2. Formation of Partial Differential equations
Partial Differential Equation can be formed either
by elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation
involving three or more variables .
SOLVED PROBLEMS
1.Eliminate two arbitrary constants a and b from
here R is known
( x - a)2 + ( y - b)2 + z2 = R2
constant .
3. (OR) Find the differential equation of all spheres
of fixed radius having their centers in x y- plane.
solution
( x - a)2 + ( y -b)2 + z2 = R2.......(1)
Differentiating both sides with respect to x and y
x a
=- -
2 2( )
y b
=- -
2 2( )
q
p z
¶
¶
¶
¶
= ¶
y
z z
x
z z
z
¶
x
y
=
¶
x a pz y b qz
¶
- =- - =-
,
,
4. By substituting all these values in (1)
2 2 2 2 2 2
p z + q z + z =
R
2
z R
2 2 1
2
+ +
Þ =
p q
or
1
2
z R
2 2
2
ö
+ ÷ ÷ø
æ
z
¶
+¶ ÷ø
ç çè
z
¶
=
æ
ö ¶
çè
y
x
5. 2. Find the partial Differential Equation by eliminating
arbitrary functions from
z = f (x2 - y2 )
2 2
SOLUTION
z f x y
( )..........(1)
d . wr . . to .
xandy
' 2 2
( ) 2 ......(2)
' 2 2
f x y y
( ) 2 ......(3)
z
¶
z
¶
y
f x y x
x
= - ´-
¶
= - ´
¶
= -
6. (2)
By (3)
x
=-
z
¶
ö çè
ö
÷ ÷ø
x
æ
¶
x
z
=- Þ + =0
ç çè
¶
÷ø
æ
¶
py qx
y
p
q
y
y
7. 3.Find Partial Differential Equation
by eliminating two arbitrary functions from
z = yf (x) +xg( y)
SOLUTION
z = yf (x)+xg( y)......(1)
Differentiating both sides with respect to x and y
( ) ( )........(2)
f x xg y
( ) ( )........(3)
z
¶
z
¶
y
yf x g y
x
= + ¢
¶
= ¢ +
¶
8. Again d . w .r. to x and yin equation (2)and(3)
= ¢ + ¢
f x g y
z
¶
x y
( ) ( )
¶ ¶
x (2) y (3)...... to ...
get
2
´ + ´
9. =
y z
+ ¶
¶
x z
xg y yf x xy f x g y
( ) ( ) ( ( ) ( ))
( )
ö
÷ ÷ø
æ
¶ ¶
¶
= + ¢+ ¢
z xy z
= + ¶
ç çè
z xy f g
y z
+ ¶
¶
x z
Þ ¶
¶
+ + ¢ + ¢
¶
x y
y
x
y
x
2
10. Different Integrals of Partial Differential
Equation
1. Complete Integral (solution)
Let
F x y z ¶ z
¶
z
F x y z p q
( , , , , ) = ( , , , , ) = 0......(1)
y
¶
x
¶
be the Partial Differential Equation.
The complete integral of equation (1) is given
by
f (x, y, z, a,b) = 0..........(2)
11. 2. Particular solution
A solution obtained by giving particular values to
the arbitrary constants in a complete integral is
called particular solution .
3.Singular solution
The eliminant of a , b between
=
x y z a b
f f
f
( , , , , ) 0
= ¶
=
¶
0, 0
¶
¶
a b
when it exists , is called singular solution
12. 4. General solution
In equation (2) assume an arbitrary relation
of the form . b = f (a) Then (2) becomes
f (x, y, z, a, f (a)) = 0.........(3)
Differentiating (2) with respect to a,
¶ f a
a b
¢( ) = 0..........(4)
+ ¶
¶
f f
¶
The eliminant of (3) and (4) if exists,
is called general solution
13. Standard types of first order equations
TYPE-I
The Partial Differential equation of the form
f ( p,q) = 0
has solution
z = a x + b y + c with
f (a,b) = 0
TYPE-II
The Partial Differential Equation of the form
z = px + qy + f ( p, q)
is called Clairaut’s form
of pde , it’s solution is given by
z = ax + by + f (a,b)
14. f (z, p,q) = 0
TYPE-III
If the pde is given by
then assume that
z = f
x +
ay
( )
u = x +
ay
z f
u
( )
=
15. dz
z
= ¶
¶
.1
a a dz
u
du
z
u
u
= ¶
¶
u
u
= ¶
¶
y
z
p z
q z
= ¶
¶
y
du
u
x
z
x
=
¶
¶
¶
= ¶
=
¶
¶
¶
= ¶
.
The given pde can be written as
f ( z , dz , a dz
) = 0
.And also this can
dx
dy
be integrated to get solution
16. TYPE-IV
The pde of the form f (x, p) = g( y,q)
can be
solved by assuming
f ( x , p ) = g ( y , q )
=
a
f x p = a Þ p =
f
x a
( , ) ( , )
g y q = a Þ q = Y
y a
( , ) ( , )
dz = ¶
z
dx + ¶
z
dy
¶
x
¶
y
= f
( , ) +Y( , )
dz x a dx y a dy
Integrate the above equation to get solution
17. SOLVED PROBLEMS
1.Solve the pde p2 - q = 1
and find the complete
and singular solutions
Solution
Complete solution is given by
z = ax + by + c
2
Þ = -
1
1
2
a b
- =
b a
with
18. z =ax +(a2 -1) y +c
d.w.r.to. a and c then
2
= =
1 0
z
¶
z
¶
¶
= +
¶
c
x ay
a
Which is not possible
Hence there is no singular solution
pq + p +q =0
2.Solve the pde and find the
complete, general and singular solutions
19. Solution
The complete solution is given by
z = ax +by +c
with
ab a b
+ + =
a b
1
0
= -
+
b
.......(1)
z b + +
1
x by c
= -
b
+
20. = -
1
( )
1 0
0
1
2
= =
z
¶
z
¶
¶
+ =
+
¶
c
x y
b b
no singular solution
To get general solution assume that
c = g(b)
( ).......(2)
From eq (1)
z b + +
1
x by g b
= -
b
+
21. z = -
1
+ + ¢
( ) ( ).......(3)
1
2 x y g b
¶
c b
+
¶
Eliminate from (2) and (3) to get general
solution
3.Solve the pde
z = px + qy + 1+ p2 + q2
and find the complete and singular solutions
Solution
The pde
z = px + qy + 1+ p2 + q2
is in Clairaut’s form
22. complete solution of (1) is
z =ax+by + 1+a2 +b2 .......(2)
d.w.r.to “a” and “b”
ö
........(3)
0
x a
y b
1
0
1
2 2
2 2
÷ ÷ ÷ ÷ ÷
ø
=
+ +
= +
z
¶
z
¶
¶
=
+ +
= +
¶
a b
b
a b
a
23. y b
2 2
2
From (3)
x a
x y a b
1 ( )
+ = +
1
1
2
1
1
,
1
2 2
2 2
2 2
2 2
2 2
2
2 2
2
x y
a b
a b
a b
a b
= - +
+ +
Þ
+ +
+ +
=
+ +
=
24. 2
2 2
ax a
a b
+ +
2 2
=
=
by b
a b
+ +
1 1
= Þ = - +
0 1 ( )
1
1
1
0
1
0
1
0
1
+ +
2 2 2
2 2 2
2 2
2 2
2 2
2
+
+
-
Þ + + =
=
+ +
+ + + + -
x y z
z x y
a b
z
a b
ax by a b
is required singular solution
25. 4.Solve the pde(1- x) p + (2 - y)q = 3- z
Solution
pde
- x p + - y q = -
z
= + + - -
(1 ) (2 ) 3
z px qy p q
(3 2 )
Complete solution of above pde is
z =ax +by +(3-a -2b)
5.Solve the pde p2 + q2 = z
Solution
Assume that z =f(x+ay)
26. u = x +
ay
z =f
(u)
dz
z
= ¶
¶
.1
a a dz
u
du
z
u
u
= ¶
¶
u
u
= ¶
¶
y
z
p z
q z
= ¶
¶
y
du
u
x
z
x
=
¶
¶
¶
= ¶
=
¶
¶
¶
= ¶
.
2
2
÷ø
p q z dz + a 2
æ ÷ø
dz
= 2
+ = Þ æ
2 2 z
du
du
ö çè
ö çè
From given pde
27. du
dz
z a
z
z
2
a
dz
ö çè
dz
ö du
çè
a
du
2 2
2
1
1
1
1
+
Þ =
+
= ÷ø
æ
+
= ÷ø
æ
Integrating on both sides
b
2
z x ay
a
b
a
z u
+
= +
+
+
+
=
2
1
2
1
2
28. 6. Solve the pde zpq = p + q
Solution
Assume
q = ap
Substituting in given equation
dy
zpap = p +
ap
q a
p a
= + = +
1 , 1
dx +¶
z
x
¶
dx a
z
=¶
¶
dz a
az
dy
y
dz z
z
az
Þ = 1 + + 1
+
29. zadz =(1+a)(dx+ady)
Integrating on both sides
a z 2
=(1+a)(x+ay)+b
2
7.Solve pde
pq xy
z
z
¶
¶
(or) xy
y
x
=
¶
¶
=
( )( )
Solution
q
y
p =
x
30. Assume that
y
= =
p ax q y
p
= =
,
dz pdx qdy axdx y
dy
a
a
a
q
x
= + = +
Integrating on both sides
2 2
z =a x + y +
b
a
2 2
31. 8. Solve the equation p2 + q2 = x + y
Solution
2 2
p - x = y - q =
a
p = a + x ,
q = y -
a
dz = pdx + qdy = a + xdx + y -
ady
integrating
3
3
z = a + x + y - a 2 + b
( ) 2
( )
3
2
32. Equations reducible to the standard forms
(i)If and occur in the pde as in
(xm p) ( ynq)
F(xm p, ynq) = 0 Or in F(z, xm p, ynq) = 0
Case (a) Put and
x1-m = X y1-n = Y
m ¹ 1 n ¹ 1
if ;
n
m
(1 )
n y
z
= ¶
¶
z
= ¶
¶
Y
X
z
= ¶
¶
z
Y
Y
x
p z
q z
= ¶
¶
y
m x
X
x
X
x
-
-
-
¶
¶
¶
= ¶
-
¶
¶
¶
= ¶
(1 )
33. (1 ) (1 )
n Q n
(1 ) (1 )
x m
p = ¶
z
y q = ¶
z
Y
m P m
X
n
- = -
¶
- = -
¶
z =
P z
¶ ,
= ¶
where Q
Y
X
¶
¶
Then F ( x m p , y n q ) = 0 reduces to F(P,Q) = 0
F(z, xm p, ynq) = 0 F(z, P,Q) = 0
Similarly reduces
to
34. case(b)
m = 1 n = 1
log x = X ,log y = Y
If or
put
qy Q
p = ¶
z
q =¶
z
1
Y y
px P
X x
Þ =
¶
Þ =
¶
1
(zk p) (zkq) F(zk p, zkq)
(ii)If and occur in pde as in
( , ) ( , ) 1 2 Or in f x zk p = f y zkq
35. Case(a) Put z 1 + k = Z if k ¹ -1
= + ¶
¶
k k
- - -
`1 `1
(1 ) (1 )
= + ¶
¶
- - -
z q k Q
z k Z
z k Z
Þ = +
¶
y
Z
Z
= ¶
¶
z
Z
Z
y
z
z
= ¶
¶
y
z p k P
x
x
z
x
k k
`1 `1
(1 ) (1 )
¶
¶
¶
Þ = +
¶
¶
¶
¶
Z =
Q
P Z
¶ , where
= ¶
y
x
¶
¶
Given pde reduces to
F(P,Q) and
f ( x , P ) = f ( y , Q
) 1 2
36. Case(b) if k = -1 log z = Z
-
1
z q Q
z Z
= ¶
¶
z Z
y
Z
Z
= ¶
¶
Z
Z
= ¶
¶
y
z
z
z
= ¶
¶
y
z p P
x
x
z
x
Þ =
¶
¶
¶
¶
Þ =
¶
¶
¶
¶
-
1
Solved Problems
1.Solve
p2x4 +q2 y4 = z2
2 2 2 2
ö
æ
+ ÷ ÷ø
æ
qy
px
Solution 1.......(1)
= ÷ ÷ø
ç çè
ö
ç çè
z
z
37. m n
= =
k
2, 2
=-
1
x-1 = X y-1 = Y log z = Z
= - ¶
¶
- -
2 2
= - ¶
¶
- -
zy Q
zx Z
zy Z
Y
X
X
¶
¶
Y
Y
y
z
Z
Z
p = ¶
z
q z
y
zx P
X
x
z
Z
Z
x
2 2
= -
¶
¶
¶
¶
¶
= ¶
= ¶
¶
= -
¶
¶
¶
= ¶
¶
Z =
Q
P Z
¶ , where
= ¶
Y
X
¶
¶
38. Q
px = - = -
P qy
z
z
2 2
,
( ) ( )
2 2
+ =
P Q
- + - =
P Q
1
1
2 2
(1)becomes
Z aX bY c
= + +
a 2 + b 2 = 1, b = 1
-
a
2
log 1
2 2 2
z = ax + - a y +
c
39. 2. Solve the pde p2 + q2 = z2 (x2 + y2 )
SOLUTION
p = + ÷ø
( 2 2 ).....(1)
2 2
x y
q
z
z
æ + ÷ø
ö çè
çè
æ
ö k =-1 log z = Z
-
1
z q Q
z Z
= ¶
¶
z Z
y
Z
Z
= ¶
¶
y
z
z
= ¶
¶
= ¶
¶
Z
z
z
y
z p P
x
x
Z
x
Þ =
¶
¶
¶
¶
Þ =
¶
¶
¶
¶
-
1
40. Eq(1) becomes
P 2 + Q 2 = ( x 2 +
y
2 ).....(2)
P 2 - x 2 = y 2 - Q 2 =
a
2
x a x
1 2 2
b
2
z a x
ö çè
y y a a y
ö a
çè
a
+ ÷ø
- æ
-
+ + + ÷ø
= æ
-
-
1
2 2 2
cosh
( )
2 2
( )
2
sinh
2
log
41. Lagrange’s Linear Equation
Def: The linear partial differenfial equation
of first order is called as Lagrange’s linear Equation.
This eq is of the form Pp + Qq = R
Where P , Q and R are functions x,y and z
The general solution of the partial differential
equation P p + Q q = R is F(u,v) = 0
Where is arbitrary function of
and
F 1 u(x, y, z) = c
2 v(x, y, z) = c
42. Here u = c and v = 1 c2 are independent solutions
dz
of the auxilary equations
R
dx = dy
=
Q
P
Solved problems
1.Find the general solution of x2 p + y2q = (x + y)z
Solution
dx
dy
dz
auxilary equations are = =
x
2 y
2 ( x + y )
z
43. dy
dx
=
Integrating on both sides
2 y
2
u x y c
( - 1 -
1
) 1
x
= - =
2 2 ( )
x y z
( )
( )( ) ( )
dz
z
dx -
dy
d x -
y
d x -
y
( )
x y
dz
x y z
x y x y
dz
x y
=
-
+
=
- +
+
=
-
( )
Integrating on both sides
44. x - y = z +
c
= - =
log( ) log log
v x y z c
2
1
2
( )
-
The general solution is given by F(u,v) = 0
F(x-1 - y-1,(x - y)z-1) = 0
2.solve x2 (y - z) + y2 (z - x)q = z2 (x - y)
solution Auxiliary equations are given by
dz
dy
dx
2 ( ) y 2 ( z x
) z2 (x y)
x y z
-
=
-
=
-
45. dz
dy
dx
2 2 2
z
y
x
( ) ( ) ( )
dy
+ +
dz
dx
2 2 2
y - z + z - x + x -
y
( ) ( ) ( )
0
dz
dy
+ + =
dx
2 2 2
-
=
-
=
-
z
y
x
z
y
x
x y
z x
y z
Integrating on both sides
46. u =1 +1 +1 =
a
x y z
- - -
1 1 1
z dz
z x y
y dy
y z x
x dx
x y z
( ) ( ) ( )
1 1 1
+ +
x y - z + y z - x + z x -
y
( ) ( ) ( )
0
dz
dy
+ + =
-
=
-
=
-
- - -
z
y
dx
x
x dx y dy z dz
Integrating on both sides
v = xyz =b
47. The general solution is given by
F(x-1 + y-1 + z-1, xyz) = 0
HOMOGENEOUS LINEAR PDE WITH
CONSTANT COEFFICIENTS
Equations in which partial derivatives
occurring are all of same order (with degree
one ) and the coefficients are constants ,such
equations are called homogeneous linear PDE
with constant coefficient
48. a z
n
a z
a z
+ ¶
+ ¶
+ ¶
........ ( , ) 1 1 2 2 2 F x y
¢ = ¶
= ¶
¶
Assume that , .
y
D
x
D
¶
¶
nth
then order linear homogeneous equation is
given by
n + n- ¢ + n- ¢ + + ¢ =
1 D a D D a D D a D n z F x y
n
( 2 2 ......... ) ( , )
2
1
or
f (D,D¢)z = F(x, y).........(1)
y
x y
x y
x
z
n
n
n n
n
n
n
n
=
¶
¶ ¶
¶ ¶
¶
- -
49. The complete solution of equation (1) consists
of two parts ,the complementary function and
particular integral.
The complementary function is complete
solution of equation of f (D,D¢)z = 0
Rules to find complementary function
Consider the equation
k z
0 2
2
+ ¶
2
2
k z
+ ¶
2 1
2
=
¶
¶ ¶
¶
¶
y
x y
x
z
or
D2 + k DD¢ + k D¢ z =
( 2 ) 0.............(2)
1 2
50. The auxiliary equation for (A.E) is given by
D2 + k DD¢ + k D¢ =
D = m,D¢ =1
And by giving
2 0
1 2
m2 + k m+ k =
The A.E becomes 0....(3) 1 2
Case 1
If the equation(3) has two distinct roots 1 2 m ,m
The complete solution of (2) is given by
( ) ( ) 1 1 2 2z = f y + m x + f y + m x
51. Case 2
If the equation(3) has two equal roots i.e 1 2 m = m
The complete solution of (2) is given by
( ) ( ) 1 1 2 1z = f y +m x +xf y +m x
Rules to find the particular Integral
Consider the equation
D2 + k DD¢+ k D¢ z = F x y
( 2 ) ( , )
1 2
f (D,D¢)z = F(x, y)
52. Particular Integral (P.I)
F x y
( , )
f D D
¢
( , )
=
Case 1 If
F(x, y) = eax+by
then P.I=
¢
ax +
by
= ¹
, ( , ) 0
1
( , )
1
( , )
+
e f a b
f a b
e
f D D
ax by
D - a ¢
f (a,b) = 0 ( D )
If and is
b
factor of f (D,D¢)
then
53. P.I =xeax+by
If and D - a ¢
is
factor of
2
then P.I
F(x, y) = sin(mx +ny)or cos(mx +ny)
= +
sin( )
mx ny
= +
2 2 f m2 mn n2
( , , )
sin( )
( , , )
mx ny
f D DD D
- - -
¢ ¢
Case 2
P.I
f (a,b) = 0 ( D )2
b
f (D,D¢)
= x eax+by
2
54. Case 3
F(x, y) = xm yn
1 = ¢ -
xm yn [ f D D ] xm yn
f D D
( , ) 1
¢
( , )
P.I =
[ ( , )] 1 Expand f D D ¢ - in ascending powers of
D or D ¢ and operating on x m y n term by term.
Case 4 when is any function of x
and y.
P.I=
F(x, y)
1 F x y
( , )
f D D¢
( , )
1 ( , ) ( , )
=ò -
- ¢
F x y F x c mx dx
D mD
55. (D -mD¢) f (D,D¢)
Here is factor of
(y + mx)
Where ‘c’ is replaced by after integration
Solved problems
1.Find the solution of pde
(D3 - D¢3 + 3DD¢2 - 3D2D¢)z = 0
Solution
The Auxiliary equation is given by
56. Solution
The Auxiliary equation is given by
m3 -1+3m-3m2 = 0
By taking
D = m,D¢ = 1
m =1,1,1.
2
1 2 = f y + x + xf y + x + x f y + x
Complete solution ( ) ( ) ( ) 3
2. Solve the pde (D3 + 4D2D¢ - 5DD¢)z = 0
Solution
The Auxiliary equation is given by
57. 3 2
m m m
+ - =
4 5 0
= -
0,1, 5
z f y f y x f y x
( ) ( ) ( 5 )
1 2 3
m
= + + + -
3. Solve the pde (D2 + D¢2 )z = 0
Solution
the A.E is given by m2 +1=0
m = ±i
( ) ( ) 1 2 z = f y + ix + f y - ix
58. 4. Find the solution of pde
(D2 + 3DD¢ - 4D¢2 )z = e2x+ 4 y
Solution
Complete solution =
Complementary Function + Particular Integral
m2 + 3m- 4 = 0
The A.E is given by
m =-4,1
. ( ) ( 4 ) 1 2 C F = f y + x +f y - x
59. 2 4
P I e
3 4 36
.
2 4
2 2
-
=
+ ¢- ¢
=
x+ y e x+ y
D DD D
Complete solution
= C.F + P.I
e x y y x y x
36
( ) ( 4 )
2 4
1 2
+
= f + +f - -
60. 5.Solve (D3 - 3DD¢ + 2D¢3 )z = e2x- y + ex+ y
Solution
3
A E m m
= - +
. 3 2
= -
1,1, 2.
C F y x x y x y x
. ( ) ( ) ( 2 )
1 2 3
m
= f + + f + + f
-
x y 2
x y
2
e
P I e
=
. 2 2
1 D 3 - DD ¢ 2 + D
¢
3
D D D D
- ¢ + ¢
3 2 ( ) ( 2 )
=
- -
61. 2
P I e
- -
( ) ( 2 ) 9
.
2
2
1
x y xe x y
D D D D
=
- ¢ + ¢
=
x y x y
P I e
2 3 2 3 2 2
x y
P I x e
e
D D D D
D DD D
+
+ +
=
- ¢ + ¢
=
- ¢ + ¢
=
6
.
3 2 ( ) ( 2 )
.
2
2
1 2 z = C.F + P.I + P.I
62. x y
x -
y
z = y + x + x y + x + y - x + xe +
x e +
9 6
( ) ( ) ( 2 )
2 2
1 2 3 f f f
6.Solve (D2 -DD¢)z =cos x cos 2y
Solution
(D2 -DD¢)z = 1 x + y + x - y
[cos( 2 ) cos( 2 )]
2
2
A E m m
. 0
0,1
C F y x y
. ( ) ( )
1 2
m
=f + +f
=
= - =
63. P I x y = +
. cos( 2 ) 1 2 x y x y
cos( 2 )
= +
cos( 2 )
(( 1) ( 2))
= +
D DD
( )
- - -
- ¢
P I x y
= - = x -
y x y
D DD
. cos( 2 ) 2 2 -
cos( 2 )
3
cos( 2 )
(( 1) (2))
( )
= -
- -
- ¢
( ) ( ) cos( 2 ) 1 1 2 z = f y + x +f y - x + x + y - x - y
cos( 2 )
3
7.Solve (D2 + DD¢ - 6D¢2 )z = x2 y2
Solution
. 2 6 0
= -
A E m m
= + - =
2, 3.
m
64. . ( 2 ) ( 3 ) 1 2 C F = f y + x +f y - x
2 2
2
2
æ ¢
2
2
2
æ ¢
æ ¢
2
P I x y
D
é
1 1 6
é
D 2
D
2 2
1
2
2
2 2
2 2
6
D
D
D
ù
ö
ö
1 6 6
.
x y
D
D
D
D
D
x y
D
D
D
D DD D
ù
ú ú
û
ê ê
ë
ö
÷ ÷ø
ç çè
-
¢
+ ÷ ÷ø
ç çè
-
¢
= -
úû
êë
÷ ÷ø
ç çè
-
¢
= +
+ ¢ - ¢
=
-
-
65. é ¢
ö
D D
1 6
2
D
D
2 2
2
x
D x y x y
2 6 2 2
2
x
3 4 4
D x y x y 6 2
x x
3 4
D x y x y 8 2
x
ù
úû
é
é
2 2 2
é
æ ¢
æ
-
-
-
2 2 2
= + +
é
= + +
êë
ù
ö
ö
ù
úû
êë
ù
ù
úû
êë
2
+ ÷ ÷ø
ç çè æ
= - -
úû
êë
+ ÷ ÷ø
ç çè
= - -
úû
êë
+ ÷ ÷ø
ç çè
-
¢
= -
-
2
90
2
60
12
12
3
2
12
12
3
2
4 2 5 6
2
2
2 2 2
2 2
2
2
2
x y x y x
D
D
D
x y
D
D
D
66. 7.Solve (D2 - 5DD¢ + 6D¢2 )z = y sin x
Solution
A.E is m2 - 5m+ 6 = 0
m =3,m = 2.
C . F =P . I f ( y y sin x
+ 3 x ) +f ( y + 2 x
) 1 2 2
2 y x
y x
sin
sin
ù
é
[ ]
[ x x x y x x]
1
1
1
1
D D
a x x x x
D D
a x xdx
D D
D D
D D
D D D D
D DD D
2 cos 2sin ( 2 ) cos
( 3 )
cos 2( cos sin )
( 3 )
( 2 )sin
( 3 )
( 2 )
( 3 )
( 3 )( 2 )
5 6
- - +
- ¢
=
- - - +
- ¢
=
-
- ¢
=
úû
êë
- ¢ - ¢
=
- ¢ - ¢
=
- ¢ + ¢
=
ò
67. P I y x
. sin 2 2
y x
y x
sin
sin
ù
é
here
= +2
[ ]
[ x x x y x x]
1
1
1
1
D D
a x x x x
D D
a x xdx
D D
D D
D D
D D D D
D DD D
2 cos 2sin ( 2 ) cos
( 3 )
cos 2( cos sin )
( 3 )
( 2 ) sin
( 3 )
( 2 )
( 3 )
( 3 )( 2 )
5 6
- - +
- ¢
=
- - - +
- ¢
=
-
- ¢
=
úû
êë
- ¢ - ¢
=
- ¢ - ¢
=
- ¢+ ¢
=
ò (a y x)
68. [ y x x
]
1
=
D D
= ò ( - ( b - 3 x )cos x -
2sin x )
dx
(b y x)
b x x x x x
= - + + +
sin 2cos 3( sin cos )
y x x x x x x
= - + + + +
( 3 )sin 2cos 3( sin cos )
x y x
5cos sin
cos 2sin
( 3 )
= -
- -
- ¢
here
= +3
69. Non Homogeneous Linear PDES
If in the equation f (D,D¢)z = F(x, y)............(1)
the polynomial expression f (D,D¢)
is not
homogeneous, then (1) is a non- homogeneous
linear partial differential equation
Ex (D2 + 3D + D¢ - 4D¢2 )z = e2x+3 y
Complete Solution
= Complementary Function + Particular Integral
To find C.F., factorize
into factors of the form
f (D,D¢)
(D -mD¢-c)
70. If the non homogeneous equation is of the form
- ¢ - - ¢ - =
D m D c D m D c z F x y
( )( ) ( , )
1 1 2 2
= c x + + c x +
C F e 1 f y m x e 2 f
y m x
. ( ) ( )
1 2
1.Solve (D2 -DD¢+D)z = x2
Solution
f (D,D¢) = D2 - DD¢ + D = D(D - D¢ +1)
. ( ) ( ) 1 2 C F =e-xf y +x +f y
71. 2
D
P I x
. 1 1 ( 1)
x D
x D
1 ( 1) ( 1) ......
ù
úû
é ¢ +
= -
é ¢ +
+ úûù
é
ù
+ + = úû
êë
ù
é
é
é ¢ +
= + +
êë
ù
ú úû
ê êë
ù
+ úû
êë
êë
= +
úû
êë
- ¢ +
=
-
3 12 3.4 3.4.5 12.5.6
1
3 4 4 5 6
2
2
2
2
2 2
2
2
1
2 2
x x x x x x
D
x
D
D
D
x
D
D DD D D
72. 2.Solve (D +D¢-1)(D + 2D¢-3)z = 4
Solution
( ) ( 2 ) 4 1
3
z =exf y -x +e 3
xf y - x +
1
