Post on 18-Mar-2020
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
MAHALAKSHMI ENGINEERING COLLEGE,TRICHY.
PARTIAL DIFFERENTIAL EQUATIONS-MA2211
PART-A
1. Form the p.d.e from (x-a)2 + (y-b)
2 + z
2 = r
2 (AU May 2013)
Solution:
Given that (x-a) 2
+(y-b) 2
+z2 = r
2 ---------- (1)
d.p.w.r to x,
2(x-a) + 2z
= 0 [ z is a fun of x and y]
(x-a) + zp = 0 ---------- (2)
d.p.w.r. to y, p=
2 (y-b) + 2z
= 0 q=
= (y-b) + zq = 0 ---------- (3)
Eliminating a and b from 1, 2 and 3
2 x – a = -zp
3 y – b = -zq
(1) (-zp) 2
+(-zq) 2
+z2 = r
2
z2p
2 + z
2q
2 + z
2 = r
2
z2 (p
2+q
2+1)=r
2
which is the required p.d.e
2. Find the p.d.e of all spheres having their centres on the z-axis (AU Dec 2012)
Solution:
Let the Centre of the sphere be (0, 0, c) point on the Z –axis and ‘r’ it’s radius.
(x-0) 2
+(y-0) 2
+(z-c) 2
=r2
[Since centre lies on Z axis]
ie, x2+y
2+(z-c)
2 = r
2 ------- 1
d (1) p.w.r. to x, [c&r arbitrary constants]
2x + 2(z-c)
= 0
x + p (z – c) = 0 ---------(2)
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
d (1) p.w.r. to y,
2y + 2(z-c)
= 0
y + q (z – c) = 0 ---------(3)
From( 2) and (3)
(2) 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= (x2+a
2) (y
2+b
2)
Solution: AU Dec 2010
G.T. z=(x2+a
2) (y
2+b
2) ----------- (1)
d (1) p w.r to x
= p = 2x (y
2+b
2)
= y
2 + b
2 ----------- (2)
d (1) p w.r to y,
= q = 2y (x
2+a
2)
= x
2 + a
2 ----------- (3)
Substitute
(2) & (3) in (1)
z =
pq = 4xyz
4. Eliminate the arbitrary function ‘f’ from z = f(y/x) and form a p.d.e (AU Dec 2012)
Solution:
Given that z =f(y/x) -----------(1)
d 1 p.w. r to x, p =
= f
1 (y/x) (-y/x
2) ---------- (2)
d 1 p.w.r to y, q =
=f
1(y/x) (1/x) -----------(3)
Now,
=
(
)
(
)
=
=
px = -qy
is, px + qy = 0 is the required p.d.e.
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
5. Form the p.d.e by eliminating the arbitrary function from z2-xy = f(x/z)
(AU June 2012)
Solution:
G.T z2 – xy = f(x/z) -------- (1)
d (1) p.w.r. to x
2z
-y = f
1 (x/z) ⌈
⌉
2z p-y = f1(x/z) ⌈
⌉ -------------(2)
d (1 ) p.w.r. to y
2z
-x = f
1(x/z) ⌈
⌉
2z q-x = f1(x/z) ⌈
⌉ -------------(3)
(-xq) (2zp-y) = (2zq-x)(z-xp)
-2xzpq + xyq = 2z2q – 2xzpq – xz + x
2p
xyq = 2z2q – xz + x
2p
x2p + 2z
2q – xyq = xz
x2p –(xy – 2z
2)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 Dec 2009)
Solution:
The equation of such plane is
x/a + y/a + z/b = 1 --------- 1 (x and y have equal intercepts)
p.d.w.r. to x
, p= -
-------------------- 2
p.d.w.r. to y
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
q= -
-------------------- (3)
From (2) and (3)
p = q
p –q = 0 is the required p.d.e.
7. Find the complete integral of p + q – pq (AU May 2013)
Solution:
Given p + q = pq --------------(1)
It is of the form F (p,q) = 0 --------------(2)
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(1-a) = -a
=> b = -a / 1-a
Hence the complete soln is
z =
8. Solve : pq = x (AU May 2010)
Solution:
It is of the form f1 (x,p) = f2
(y,q)
Let p/x = 1/q = k
=> p/x = k => p = kx
z = ∫
= ∫ ∫
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
z=k
9. Solve : (D-D1)3 z = o (AU Dec 2011)
Solution:
The A.E. is (m-1)3=0
m = 1, 1, 1
The C.F = f1 (y+x) + x f2(y+x) + x2f3(y+x)
10. Solve: (D3-2D
2D
1) Z = 0 (AU Dec 2009)
Solution:
The A.E. is m3 – 2m
2 = 0
=>m2 (m-2) =0
= >m =0, 0, 2
z = f1(y+0x) + xf2(y+0x) + f3(y+2x)
= f1(y) + xf2(y) + f3(y+2x)
11. Solve :(D2-7DD
1+6D
12) Z= 0 (AU June 2012)
Solution:
A.E is m2-7m + 6 = 0
=> (m-6) (m-1) = 0
=> m = 6,1
C.F. = f1(y+x) + f2(y+6x)
12. Find the P.I. of (D2-2DD
1+D
12) Z= e
x y (AU Dec 2010)
Solution:
a=1, b= - 1
P.I. =
=
2e
x-y
. =
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
13. Solve :(D-1) (D-D1+1) z = 0 (AU Dec 2012)
Solution:
It is of the form
(D-m1 D
1-c1) (D-m2D
1-c
2) z=0
Compare with general form the given eqn can be written as
(D-OD1-1) (D-1D
1-(-1))Z=0
Here m1=0, m2=1
C1=1, C2=-1
z = exf1(y+ox)+e
-xf2(y+x)
= exf1(y)+e
-xf2(y+x)
14. Form the p.d.e by eliminating the function from z=f(x+t) + g(x-t)
Solution : (AUN/D 2010)
d.p.w.r. to x, p =
=f
1(x+t)+g
1(x-t) ------------- (1)
d.p.w.r to t, q=
=f
1(x+t)-g
1(x-t) ------------- (2)
=f” (x+t) +g”(x-t) --------------( 3)
=f” (x+t) +g”(x-t) -------------- (4)
15. Form a p.d.e by eliminating the arbitrary
constants a and b from z = (x+a) (y+b) --------(1) (AUM/J 2008)
Solution:
d(1) p.w. r to x,
z/ x = 1(y+b) => p = y+b - ------(2)
d. p.w. r to y
z/ y= (x+a) 1 => q = x+a --------(3)
Sub (2) and (3) in (1), z = qp
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
16. Solve:(D3-3DD
12+2D
13)z=0 (AU A/M 2010)
Solution:
m3-3m+2 = 0=> m = 1,2
C.F = 1(y+x) + 2 (y+2x)
17. Form the p.d.e by eliminating a and b from z =(x+a)2
+ (y+b)2
(AU A/M 2008)
Solution:
Given that z=(x+a)2+(y+b)
2 --------- (1)
d (1) p.w.r to x z/ x =2(x+a) 1 = p=2(x+a) = P/2 = x+a ---------(2)
d (1) p.w.r. to y z/ y =2(y+b) 1 = q = 2(y+b)= q/2 = y+b --------- (3)
1=> z (p/2)2
+ (q/2)2 = p
2/4 + q
2/4=
=> 4z = p2+q
2
18. Form a p.d.e by eliminating the arbitrary Constants a and b from the equation
(x2+a
2)(y
2+b
2) = z (AU Dec 2010)
Solution:
z=(x2+a
2)(y
2+b
2) -------(1)
d.p.w.r to x
=p=2x(y
2+b
2) --------(2)
d.p.w.r to y,
=q=2y(x
2+a
2) ------- (3)
(2) = >
= y
2+b
2 -- ------(4)
( 3) = >
= x
2+a
2 - - - - (5)
Sub (4) and (5) in (1) we have
z =
z =
4xyz = pq.
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
PART – B
1. Form the p.d.e by eliminating the arbitrary function from
2 2 2x +y +z ,ax+by+cz 0 (AUC Dec 2010)
Solution:
Given 2 2 2x +y +z ,ax+by+cz =0 ---------1
Let u = 2 2 2x +y +z ---------2
v = ax+by+cz ---------3
Eqn (1) u,v =0 ---------4
Elimination of from (4) gives
u v x x
= 0u v
y y
--------5
u2x+2zp
x
u 2y+2zq
y
v=a+cp
x
v =b+cq
y
--------6
Now (6) in (5)
2x+2zp a+cp
=02y+2zq b+cq
2x+2zp b+cq - a+cp 2y+2zq =0
2 x+zp b+cq - a+cp 2 y+zq =0
2 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-cpzq=0
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
p(zb-cy)+q(cx-az)=ay-bx
2. Solve : (mz-ny)p+(nx- x)q=y-mxl (AUC Apr/May 2010)
Solution:
p qP +Q =R
P=mz-ny, Q=nx-lz, R = y-mxl
The Lagrange’s subsidiary equations are dx dy dz
= =P Q R
ie, dx dy dz
= =mz-ny nx-lz 1y-mx
Using lagrangian multipliers as l, m, n each of ratio is equal to
x+mdy+ndz dx+mdy+ndz=
(mz-ny)+m(nx- z)+n( y-mx) 0
ld l
l l l
dx+mdy+ndz=0l
Integrating , 1x+my+nz=Cl
Choosing another set of multipliers x, y, z
xdx+ydy+zdz xdx+ydy+zdz=
x(mz-ny)+y(nx-lz)+z(1y-mx) 0
xdx+ydy+zdz=0
22 22Cyx z+ + =
2Integrati ,
2 2ng
2
2 2 2
2x +y +z =C
The general solution
2 2 2x +y +z , lx+my+nz =0 ( is arbitrary)
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
3. Solve: 2 2 2x (y-z)p+y z-x q=z x-y (AUC Dec 2010/June 2012)
Solution :
2 2 2x (y-z)p+y (z-x)q = z (x-y) ---------1
Lagrange’s equation is
p qP +Q =R
Here 2 2 2P x y-z ,Q y z-x ,R z x-y
The S.E is dx dy dz
= =P Q R
2 2 2
dx dy dz= =
x y-z y (z-x) z x-y
Choosing 2 2 2
1 1 1, ,
x y zas multipliers each ratio is equal to
22 2 2 22 2
ddy dydxdx dz + +y x y zx z= = =
y-z z-x x-y y-z+z-x+x-y
2 2 2
dydx dz+ + =0x y z
Integrating we have
2 2 2
1 1 1dx dy dz=0
x y z
-2 -2 -2x dx y dy z dz 0
-2 -2+1
1
x +1 y-2+1 z+ + =C
-2+1 -2+1 -2+1
11 1 1- - - =C
x y z
11 1 1+ + =C
x y z 1 1 1u= + +
x y z
Similary choosing 1 1 1+ +x y z
as Lagrange’s multipliers we get
v=x y z
1 1 1+ + , xyz =0x y z
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
4. Solve : 2 33 2 1 1 1D +D D +4DD +4D z=cos(2x+y) (AUC Jun’2012)
Solution:
The A.E is m3+m
2+4m+4=0
(m+1) (m2+4) =0
m=-1, m = ±2i
synthetic division
1
1 2 3
1 1 4 4
0 -1 0 -4 C.F= y-x y+2ix y-2ix
1 0 4 0
Now,
2 33 2 1 1 1
1P.I=
D +D D +4DD +4cos
D2x+y
1 1
1=
-4D-4D -4D-cos
4D2x+y D
3is D
2.D
=1
1 1- Cos(2x+y)8 D+D
2 12 1Re D 4, 1, 2place D DD
1
1 D Cos (2x+y)-8 D(D+D )
(X and ÷ by D)
2 1
1 D Cos (2x+y)=-
8 D +DD
1 -2 sin (2x+y)=-
8 -4-2
2 sin (2x+y)
48
= 1 sin(2 x y)24
1 2 31z y-x y+2ix y-2ix sin(2x+y)
24
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
5. Solve the p.d.e. ( ) ( ) ( )x y z p y z x q z x y (AUC Dec 201
Solution :
Lagrange’s type Pp+Qq=R
The S.E is dydx 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)
Choosing 1,1,1 as lagrange’s multipliers, each of above ratio is equal
to
dx+dy+dz dx+dy+dz
=xy-xz+yz-yx+zx-zy 0
dx+dy+dz=0
Integrating, 1d(x+y+z)=0 x+y+z =c u=x+y+z
choosing 1 1 1
, ,x y z
as Lagrange’s multipliers
1 1 1 1 1 1dx+ dy+ dz dx+ dy+ dz
x y z x y z=
y-z+z-x+x-y 0
dydx dz+ + =0x y z
Integrating, 2log x +log y +log z =log C
2
2
log(xyz)=logC
(xyz)=C v=xyz
x+y+z, xyz =0
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
6. Solve: x-2z + 2z-y =p q y-x
Solution:
Pp+Qq=R
The equation is of the form Pp+Qq=R
P = x-2z, Q = 2z-y, R=y-x
The S.E dx dy dz
= =P Q R
ie , dx dy dz
= =x-2z 2z-y y-x
Using multipliers as 1,1,1
Each ratio = dx+dy+dz dx+dy+dy
=x-2z+2z-y+y-x 0
ie, dx+dy+dx=0 Integrating 1x+y+z =C u = x+y+z
Next, using multipliers as y, x, 2z
Each ratio = ydx+xdy+2zdz
yx-2yz+2xz-xy+2yz-2xz
ie ydx+xdy+2zdz
0
ie, ydx+xdy+2zdz=0
d(xy)+2zdz =0
integrating , 2
22zxy+ =C
2
2 2
2xy+z =C v = z xy+
2(x+y+z, xy+z )=0
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
7. 2 2 2x yz p y zx q z xy (AUC May 2010)
Solution:
The equation is of the form Pp+Qq=R
P = 2 2 2x -yz,Q=y -zx,R=z -xy
Lagrange’s subsidiary equations are
dydx dz= =P Q R
ie,2 2 2
dx dy dz= =
x -yz y -zx z -xy
Using lagrange’s multipliers x,y,z we have
3 3 3 2 2 2
xdx+ydy+zdz dx+dy+dz=
x +y +z -3xyz x +y +z -xy-yz-zx
2 2 2
xdx+ydy+zdz
x+y+z (x +y +z -xy-yz-zx)
2 2 2
dx+dy+dz
x +y +z -xy-yz-zx
xdx+ydy+zdz dx+dy+dz
=x+y+z 1
xdx+ydy+zdz=(x+y+z)(dx+dy+dz)
Integrating
222 2 x+y+zyx z+ + =
2 2 2 2
2 2 2 2 2 2x +y +z =x +y +z +2xy+2yz+2zx
2(xy+yz+zx)=0
1u=xy+yz+zx=C
u (x,y,z) = xy+yz+zx -------------1
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
Now,
2 2 2 2
dx-dy dy-dz=
x -yz - y -zx y -zx - z -xy
2 2
d y-zdx-dy=
y+z y-z +x y-zx -y +z(x-y)
d(x-y) d(y-z)
=x-y (x+y+z) (y-z)(x+y+z)
d(x-y) d(y-z)
x-y y-z
2log(x-y)=log(y-z)+logC
2
x-ylog =logC
y-z
2
x-y=C
y-z, ie,
x-yv=
y-z
The general solution is , 0x y
xy yz zxy z
8. Solve : 2 2 2 2 2 2x(y -z )p+y(z -x )q=z x -y (AUC May 2013)
Solution:
It is of the form Pp+Qq=R
Here 2 2 2 2 2 2P=x y -z , Q=y z -x , R= z (x -y )
The S.E is dydx dz= =p Q R
2 2 2 2 2 2
dx dy dz= =
x(y -z ) y z -x z x -y
Use lagrange’s multipliers x,y,z
Each ratio 2 2 2 2 2 2 2 2 2
xdx+ydy+zdz xdx+ydy+zdz= =
0x (y -z )+y (z -x )+z x -y
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
ie, xdx +ydy+zdz =0 integrating, 22 2y ax z+ + =
2 2 2 2
2 2 2x +y +z =a ie, 2 2 2u= x +y +z
Similarly, taking 1 1 1
, ,x y z
as L.M we get
Each ratio = 2 2 2 2 2 2
1 1 1dx dy dz
x y z
y z z x x y
=
1 1 1
0
dx dy dzx y z
ie, 1 1 1
0dx dy dzx y z
, integrating, logx+logy+logz = log b
ie, log (xyz) = logb b=xyz ie, v=xyz
2 2 2(x +y +z ,xyz)=0
9. (D3-2D
2D
1)z = 2e
2x+3x
2y (AUC Dec 2011)
Solution:
The A.E is m3-2m
2 =0 m
2 (m-2) =0
m=0,0, 2
C.F = 1 2 3(y+ox) x (y+ox) (y+2x)
2x+oy
1 3 2 1
2eP.I =
D -2D D
= 2x1 2e
8-0 a=2, b=0
=2xe
4 Replace
1
2
0
D
D
2
2 3 2 1
1P.I = 3x y
D -2D D
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
2
3 2 13
3
1= .3x y
D -2D DD
D
2
13
3.x
21
yD
DD
-11
2
3
3 2D= 1- x y
D D
1 122
3 1
3 2D 4D= 1+ + +.... x y
D D D
22
3
3 2xx y+
D D
2 2
3 4
3 1= x y +6 (x )
D D
5 6
. 660 360
x xy
5 6x y x= +
60 60
2x 5 6
1 2 3
e x y xz=f y+ox +xf y+ox +f y+2x + + +
4 60 60
10. Solve 3 3
x+2y
3 3
z z2 e 4sin ( )
x x yx y
Solution:
The given equation can be written as
(D3-2D
2D
1) z=e
x+2y+4sin (x+y)
The A.E is m3-2m
2=0 m
2(m-2)=0
m=0,0 (or) m=2
C.F = f1(y) + xf2(y)+f3(y+2x)
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
P.I= x+2y
3 2 1 3 2 1
1 1e + 4sin (x+y)
D -2D D D -2D D a=1, b=2 m=1, n=1
x+2y
1
1 1= e + 4sin(x+y)
1-4 -D+2D Replace
1
1
2
D
D
2
2
1
1
1
1
1
D
D
DD
x+2y
1
1 4sin(x+y)=- e +
3 -D+2D
x+2y
2 1
1 4D(sin(x+y))=- e +
3 -D +2DD
x+2y1 4cos(x+y)=- e +
3 1-2
x+2y1=- e -4cos(x+y)
3
x+2y
1 2 3
1z=f (y)+xf (y)+f (y+2x)- e -4cos(x+y)
3
11. Solve: (D2+DD
1-6D
12
) z=y cosx (AUC May 2013)
Solution: m2+m-6=0 m =2, -3
C.F= 1 2(y+2x) (y-3x)
P.I = 22 1 12
1 12
2 2
1 1ycosx = ycosx
D +DD -6D DD 6DD 1+ -
D D
=2
-1-11 12 1 1
2 2 2 2 2
1 DD 6D 1 D 6D1+ - ycosx= 1+ - ycosx
D D D D D D
=
21 1
2 2 2
1 D 6D1- - +.... ycosx
D D D
21 1
2 2
1 D 6D= ycosx - ycosx + ycosx
D D D
2 2
1 1 6= ycosx - cosx + (0)
D D D
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
2
1= ycosx -sinx
D
1 1 1
= (ycosx -sinx) = ysinx+cosxD D D
= -y cos x + sin x
C.F+P.Iz
12. Solve: P (1+q) = qz
Solution:
Give that p(1+q) = qz -------------- 1
It is of the form f (z,p,q) =0
Let u = x+ay
Now u u
=1, =ax y
dz dz
p= =,q=adu du
dz dz dz(1) 1+a =a .z
du du du
dz
1+a =azdu
dz=az-1
dua
dz az-1=
du a
adu =dz az-1
adzdu=
az-1
Integrating on b.s
u = log (az-1) +logc
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
Hence the complete solution is (since the number of a.c= no.of. I.V)
x+ay=log [c(az-1)]
13. Solve 2 2z=px+qy+ p +q +1 (AUC Dec 2011, May 2013)
Solution:
Given that 2 2z=px+qy+ p +q +1
It is of the form z = px+qy+f(p,q) (Clairaut’s form)
Hence the complete integral is 2 2z=ax+by+ a +b +1
(a and b are arbitrary constants)
To find singular solution:
2 2z=ax+by+ a +b +1 --------1
d (1) p.w.r. to a,
2 2 2 2
a -ao=x+ x=
a +b +1 a +b +1 --------2
d (1) p .w.r. to b,
2 2 2 2
b -bo=y+ y=
a +b +1 a +b +1 --------3
2 22 2
2 2
a +bx +y =
1+a +b
2 2
2 2
2 2
a +b1- x +y =1-
1+a +b
2 2
2 2
11-x -y =
1+a +b
2 2
2 2
11-x -y =
1+a +b ---------i
2 2
2 2
11+a +b =
1-x -y --------ii
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
(2) 2 2x=-a 1-x -y by (i)
2 2(3) y=-b 1-x -y by (ii)
Now 2 2 2 2
-x -ya= , b=
1-x -y 1-x -y
Substitute in (1)
2 2
2 2 2 2 2 2
-x y 1z= - +
1-x -y 1-x -y 1-x -y by(ii)
= 2 2
2 2
1-x -y
1-x -y
2 2 2 2 2z= 1-x -y z =1-x -y
2 2 2x +y +z =1 is the singular solution
To find the general integral
Put b = (a) in (1),
22z=ax+ (a)y+ 1+a + (a) -------------4
d. (4) p.w.r to ‘a’
1
22
2a+2 (a) (a)o=x+ '(a)y+
2 1+a + (a)
-------------5)\
Eliminate ‘a’ between (4) and (5) we get the general solution
14. Solve : p2+q
2 = x
2 +y
2
Solution :
Give that p2+q
2 = x
2 +y
2
It is of the form F1(x,p) = F2 (y,q)
Let p2-x
2 = y
2-q
2 =a
2 constant
2 2 2 2 2 2 2 2p -x =a p =a +x p= x +a ------------1
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
2 2 2 2 2 y q a q y a -------------2
pdx qdyz -----------3
Substitute (1) and (2) in (3)
2 2 2 2z= x a dx y -a dy
2 22 2
2 2-1 -1y y -ax x +a ya ax= sin h + + - cos h +b
2 a 2 a2 2
Which is the complete integral
15. Solve :p (1-q2) = q(1-z) (AU Nov/Dec 2009)
Solution:
Given p(1-q2) = q(1-z) -------------1
It is of the form F (z,p,q) = 0
Let z = f (x+ay) be the solution of (1) --------------2
Put x+ay = u u u
1, ax y
(*)
Z = f(u)
z dz u dzp= = . = .1
x du x du
z dz u dzq= = . = .a
y du y du
Using (*) 3
Substituting (3) in (1)
2
2dz dz dz1-a =a 1-z
du du du
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
2
2
2
2
dz1-a =a(1-z)
du
dza = 1-a+az
du
dz 1 = az+(1-a)du a
dz 1= du
aaz+(1-a)
-1
2 1az+(1-a dz= u+b
a
-1 +12az+ 1-a 1
= x+ay +b-1 a+1 a
2
12az+ 1-a 1
= x+ay +b1 aa
2
2 1az+(1-a)= (x+ay)+b
a a which is complete integral
15.Solve: 2
2 22
2 1 12 2
z(D -5DD +6D )z = ysinx (or) -5 +6 =ysinx
x x
zzy y
Solution:
m2-5m+6=0 (m-3) (m-2) =0 m=3,2
1 2C.F= (y+2x)+ (y+3x)
P.I. = 22 1 1
1ysinx
D -5DD +6D
= 21 1
2
2 2
1ysinx
5DD 6DD 1- +
D D
= 2
-11 1
2 2
1 5D 6D1- - ysinx
D D D
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
=
21 1
2 2
1 5 61 ... y sinx
D D
D D D
=
21 1
2 2
1 5D 6Dysinx+ (ysinx)- ysinx
D D D
=2 2
1 5 6ysinx+ sinx- (0)
D D D
21
1D ysin x = sin x
0ysinxD
2
1= ysinx+5(-cosx)
D
1
= -cosx.y-5sinxD
= -ysinx+5cosx
The general solution is
Z =C.F +P.I
17. Solve 2 33 1 1D -7DD -6D z=sin(2x+y)
(AUC May 2013)
Solution:
m3-7 m-6=0
m=-1, -2, 3
1 2 3C.F= (y-x)+ (y-2x)+ (y+3x)
2 31 3 1 1
1P.I = sin(x+2y)
D -7DD -6D D
2by -1
2 =-1
2 32 1 1
sin(x+2y)=
D(D )-7DD -6D
21D by-22 =-4
1 1
sin(x+2y)=
D(-1)-7(-2)D -6(-4)D DD
1by-(1)(2)=-2
1 1
sin(x+2y)=
-D+14D +24D
1
sin(x+2y)=
-D+38D
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
2
1
2 1
-D-38D= sin(x+2y)
D -1444D
1-D-38D sin(x+2y)=
-1-1444(-4)
1
= -cos(x+2y)-38cos(x+2y).25775
1
=- cos(x+2y)+76cos(x+2y)5775
77=- cos(x+2y)
5775
1=- cos(x+2y)
75
18. Find the singular integral of z=px+qy+p2+pq+q
2 (AUC Dec 2012)
Solution:
Z =px+qy+p2+pq+q
2
It is of the form z=px+qy+f(p,q) (clairaut’s form)
Hence the complete soln is z =ax+by+a2+ab+b
2 -------------1
(since the number of a.c = number of I.V)
To find the singular integral
d.(1). p.w.r.to ‘a’ and ‘b’ we get
0 = x+2a+b -------------2
0 = y+2b+a -------------3
(2) x 1 2a+b+x =0 -------------4
(3)x2 2a+4b+2y=0 -------------5
(4) –(5) -3b+x-2y =0
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
3b =x-2y b = 2
3
x y
Similarly,
2y(3) 0=y+2 x- +a
3
2x-y y-2x a=- a=
3 3
Substitute a,,b in (1)
2 2
y-2x x-2y y-2x y-2x x-2y x-2yz= x+ y+ + +
3 3 3 3 3 3
Simplifying,
2 2 2 213z= -x -y +xy z= -x -y +xy9 3
2 2 2 23z=-x -y +xy 3z+x +y -xy=0 which is the singular integral.
19.22 1 1 1 y(2D -DD -D +6D+3D )z=xe (AU June 2012’)
Solution:
Given 22 1 1 1(2D -DD -D +6D+3D )=0
1 11D+ D D-D +3 =0
2
The solution of 1 1
1 1 2 2D-m D -α D-m D -α z=0 is
1 2α x α x
1 1 2 2z=e f (y+m x)+e f y+m x
Here 1 1
2 2
1α =0, m = -
2
α =-3, m = 1
ox -3x
1 2
1C.F=e f y- x +e f y+x
2
1 1
D D+O
D D 1
2
y
2 1 1 1
1P.I= xe
2D -DD -D +6D+3D
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
ox+y
2 1 1 2 1
1=e x
2(D+0) -(D+0)(D +1)-(D +1) +6(D+0)+3(D +1)
2
y
2 1 1 1 1
1=e x
2D -DD -D-D -1-2D +6D+3D +3)
2
y
2 1 1 1
1=e x
2D -DD +5D+D -D +2
2
y
2 1 1 1
1=e x
2D -DD +5D+D -D2 1+
2
2 -12 1 1 1
y 2D -DD +5D+D -De= 1+ .x2 2
22 1 1 1y 2D -DD +5D+D -De= 1- +... x
2 2
ye 5= x-2 2
z= C.F +P.I
20. Solve : 22 1 1 1 2x-y(D -2DD +D -3D+3D +2)z=e (AU Dec 2011)
Solution:
The given equation can be written as
1 1 2x-yD-D -1 D-D -2 z=e
C.F:
1 1
1 1 2 2D-m D -α D-m D -α z=0 is
1 2α x α x
1 1 2 2z=e f (y+m x)+e f (y+m x)
1 2 1 2Here α =1, α =2, m =1, m =1
x 2x
1 2C.F=e f (y+x)+e f (y+x) 1
Replace D 2
D 1
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
2x-y
1 1
1P.I= e
D-D -1 D-D -2
2x-y1
= e2+1-1 (2+1-2)
= 2x-ye
2
z = C.F +P.I
=
2x-yx 2x
1 2
ee f (y+x)+e f (y+x)+
2
21. 22 1 13 4 cos(2 )D DD D z x y xy
(AU Nov/Dec 2012)
Solution:
Give that
22 1 1D +3DD -4D z=0
The A.E is m2+3m-4=0 (m-1) (m+4) =0
1, -4m
1 2C.F (y-4x) (y+x)
21 2 1 1
1P.I = cos(2x+y)
D +3DD -4D
1cos(2x+y)
-4+3(-2)-4(-1)
1=- cos(2x+y)
6
2 22 2 1 1 1 12
2
1 1P.I = xy= xy
D +3DD -4D 3DD -4DD 1+
D
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
21 12
2
1= xy
3DD 4DD 1+ -
D D
21 12
2
1= xy
3D 4DD 1+ -
D D
2-1
1 1
2 2
1 3D 4D= 1+ - xy
D D D
21 1
2 2
1 3D 4D= 1- - +... xy
D D D
1
2
1 3D= xy- xy
D D
2
1 3= xy- x
D D
2
2
1 3x= xy-
D 2
21 1 3x= xy-
D D 2
2 3 2 31 x 3x 1 x y x= y- = -
D 2 6 D 2 2
= 3 4x x
y-6 8
22. Solve: (D2-DD
1+2D)z = e
2x+y+4 (AU Dec 2012)
Solution:
Here m1 = 0, c1 =0
m2 =1, c2 = -2
C.F. =f1(y)+e-2x
f2(y+x)
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
P.I1 = 2x+y 2x+y
2x+y
2 1
e ee = =D -DD +2D 4-2+4 6
ox+oy
2 2 1
1P.I = 4e
D -DD +2D
ox+oy
1
1= 4 e
2D-D +2
ox+oyx= 4 e
2
= 2 x
z= C.F+P.I
-2x 2x+y
1 2
1= f (y)+e f (y+x)+ e +2x
6
23.Find the p.d.e of all planes which are at a Constant distance from the
origin (AU May 2010)
Solution:
The equation of a plane which is at a distance ‘k’ from the origin is
x+my+nz=kl where 2 2 2+m +n =1l ----------------1
Let l =a ; m =b ; n=c we get
ax+by+cz=k ---------------2
2 2 2 2 2 2
(1)
a +b +c =1 c =1-a -b
2 2c= 1-a -b ---------------3
Substitute (3) in (2)
2 2ax+by+ 1-a -b z=k ---------------4
d.p.w.r to ‘x’ 2 2 za+ 1-a -b 0
x
Prepared by Mrs.S.A.Vasantha Gowri, M.Sc., M.Phil., M.C.A., Asst.Prof/Mathematics
2 2
2 2
-a1-a -b p p=
1-a -ba
d (4) p.w.r to ‘y’
2 2 zb+ 1-a -b 0
y
2 2
-bq=
1-a -b
Now,
2 2a b= =- 1-a -b (say)
p q
2 2a=p ; b=q and 1-a -b =-λ
2 2 2 21-p λ -q λ =-λ
Squaring on b.s
2 2 2 2 21-p λ -q λ =λ
2 2 2 21- p +q λ =λ
2 2 2 2 2 2 21=λ +λ (p +q )=λ 1+p +q
2
2 2 2 2
1 1λ = =±
1+p +q 1+p +q
2 21- = 1+P +q λis-veλ
(*)
(4) pλx+qλy-λz=k
2 2
(px+qy-z)λ=k
kpx+qy-z=
λ
kz=px+qy-λ
=px+qy+k 1+p +q
[by (*)]
*****