of the string will depend on the initial deflection (deflection at
t = 0) and on the initial velocity (velocity at t = 0). Denoting the initial deflection by f(x) and the initial velocity by g(x), the two initial conditions are
u( x , 0) = f ( x ) ∂u∂ t
|t=0=g( x ) (14marks)
5. Given xp + yq=pq Find:
a. the initial element if x=xo , y=o and z=
xo2 z
( x ,o )=x 2 (5marks)
b. the characteristics stripe containing the initial elements (5marks)
c. the integral surface which contain the initial element. (4marks)
6a. Form the PDEs whose general solutions are as follow:
(i) z=Ae−p2 tcos px (6marks)
b. Separate ux+2utx−10u tt=0 and the boundary conditions u (0 , t )=0 , ux (L , t )=0
For 0<x<L and ∀ t hence, solve completely. Hint: Let u ( x , y )=X (x )T ( t ) (8marks)
7a. Reduce the equation uxx+5uxy+6u yy=0 to canonical form and find its general solution
7marks
b. Prove that u=F ( xy )+xG( yx ) is the general solution of x2uxx− y2uyy=0 (7mark)