MAHALAKSHMI ENGINEERING COLLEGE,TRICHY.

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1 MAHALAKSHMI ENGINEERING COLLEGE,TRICHY. PARTIAL IFFERENTIAL EQUATIONS-MA PART-A. Form the p.d.e from (x-a) + (y-b) + z r (AU May 03) Given that (x-a) +(y-b) +z r () d.p.w.r to x, (x-a) + z 0 [ z is a fun of x and y] (x-a) + zp () d.p.w.r. to y, p (y-b) + z 0 q (y-b) + zq (3) Eliminating a and b from, and 3 x a -zp 3 y b -zq () (-zp) +(-zq) +z r z p + z q + z r z (p +q +)r which is the required p.d.e. Find the p.d.e of all spheres having their centres on the z-axis (AU ec 0) Let the Centre of the sphere be (0, 0, c) point on the Z axis and r it s radius. (x-0) +(y-0) +(z-c) r [Since centre lies on Z axis] ie, x +y +(z-c) r d () p.w.r. to x, [c&r arbitrary constants] x + (z-c) 0 x + p (z c) ()

2 d () p.w.r. to y, y + (z-c) 0 y + q (z c) (3) From( ) and (3) () z c - x/p (3) z c -y/q -x/p - y/q qx py, which is the required p.d.e 3. Form the p.d.e by eliminating the constants a and b from z (x +a ) (y +b ) AU ec 00 G.T. z(x +a ) (y +b ) () d () p w.r to x p x (y +b ) y + b () d () p w.r to y, q y (x +a ) x + a (3) Substitute () & (3) in () z pq 4xyz 4. Eliminate the arbitrary function f from z f(y/x) and form a p.d.e (AU ec 0) Given that z f(y/x) () d p.w. r to x, p f (y/x) (-y/x ) () d p.w.r to y, q f (y/x) (/x) (3) Now, ( ) ( ) px -qy is, px + qy 0 is the required p.d.e.

3 5. Form the p.d.e by eliminating the arbitrary function from z -xy f(x/z) (AU June 0) G.T z xy f(x/z) () d () p.w.r. to x z -y f (x/z) z p-y f (x/z) () d ( ) p.w.r. to y z -x f (x/z) z q-x f (x/z) (3) (-xq) (zp-y) (zq-x)(z-xp) -xzpq + xyq z q xzpq xz + x p xyq z q xz + x p x p + z q xyq xz x p (xy z )q xz is the required p.d.e 6. Form the p.d.e of all planes cutting equal intercepts from the x and y axes (AU ec 009) The equation of such plane is x/a + y/a + z/b (x and y have equal intercepts) p.d.w.r. to x p.d.w.r. to y, p

4 q (3) From () and (3) p q p q 0 is the required p.d.e. 7. Find the complete integral of p + q pq (AU May 03) Given p + q pq () It is of the form F (p,q) () Hence the trial saln is z ax + by + c (3) To get the complete integral of solution 3 we have to eliminate any one of the arbitrary constants. Now (3) > > b - ab - a > b(-a) -a > b -a / -a Hence the complete soln is z 8. Solve : pq x (AU May 00) It is of the form f (x,p) f (y,q) Let p/x /q k > p/x k > p kx z

5 zk 9. Solve : (- ) 3 z o (AU ec 0) The A.E. is (m-) 3 0 m,, The C.F f (y+x) + x f (y+x) + x f 3 (y+x) 0. Solve: ( 3 - ) Z 0 (AU ec 009) The A.E. is m 3 m 0 >m (m-) 0 >m 0, 0, z f (y+0x) + xf (y+0x) + f 3 (y+x) f (y) + xf (y) + f 3 (y+x). Solve :( ) Z 0 (AU June 0) A.E is m -7m > (m-6) (m-) 0 > m 6, C.F. f (y+x) + f (y+6x). Find the P.I. of ( - + ) Z e x y (AU ec 00) a, b - P.I. ex-y.

6 3. Solve :(-) (- +) z 0 (AU ec 0) It is of the form (-m -c ) (-m -c ) z0 Compare with general form the given eqn can be written as (-O -) (- -(-))Z0 Here m 0, m C, C - z e x f (y+ox)+e -x f (y+x) e x f (y)+e -x f (y+x) 4. Form the p.d.e by eliminating the function from zf(x+t) + g(x-t) Solution : (AUN/ 00) d.p.w.r. to x, p f (x+t)+g (x-t) () d.p.w.r to t, q f (x+t)-g (x-t) () f (x+t) +g (x-t) ( 3) f (x+t) +g (x-t) (4) 5. Form a p.d.e by eliminating the arbitrary constants a and b from z (x+a) (y+b) () (AUM/J 008) d() p.w. r to x, z/ x (y+b) > p y+b () d. p.w. r to y z/ y (x+a) > q x+a (3) Sub () and (3) in (), z qp

7 6. Solve:( )z0 (AU A/M 00) m 3-3m+ 0> m, C.F (y+x) + (y+x) 7. Form the p.d.e by eliminating a and b from z (x+a) + (y+b) (AU A/M 008) Given that z(x+a) +(y+b) () d () p.w.r to x z/ x (x+a) p(x+a) P/ x+a () d () p.w.r. to y z/ y (y+b) q (y+b) q/ y+b (3) > z (p/) + (q/) p /4 + q /4 > 4z p +q 8. Form a p.d.e by eliminating the arbitrary Constants a and b from the equation (x +a )(y +b ) z (AU ec 00) z(x +a )(y +b ) () d.p.w.r to x px(y +b ) () d.p.w.r to y, qy(x +a ) (3) () > y +b (4) ( 3) > x +a (5) Sub (4) and (5) in () we have z z 4xyz pq.

8 PART B. Form the p.d.e by eliminating the arbitrary function from x +y +z,ax+by+cz 0 Given (AUC ec 00) x +y +z,ax+by+cz Let u x +y +z v ax+by+cz Eqn () u,v Elimination of from (4) gives u v x x 0 u v y y u v x+zp a+cp x x u v y+zq b+cq y y Now (6) in (5) x+zp a+cp 0 y+zq b+cq x+zp b+cq -a+cp y+zq 0 x+zp b+cq - a+cp y+zq 0 x+zp b+cq - a+cp y+zq 0 x+zp b+cq -a+cp y+zq 0 bx+cqx+zpb+zcpq-ay-azq-cpy-cpzq0

9 p(zb-cy)+q(cx-az)ay-bx. Solve :(mz-ny)p+(nx- x)qy-mx l (AUC Apr/May 00) P +Q R p q Pmz-ny, Qnx-lz, R l y-mx The Lagrange s subsidiary equations are dx dy dz P Q R ie, dx dy dz mz-ny nx-lz y-mx Using lagrangian multipliers as l, m, n each of ratio is equal to ldx+mdy+ndz ldx+mdy+ndz l(mz-ny)+m(nx- lz)+n( ly-mx) 0 l dx+mdy+ndz0 Integrating, l x+my+nzc Choosing another set of multipliers x, y, z xdx+ydy+zdz xdx+ydy+zdz x(mz-ny)+y(nx-lz)+z(y-mx) 0 xdx+ydy+zdz0 Integrati ng, x x +y +z C + y + z The general solution C x +y +z, lx+my+nz 0 ( is arbitrary)

10 3. Solve: x (y-z)p+y z-x qz x-y (AUC ec 00/June 0) Solution : x (y-z)p+y (z-x)q z (x-y) Lagrange s equation is P +Q R p q Here P x y-z,q y z-x,r z x-y dx dy dz P Q R dx dy dz x y-z y (z-x) z x-y The S.E is x y z dx dy dz dx dy d + + y x z x y z y-z z-x x-y y-z+z-x+x-y dx dy + dz + 0 x y z Choosing,, as multipliers each ratio is equal to Integrating we have dx dy dz0 x y z x dx y dy z dz x + y-+ z + + C C x y z + + C x y z u + + x y z Similary choosing + + x y z vx y z + +, xyz 0 x y z as Lagrange s multipliers we get

11 Solve : zcos(x+y) (AUC Jun 0) The A.E is m 3 +m +4m+40 (m+) (m +4) 0 m-, m ±i synthetic division C.F y-x y+ix y-ix Now, P.I cos x+y cos x+y 3 is. - Cos(x+y) 8 + Re place 4,, Cos (x+y) (X and by ) - 8 (+ ) Cos (x+y) sin (x+y) sin (x+y) 48 sin( x y) 4 z y-x y+ix 3 y-ix sin(x+y) 4

12 5. Solve the p.d.e. x( y z) p y( z x) q z( x y) (AUC ec 0 Solution : Lagrange s type P p +Q q R The S.E is dx dy dz P Q R P x(y-z) Q y(z-x) R z(x-y) dx dy dz x(y-z) y(z-x) z(x-y) to Choosing,, as lagrange s multipliers, each of above ratio is equal dx+dy+dz dx+dy+dz xy-xz+yz-yx+zx-zy 0 dx+dy+dz0 ux+y+z Integrating, d(x+y+z)0 x+y+z c choosing,, x y z as Lagrange s multipliers dx+ dy+ dz dx+ dy+ dz x y z x y z y-z+z-x+x-y 0 dx + dy + dz 0 x y z Integrating, log x +log y +log z log C l og(xyz)logc (xyz)c x+y+z, xyz 0 vxyz

13 x-z + z-y q y-x 6. Solve: p P p +Q q R The equation is of the form P p +Q q R P x-z, Q z-y, Ry-x The S.E dx dy dz P Q R ie, dx dy dz x-z z-y y-x Using multipliers as,, Each ratio dx+dy+dz dx+dy+dy x-z+z-y+y-x 0 ie, dx+dy+dx0 Integrating x+y+z C u x+y+z Next, using multipliers as y, x, z Each ratio ydx+xdy+zdz yx-yz+xz-xy+yz-xz ie ydx+xdy+zdz 0 ie, ydx+xdy+zdz0 d(xy)+zdz 0 integrating, xy+ z C xy+z C v xy+ z (x+y+z, xy+z )0

14 7. x yz p y zx q z xy (AUC May 00) The equation is of the form P p +Q q R P x -yz,qy -zx,rz -xy Lagrange s subsidiary equations are dx dy dz P Q R dx dy dz ie, x -yz y -zx z -xy Using lagrange s multipliers x,y,z we have xdx+ydy+zdz dx+dy+dz x +y +z -3xyz x +y +z -xy-yz-zx xdx+ydy+zdz x+y+z (x +y +z -xy-yz-zx) dx+dy+dz x +y +z -xy-yz-zx xdx+ydy+zdz dx+dy+dz x+y+z xdx+ydy+zdz(x+y+z)(dx+dy+dz) Integrating x + y + z x+y+z x +y +z x +y +z +xy+yz+zx (xy+yz+zx)0 uxy+yz+zxc u (x,y,z) xy+yz+zx

15 Now, dx-dy dy-dz x -yz -y -zx y -zx -z -xy d y-z dx-dy x -y +z(x-y) y+z y-z +x y-z d(x-y) x-y (x+y+z) d(x-y) x-y d(y-z) y-z d(y-z) (y-z)(x+y+z) log(x-y)log(y-z)+logc x-y log logc y-z x-y C, ie, y-z x-y v y-z x y The general solution is xy yz zx, 0 y z 8. Solve : x(y -z )p+y(z -x )qzx -y It is of the form P p +Q q R (AUC May 03) Here Px y -z, Qy z -x, R z (x -y ) The S.E is dx dy dz p Q R dx dy dz x(y -z ) y z -x z x -y Use lagrange s multipliers x,y,z Each ratio xdx+ydy+zdz xdx+ydy+zdz x (y -z )+y (z -x )+z x -y 0

16 ie, xdx +ydy+zdz 0 x +y +z a ie, u x +y +z integrating, x + y + z a Similarly, taking,, x y z as L.M we get dx dy dz x y z y z z x x y Each ratio ie, dx dy dz x y z 0 dx dy dz 0, integrating, logx+logy+logz log b x y z ie, log (xyz) logb bxyz ie, vxyz (x +y +z,xyz)0 9. ( 3 - )z e x +3x y (AUC ec 0) The A.E is m 3 -m 0 m (m-) 0 m0,0, C.F (y+ox) x (y+ox) 3 (y+x) x+oy e P.I - 3 e x 8-0 e x 4 a, b0 Replace 0 3 P.I - 3x y

17 .3x y x y x y x y 3 3 x x y+ 3 3 x y +6 (x ) x x y x y x x 5 6 e x y x zf y+ox +xf y+ox +f3 y+x Solve z z x x y 3 3 x+y e 4sin ( x y) 3 3 The given equation can be written as ( 3 - ) ze x+y +4sin (x+y) The A.E is m 3 -m 0 m (m-)0 m0,0 (or) m C.F f (y) + xf (y)+f 3 (y+x)

18 P.I x+y e + 4sin (x+y) a, b m, n x+y e + 4sin(x+y) 4sin(x+y) 3 -+ x+y - e + 4(sin(x+y)) x+y - e + x+y 4cos(x+y) - e x+y - e -4cos(x+y) x+y zf (y)+xf (y)+f 3(y+x)- e -4cos(x+y) 3 Replace. Solve: ( + -6 ) zy cosx (AUC May 03) m +m-60 m, -3 C.F (y+x) (y-3x) P.I ycosx ycosx ycosx + - ycosx ycosx 6 ycosx - ycosx + ycosx 6 ycosx - cosx + (0)

19 ycosx -sinx (ycosx -sinx) ysinx+cosx -y cos x + sin x z C.F+P.I. Solve: P (+q) qz Give that p(+q) qz It is of the form f (z,p,q) 0 Let u x+ay Now u u, a x y dz dz p,qa du du dz dz dz () +a a.z du du du dz +a az du dz a az- du dz az- du a du dz a az- adz du az- Integrating on b.s u log (az-) +logc

20 Hence the complete solution is (since the number of a.c no.of. I.V) x+aylog [c(az-)] 3. Solve zpx+qy+ p +q + (AUC ec 0, May 03) Given that zpx+qy+ p +q + It is of the form z px+qy+f(p,q) (Clairaut s form) Hence the complete integral is zax+by+ a +b + To find singular solution: (a and b are arbitrary constants) zax+by+ a +b d () p.w.r. to a, a -a ox+ x a +b + a +b + d () p.w.r. to b, b -b oy+ y a +b + a +b + a +b x +y +a +b a +b -x +y - +a +b -x -y +a +b i -x -y +a +b ii +a +b -x -y

21 () x-a -x -y by (i) (3) y-b -x -y by (ii) Now -x -y a, b -x -y -x -y Substitute in () -x y z - + -x -y -x -y -x -y by(ii) -x -y -x -y z -x -y z -x -y x +y +z is the singular solution To find the general integral Put b (a) in (), zax+ (a)y+ +a + (a) d. (4) p.w.r to a ox+ '(a)y+ a+ (a) (a) )\ +a + (a) Eliminate a between (4) and (5) we get the general solution 4. Solve : p +q x +y Solution : Give that p +q x +y It is of the form F (x,p) F (y,q) Let p -x y -q a constant p -x a p a +x p x +a

22 y q a q y a z pdx qdy Substitute () and () in (3) z x a dx y -a dy x x x +a y y -a a y - a - sin h cos h +b a a Which is the complete integral 5. Solve :p (-q ) q(-z) (AU Nov/ec 009) Given p(-q ) q(-z) It is of the form F (z,p,q) 0 Let z f (x+ay) be the solution of () Put x+ay u Z f(u) u u, a x y (*) z dz u dz p.. x du x du z dz u dz q..a y du y du Using (*) 3 Substituting (3) in () dz dz dz du du du -a a -z

23 dz -a a(-z) du dz a -a+az du dz du a az+(-a) dz du az+(-a) a - az+(-a dz u+b a - + az+ -a x+ay +b - + a a az+ -a x+ay +b a a az+(-a) (x+ay)+b a a which is complete integral 5.Solve: z ( )z ysinx (or) z z ysinx x xy y m -5m+60 (m-3) (m-) 0 m3, C.F (y+x)+ (y+3x) P.I ysinx ysinx ysinx -

24 y sinx 5 6 ysinx+ (ysinx)- ysinx 5 6 ysinx+ sinx- (0) ysin x sin x ysinx 0 ysinx+5(-cosx) -cosx.y-5sinx -ysinx+5cosx The general solution is Z C.F +P.I 3 7. Solve zsin(x+y) m 3-7 m-60 m-, -, 3 (AUC May 03) C.F (y-x)+ (y-x)+ 3(y+3x) P.I sin(x+y) by - - sin(x+y) ( ) by- -4 sin(x+y) (-)-7(-) -6(-4) by-()()- sin(x+y) sin(x+y) -+38

25 --38 sin(x+y) sin(x+y) --444(-4) -cos(x+y)-38cos(x+y) cos(x+y)+76cos(x+y) cos(x+y) cos(x+y) Find the singular integral of zpx+qy+p +pq+q (AUC ec 0) Z px+qy+p +pq+q It is of the form zpx+qy+f(p,q) (clairaut s form) Hence the complete soln is z ax+by+a +ab+b (since the number of a.c number of I.V) To find the singular integral d.(). p.w.r.to a and b we get 0 x+a+b y+b+a () x a+b+x (3)x a+4b+y (4) (5) -3b+x-y 0

26 3b x-y b x y 3 Similarly, y (3) 0y+ x- +a 3 x-y y-x a- a 3 3 Substitute a,,b in () y-x x-y y-x y-x x-y x-y z x+ y Simplifying, z -x -y +xy z -x -y +xy 3z-x -y +xy 3z+x +y -xy0 which is the singular integral. y 9. ( )zxe (AU June 0 ) Given ( ) The solution of -m -α -m -α z0 is αx αx ze f (y+m x)+e f y+m x Here α 0, m - α -3, m ox -3x C.Fe fy- x +e fy+x P.I xe y +O

27 (+0) -(+0)( +)-( +) +6(+0)+3( +) ox+y e x y e x ) y e x y e x 0. Solve : y e x - e y x y e 5 x- z C.F +P.I x-y ( )ze (AU ec 0) The given equation can be written as ze C.F: x-y -m -α -m -α z0 is αx αx ze f (y+m x)+e f (y+m x) Here α, α, m, m Replace x x C.Fe f (y+x)+e f (y+x)

28 P.I x-y e +- (+-) x-y e z C.F +P.I x-y e x-y x x e f (y+x)+e f (y+x)+ e. 3 4 z cos( x y) xy Give that +3-4 z0 (AU Nov/ec 0) The A.E is m +3m-40 (m-) (m+4) 0 m, -4 C.F (y-4x) (y+x) P.I +3-4 cos(x+y) cos(x+y) -4+3(-)-4(-) - cos(x+y) 6 P.I xy xy

29 xy xy xy xy 3 xy- xy 3 xy- x 3x xy- 3x xy- 3 3 x 3x x y x y x x y Solve: ( - +)z e x+y +4 (AU ec 0) Here m 0, c 0 m, c - C.F. f (y)+e -x f (y+x)

30 P.I e e x+y x+y e x+y P.I 4e e - + x 4 e ox+oy x z C.F+P.I ox+oy ox+oy -x x+y f (y)+e f (y+x)+ e +x 6 3.Find the p.d.e of all planes which are at a Constant distance from the origin (AU May 00) The equation of a plane which is at a distance k from the origin is l x+my+nzk where l +m +n Let l a ; m b ; nc we get ax+by+czk () a +b +c c -a -b c -a -b Substitute (3) in () ax+by+ -a -b zk d.p.w.r to x z a+ -a -b 0 x

31 -a -a -b p a p -a -b d (4) p.w.r to y z b+ -a -b 0 y -b q -a -b Now, a p b q - -a -b (say) ap ; bq and -a -b -λ -p λ -q λ -λ Squaring on b.s -p λ -q λ λ - p +q λ λ λ +λ (p +q )λ +p +q +p +q +p +q λ ± λ - +P +q λis-ve (*) (4) pλx+qλy-λzk (px+qy-z)λk k px+qy-z λ zpx+qy- k λ px+qy+k +p +q ***** [by (*)]