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)

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)