NATIONAL OPEN UNIVERSITY OF NIGERIA

UNIVERSITY VILLAGE, PLOT 91, CASDASTRAL ZONE, NNAMDI AZIKIWE EXPRESS WAY, JABI, ABUJA.
FACULTY OF SCIENCES
JANUARY/FEBRUARY 2013 EXAMINATION

COURSE CODE:                               MTH 422

COURSE TITLE :                             PARTIAL DIFFERENTIAL EQUATION                              

TIME ALLOWED:                           3 HOURS

FOR WHOM:                                      400 LEVEL MATHEMATICS, COMPUTER AND MATHEMATICS AND B.ED MATHEMATICS STUDENTS.

INSTRUCTION:                               ANSWER FOUR FROM SEVEN QUESTIONS. EQUATION ONE IS COMPOUSORY.

1.  Given xp+yq=pq

            Find  a.      The initial element if  x=x₀ , y=0 and z=x_0/2 z ₌x₀/2 z  (x,0)=x/2    5marks

                     b.        The characteristics stripe containing the initial elements       5marks

                      c.       The integral surface which contain the initial element.                           4marks

2. State and Prove  CAUCHY KOVALEWASKI Theorem.                                                      14marks

3a.   Find the general solution of

      ,                                 

         By method of    langrage multiplier                                                                                                      7marks

    .3b.. Derive the solution to the Cauchy problem

utt=a2uxx+cosx, ux,0=sinx, utx,0=1+x                                                                 7marks

4. Prove that u=Fxy+xGyx  is the general solution of x2uxx-y2uyy=0 14marks
5.   A) Determine the characteristic equation, the characteristic curve and the canonical form of

x2uxx+2 xyuxy+ y2uyy+xy ux+ y2uy=0                                      7marks

             5 b) Prove that the equation in 6a above  can be solved                                              7marks

6..    By inspection, classify the following partial differential equations into the foolowing: non-linear, quasi-linear and linear. If linear, determine whether each is homogeneous or not

uxx+uyy-2u=x2 

ux2+logu=2xy 

(sinux)ux+ux=ex 

2uxx-4uxy+2uyy+3u=0                                                       3.5marks each= 14marks