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Title: Mathematical inductions
Description: This is mathematical inductions class notes.
Description: This is mathematical inductions class notes.
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Module I
Partial Differential Equations
Syllabus
Origin of partial differential equations, linear and non-linear partial equations of first order, lagrange’s
equations, Charpit’s method, Cauchy’s method of characteristics, solution of linear partial differential
equation of higher order with constant coefficients, equations reducible to linear partial differential
equations with constant coefficients
...
No
...
1
...
2
Order of a Partial Differential Equation
3
1
...
4
Linear and Non-linear Partial differential Equation
4
1
...
6
Formation of a Partial Differential Equation-
5
1
...
8
Non-linear Partial Differential Equations of Order One
14
1
...
10
Partial differential equation of higher order
23
1
Applications
Partial differential equations are used to mathematically formulate the solution of physical and other problems
involving function of several variables, such as the propagation of heat or sound, fluid flow, Waves, elasticity,
electrostatic, electrodynamics etc
...
For example,
Fluid mechanics is used to understand how the circulatory system works, how to get rockets
and planes to fly, and even some extent how the weather behaves
...
Partial differential equations in Stokes flow-
Solid sphere
vR
vθ
a
Z -axis
Approaching fluid
Uniform flow
Figure 1: Newtonian fluid around a solid sphere
The governing equation for the creeping viscous flow of an incompressible fluid are
=
(1)
and
...
Introducing the stream function
spherical coordinate system ( , , ) by
v ( , ) ̂ +v ( , ) ̂ =− ×
̂
( , ), which is related to velocity in
(3)
Using equation (1) and (3), we get fourth order partial differential equation
!
=0
(4)
where
=
"
"
+
(#$% " )
"
"
%"
, & = cos
Applying the method of separation of variable we get the general solution of equation (4) and after ignoring all terms
which are not applied in presented model (Figure 1), we get the stream function solution for the flow field for the
Figure 1, which is in the form of partial differential equation
...
1
...
Or
An equation involving derivatives or differentials of one or more dependent variables w
...
t
...
/
+ / = 0,
...
/
...
0
...
r
...
a single independent variable is called an ordinary differential
equation
Example
...
2
...
2 !
...
0
Partial Differential Equation
6
A differential equation involving partial derivatives with respect to more than one independent variables is called
partial differential equation
...
2Order of a Partial Differential Equation
Order of a partial differential equation is defined as the order of the highest partial derivative occurring in the partial
differential equation
...
3Degree of a Partial Differential Equation
The degree of a partial differential equation is a degree of highest order derivative occurs in it
...
For example,
>
?
>
@
+
>
?
">
?"
/D
= = + 0/ {first degree}
A>
+
>
@
= 1+
>
?
>
@
= 20
@A
#C
>
@
+
{first degree}
{second degree}
E==
>
@
{second degree}
1
...
A partial differential equation which is not linear is called a non-linear partial
differential equation
...
Example 1: Find the order and degree of the following PDE:
(1)
(2)
>
?
">
?"
(3) 0
(4)
">
?"
+
A>
@A
= 1+
>
?
+
= 1+
=2
>
?
>
@
>
?
= 1,
==
$#C
>
@
,
Order 3, Degree 1
>
,
@
Order 2, Degree 1
Order 1, Degree 2
7= $#
7 =
⟹9
; = 31 + 4
70
7/
4
⟹ 1+
(5) 1 +
>
?
⟹ 1+
>
@
#C
=
> :
?
">
?"
=
">
@"
= 1,
Order 2, Degree 2
#C
:
">
@"
Order 2, Degree 2
1
...
6Formation of a Partial Differential Equation(1) By elimination of arbitrary constants
a) When the number of arbitrary constants is less than the number of independent variables, then the
elimination of arbitrary constants usually gives more than one partial differential equations
...
r
...
0, we get
>
?
= L (2)
>
@
=1
Differentiating (1) partially w
...
t
...
(3) does not contain arbitrary constant, so (3) is also a partial differential equation under the consideration
...
b) When number of arbitrary constants are equal to the number of independent variables, then the elimination
of arbitrary constants shall give a unique P
...
E
...
c) When number of arbitrary constants is greater than the number of independent variables, then the
elimination of arbitrary constants lead to a P
...
E
...
Example 2: Form partial differential equation from the following equations by eliminating the arbitrary constants
...
r
...
0and /, we get
5
>
?
= Land
>
@
=N
Substituting these values of L and N in (1), we get
= = 03
7=
7=
7=
7=
4+/3 4+3 4 +3 4
70
7/
70
7/
Which is the required partial differential equation
...
r
...
0
7=
= H = (/ + N)
70
Differentiating = partially w
...
t
...
r
...
0and /, we get
7= 7=
43 4
70 7/
7=
= 2(0 − L)
70
7=
= 2(/ − N)
7/
Solution(d) Given L= + N = L 0 + /
4= = 3
(1)
7=
7=
4 +3 4
70
7/
Differentiating (1) partially w
...
t
...
r
...
/, we get
#
(3)
Multiplying (2) and (3)
(2) By the elimination of arbitrary function
HI = 1
Example 3:Form the partial differential equation by eliminating the arbitrary functions from the following
(a)= = +(0 − / )(b)= =
@
+(0 + /)
6
(c)= = (0)
...
r
...
0and /, we get
>
?
>
@
= 20+′(0 − / ) (2)
= (−2/)+′(0 − / ) (3)
Dividing (2) by (3), we obtain
>
?
>
@
=−
0
/
H
0
=−
I
/
Solution (b): Given = =
@
+(0 + /)
⟹ H/ + I0 = 0
(1)
Differentiating (1) partially w
...
t
...
(/)
Differentiating (1) partially w
...
t
...
(/) (2)
= I = (0)
...
r
...
/
">
? ?
= J = ′(0)
...
(2) and (3), we get
⟹
Practice questions
7 = 7 =
+
=0
70
72
`1
...
Ans
...
Ans
...
Ans
...
Ans
...
Form partial differential equations from the following equations by eliminating the arbitrary functions:
(a)= = +(0 + / )Ans
...
Ans
...
Ans
...
7Linear Partial Differential Equations of order One
A quasi-linear partial differential equation of order one is of the form H + `I = a, where
functions of 0, /, =
...
The general Solution of the Lagrange equation
H + `I = a (1)
is
(u, v) = 0 (2)
where is an arbitrary function and
u(0, /, =) = c#andv(0, /, =) = c are two independent Solution of
Here, *# and * are arbitrary constant
...
0
=
...
Differentiating (2) partially w
...
t
...
(1), with
=
g l
> @
−
F g
,
> @
`=
g F
? >
−
=
F g
? >
F g
? @
−
g l
(6)
? @
and a =
F g
? @
−
g l
? @
∴ (u, v) = 0is the Solution of Lagrange’s equation
...
0 +
...
e = 0
70
7/
7=
7v
7v
7v
...
v = 0
70
7/
7=
By cross multiplication, we have
F g
@ >
...
0
=
...
=
−
g F
? @
...
The standard form of Lagrange’s equation
H + `I = a(1)
2
...
(1)
p?
q
=
p@
r
=
p>
s
(2)
3
...
Suppose u = L and v = N are two Solution of eq
...
5
...
Example 1:Solve the following partial differential equations
(i)/=H − 0=I = 0/(ii)H tan 0 + I tan / = tan =
Solution (i)
...
(1)with H + `I = a, we get
= /=, ` = −0=anda = 0/
The Lagrange’s auxiliary equations are
⇒
p?
[W
=
p@
$VW
=
p>
V[
...
(2)
⇒ 0
...
/ = 0 (3)
...
0
...
0
...
/
...
0
=
...
0
...
0 − cot /
...
(2)
log sin / − log sin = = log * or
Therefore, the general solution is
@
>
...
=
=
tan / tan =
=*
⇒ cot /
...
=
(4)
sin 0 sin /
3
,
4=0
sin / sin =
Example 2:Solve the following partial differential equations
(i)H + 3I = 5= + tan / − 30
(ii) H/ + I0 = 0/= (0 − / )
Solution
...
0
(2)
Taking first two fractions of (2)
Taking first and last fraction of (2)
Or
...
0
...
0
...
(3) (4)
From eq
...
(ii) Given
(/ − 30, 50 − log•5= + tan(/ − 30)€ = * ) = 0
11
H/ + I0 = 0/= (0 − / ) (1)
The Lagrange’s auxiliary equations are
p?
•
=
p@
?
...
=
=
`
a
...
/
=
y
0
⇒ 0
...
/ = 0
Integrating
0 − / = *# (3)
Taking last two fraction of (2)
⇒
p@
?
...
/
=
0/= (0 − / )
0
p>
= ?@>" U ,
ƒ
⇒ /*#
...
=
=
using (3)
2
=*
=
using (3)
2
30 − / , / (0 − / ) + 4 = 0
=
Example 3:Solve
(i)(„= − …/)H + (…0 − †=)I = †/ − „0
(ii) 0(/ + =)H − /(0 + =)I = =(0 − / )
Solution
...
0 + /
...
0 + /
...
0 + /
...
0 + „
...
0 + „
...
0 + „
...
Solution (ii)Given partial differential equation
0(/ + =)H − /(0 + =)I = =(0 − / )
The Lagrange’s auxiliary equations for the given equation are
‡V
?(@ " ~>)
‡[
p>
= $@(? " ~>) = >(?" $@" ) (1)
# # #
Choosing? , @ , > as multipliers, each fraction of (1)
=
‹1C0 Œ
...
/ + ‹1C=Œ
...
0 + ‹1C/Œ
...
=
0
⇒ ‹1C0 Œ
...
/ + ‹1C=Œ
...
0 + /
...
0 + /
...
0 + /
...
(2) and (3) the general Solution is
Practice questions
(0/=, 0 + / − 2=) = 0
Solve the following partial differential equations
1
...
(0 : − / : , 0 − = ) = 0
2
...
#
?
−@,
#
#
#
#
@
#
+> =0
3
...
(0 + /, log(0 + / + = + 20/) − 20) = 0
4
...
(0 + / + = , 0//=) = 0
5
...
(= − /)H + (0 − =)I = / − 0Ans
...
@$>
@>
H+
>$?
>?
I=
?$@
?@
,
Ans
...
(/ + = )H − 0/I = −=0,
@
,0
>
Ans
...
8 Non-linear Partial Differential Equations of Order One
Charpit’s method: General method of Solving partial differential equations of order one but of any degree
...
H
•
?
•
>
+H
=
•
@
...
=
•
Ž
−I
•
•
=
...
/
−
•
•
=
...
Transfer all terms of the given equation to the L
...
S
...
2
...
Find
• • • • •
, , , ,
? @ > Ž •
and put these in charpit’s auxiliary
...
Select two proper fraction to find at least one of H and I and then find other
5
...
= = H
...
/, which on integration gives the complete Solution of given
equation
...
Solution
...
(1),
•
?
= −H,
•
@
=
= −I,
p>
=
= 1,
•
Ž
‘’
‘’
$•
‘–
‘—
$Ž
•
>
Using eq
...
(4), we get
...
(5),
Putting value of H and I in
...
0 + I
...
I = 0 ⇒ I = N
⇒ = = L
...
/
14
⇒ = = L0 + N/ + *
⇒ = = L0 + N/ + L + N
Or
Example 2:Solve (H + I )/ = I=
...
Let + ≡ (H + I )/ − I= = 0
(1)
Charpit’s auxiliary equation
•
?
pŽ
•
@
= 0,
= H +I ,
From eq
...
H
...
I
...
(5) and (1),
I=
= −I,
‘’
‘’
~Ž
‘“
‘”
(6)
Again from (5) eq
...
= = H
...
/
⇒
...
= = H
...
/
⇒
I
L /
œ= − L /
...
/
=
=
=
...
/
œ= − L /
= L
...
Solution
...
H
0
= ‘’
=(
‘•
p•
~•
‘’
‘”
...
=
p?
$
‘’
‘–
=
−H(0/+I)−I(H+/)
(1)
p@
$
‘’
‘—
=
(2)
...
/
−(H+/)
,
(from eq
...
H = 0 ⇒ H = L
Putting H = L in eq
...
Putting these values of H and I in the equation
...
0 + I
...
= = L
...
/
(L + /)
...
0
/
=
...
= − L
...
/
= − L0
L+/
(= − L0)(/ + L)P = N
Example 4:Find a complete integral of H 0 + I / = =
@
Solution
...
H
...
=
Now, each fraction of eq
...
0
...
(1) and (2))
⇒
...
(I2 /)
I2 /
I /L + I2 / = =
⇒ I = ž=⁄(1 + L)/Ÿ1⁄2
(5)
Using (4) and (5), we have
H=D
Putting the value of H and I in
...
0 + I
...
H + H
...
I + I
...
= =
(1)
...
/,
(#~P)@
⇒ (1 + L)#⁄ = $#⁄
...
0 + / −1⁄2
...
Example 5: Find the complete integral of 2(= + H0 + I/) = /H
Solution
...
H
4H
= ‘’
=
‘•
p•
~•
...
=
p?
$
‘’
‘–
−H(20−2/H)−2/I
=
=
p@
$
‘’
‘—
(1)
(2)
...
/
−2/
On integrating, we get H/ = L
From eq
...
(1) and (2))
(3)
...
/
=
4H −2/
(4)
= L0
L
I=− − :+ !
/ /
2/
Putting the value of H and Iin
...
0 + I
...
= =
L
= L0
L
...
/
/
/ /
2/
⇒ (/
...
0 − 0
...
/ = 0
2/
/
0
L
...
3 4 − / $:
...
Example 6Find the complete integral of =HI = H + I
Solution Let +(0, /, =, H, I) = =HI − H + I = 0 (1)
Charpit’s auxiliary equations are
pŽ
•“ ~Ž•”
=•
p•
• ~••”
p>
– $••—
= $Ž•
p?
p@
= $• = $•
–
(2)
—
From eq
...
(2) and (3),
⇒ H2 I = HI2 = −H(=I−1)−I(=H−1) = 1−=I = −=H,
...
I
...
0
...
(1) and (2))
(3)
(4)
Taking first two fraction
17
pŽ
Ž" •
= Ž•" ⇒ H = IL (5)
p•
From eq
...
= = H
...
/ = •(1 + L)⁄=€
...
/
On integrating= = 2(1 + L)•0 + (1⁄L)/€ + N, where L and N are arbitrary constants
...
Given +(0, /, =, H, I) = H − / I − / + 0 = 0
Charpit’s auxiliary equations are
pŽ
‘’
‘’
~Ž
‘“
‘”
pŽ
?
=$
= ‘’
p•
‘’
~•
‘•
‘”
p•
•@$ @
=
p>
‘’
‘’
$Ž $•
‘–
‘—
= $Ž(
p>
Ž)~•"
=
p?
‘’
$
‘–
p?
Ž
=$
=
=
p@
$
p@
@"
Taking first and fourth fraction
Integrating H + 0 = L
(1)
‘’
‘—
(2)
(3)
...
0
=
⇒ H
...
0 = 0
20 −2H
(4)
Using eq
...
= = H
...
/ = (L − 0 )#⁄ + (L / $ − 1)
...
2
...
4
...
= = H0 + I/ + HI
...
= = L0 + N/ + LN
(H + I)(H0 + I/) − 1 = 0
...
=(1 + L)#/ = 2(L0 + /)#/ + N
HI = H0 + I/
...
= = (1⁄2L)(L0 + /) + N
2=0 − H0 − 2I0/ + HI
...
= = L/ + N(0 − L)
H = (I/ + =)
...
/= = L0 + 2œ(L/) + N
1
...
Equation of the form +(H, I) = 0 (i
...
equation containing H and I only)
...
(1) is = = L0 + N/ + *
(2)
whereL and N are connected by the relation +(L, N) = 0
From eq
...
Solution:The given equation is of the standard form I i
...
+(H, I) = 0
∴ Its complete integral is = = L0 + N/ + *
whereL and N are connected by the relation +(L, N) = 0
∴ We have L − N = 25 ⇒ N = √L − 25
∴ The complete integral is = = L0 + √L − 25/ + *
...
>
?
Solution: Given eq
...
£ =
∴ eq
...
0,
...
(2)
Clearly eq
...
e
...
∴ Its complete integral is
¥ = L£ + N¤ + *#
whereL + N = 1 ⟹ Nœ1 − L
...
(3) becomes:
or
¥ = L£ + œ1 − L
(3)
¤ + *#
log = = L log 0 + œ(1 − L ) log / + *#
Now putting L = cos - and *# = log *, we get
⟹ log = = cos - log 0 + sin - log / + log *
Standard form II: ^(\, ], W) = Z
⟹ = = *0 ®¯ ° /
°
...
i
...
equation of the form +(H, I, =) = 0
Working Rule
19
Step I: Given equation is of the form: +(H, I, =) = 0
...
(1)
Step III: Solve the resulting ordinary differential equation between 0 and =
...
(1) is of standard form II i
...
+(=, H, I) = 0
Putting = = +(£), where £ = 0 + L/,
H=
p>
p¨
andI = L
p>
p¨
in (1), we get
...
=
= =3 4 +L 3 4
...
£
⟹ = = (1 + L ) 3
⟹
⟹
On integrating, we get
...
£
...
£ √1 + L
...
£
√1 + L
1
√1 + L
⟹ 2 √= =
1
√1 + L
⟹ 2 √= =
£ + *#
(£ + N)
0 + L/ + N
√1 + L
⟹ (0 + L/ + N) = 4=(1 + L )
...
Solution: Given = (H = + I ) = 1
This is of the form +(=, H, I) = 0
...
e
...
Putting = = +(0 + L/) = +(£), so that H = p¨ and I = L p¨ in (1), we get
p>
= ±3
...
=
...
£
...
£
On integrating, : (= + L ):⁄ = £ + N
#
p>
⟹ 9(0 + L/ + N) = (= + L ):
...
£
20
Standard form III: ^´ (V, \) = ^Y ([, ])
...
(2) for H and I, we get
H=
# (0, L)andI
=
(/, L)
Step 3: Complete integral is
...
0 + I
...
= =
Example 1: Solve I = H0 + I
...
0
# (0, L)
...
Solution: Given H0 = I − I
Which is of the standard form III
...
e
...
Here +# (0, H) = H0 and + (/, I) = I − I
Let +# (0, H) = L and + (/, I) = L
⟹ H0 = LandI − I = L
⟹ H = andI − I + L = 0 ⟹ I =
P
?
#±√#$!P
We know that complete integral is
=
On integrating, we get
...
0 + I
...
0 +
...
¥ ⟹ ¥ =
∴ Eq
...
2
(1)
>"
7¥
7¥
3 4 +3 4 =0 +/
70
7/
21
⟹
+ ` = 0 + / ,where =
§
?
,` =
⟹
§
@
− 0 = −` + /
Which is in standard form III, i
...
+# (0, ) = + (/, `)
...
0 + `
...
¥ = œ(L + 0 )
...
/
0
L
/
L
œ(L + 0 ) + log 0 + œ(L + 0 )¡ + œ(/ − L) − log / + œ/ − L¡ + N
2
2
2
2
⟹ = = 0œ(L + 0 ) + L log 0 + œ(L + 0 )¡ + / œ(/ − L) − L log·/ + œ(/ − L)¸ + N
...
Working Rule
Step I: Given equation is of the form = = H0 + I/ + +(H, I)
...
r
...
L and N, we get
0=0+
0=/+
•
P
(2)
(3)
•
T
(4)
The singular solution is obtained by eliminating L and N from (2), (3) and (4)
...
Solution:Given = = H0 + I/ + *œ1 + H + I ,
Which is of standard form IV i
...
Clairaut’s form i
...
= = H0 + I/ + +(H, I)
...
Example 2: Find the complete and singular solution of = = H0 + I/ + log HI
...
e
...
Putting H = L and I = N in (1), we get the compete integral is
= = L0 + N/ + log LN
...
r
...
L and N, we get
0 = 0 + and0 = / +
#
P
⟹ L = − andN = −
#
?
#
T
#
@
Putting these values of L and b in eq
...
Practice Questions
Solve the following partial differential equations:
1
...
Ans
...
HI = H + I
...
= = L0 +
P
P$#
/+*
Ans
...
= (H + I + 1) = L
...
= (H 0 + I ) = 1
...
√1 + L = = ±2(†¹R 0 + L/) + *
#
5
...
Ans
...
œH + œI = 0 + /
...
= = (0 + L): + (/ − L): + N
:
:
#
#
7
...
= = L0 + N/ − 2√LN
8
...
= L0 + N/ + JQ…(L + N)
1
...
+º
+
½
+ ⋯…+
#
‰
˜
70 ‰
70 ‰$# 7/
7/ ‰
7/ ‰$#
where º˜ , º# , …
...
7=
7=
+`
+ a= = +(0, /)
70
7/
In the above PDE all the partial derivatives are of the same order is called homogenous; otherwise it is called
non-homogenous
...
− 3= = 2 cos(0 + 2/) is non-homogenous linear partial differential equation
...
F
...
Then ¿
...
‰ (/
+ „‰ 0)
Case II: The auxiliary equation has repeated roots
...
F
...
−
">
? @
−6
">
@"
= 0
...
=
# (/
Example 2:Solve (¾ + 2¾S )(¾ − 3¾S ) = = 0
...
E
...
(1) is
(1)
(„ + 2)(„ − 3) = 0
∴ ¿
...
I
...
I
...
Formula (i) when •(L, N) ≠ 0 then
...
Å (e)
...
e …
...
Formula (ii) when •(L, N) = 0, then
...
= 0
1
•′(¾, ¾′)
Â
(L0 + N/),
Example
...
¾S =
Auxiliary equation is
?~ @
@
„: − 3„ + 4 = 0
⇒ „ („ + 1) − 4„(„ + 1) + 4(„ + 1) = 0
⇒ „ = 2, 2, −1
⇒ ¿
...
I
...
e
...
e, where0 + 2/ = e
=
1
27
?~ @
∴ Complete solution is ¿
...
Solution: Given equation is ‹¾ − 3¾¾S + 2¾S Œ= =
The auxiliary equation is „ − 3„ + 2 = 0
...
(1)
⇒ „ = 1, 2
?~:@
=
#
+
+
1
‹¾ − 3¾¾S + 2¾S Œ
sin(0 − 2/)
(2)
sin(0 − 2/)
Å Å sin Ê
...
e
...
= +# (/ + 0) + + (/ + 20)
‹¾ − 3¾¾S + 2¾S Œ
 " $:ÂÂÇ ~  Ç
)(:)~ (:)"
=!
1
?~:@
#
...
= +# (/ + 0) + + (/ + 20) +
1
4
?~:@
−
1
sin(0 − 2/)
...
The particular integral (P
...
) is evaluated by expanding the symbolic function 1/+(¾, ¾S ) in an infinite series
of ascending powers of ¾or ¾′
...
Example 1: Solve ‹¾ − 2¾¾S + ¾S Œ= = 120/
...
=
= Â" (1 +
#
ÂÇ
Â
#
 " $ ÂÂÇ ~ÂÇ
"
∴ ¿
...
120/ = (Â$ÂÇ )" 120/
#
+ ⋯ )
...
120/
{by Binomial theorem}
1
2
1
1
(120/) + : (240)
3120/ +
...
120/ =
¾
¾
¾
¾
=i
1
1
0j 12/ + i : 0j 24
¾
¾
= 20 : / + 0 !
...
⇒ = = +# (/ + 0) + 0+ (/ + 0) + 20 : / + 0 !
...
#
Solution: Given (¾ ¾S − 2¾¾S + ¾S: )= = ? "
...
?
#
(1)
¿
...
= (Â$ÂÇ)" ÂS ? " = (Â$ÂÇ )" ÂS ? "
#
#
#
= /
...
3− 4 + 2
3− 4
¾ 0
¾ 0
¾
0
¾
0
26
−/ log 0 −
2
(log 0)
¾
−/ log 0 − 2(0 log 0 − 0)
= −/ log 0 − 20 log 0 + 20
Hence the complete solution is = = ¿
...
I
...
=
1
(0, /)
(¾ − „# ¾ S )(¾ − „ ¾S ) … (¾ − „‰ ¾S )
1
(0, /) = µ (0, * − „0)
...
?
„ − „ − 2 = 0 ⇒ („ − 2)(„ + 1) = 0
⇒ „ = 2, −1
∴ ¿
...
=
=
=
Or
...
= /
?
1
¾ − ¾¾ S − 2¾S
(¾ −
1
2¾ S )(¾
+ ¾S)
(/ − 1)
=
1
1
(/ − 1)
i
S
(¾ + ¾ ) (¾ − 2¾ S )
=
1
•(* − 20 − 1)
(¾ + ¾ S )
1
µ(* − 20 − 1)
(¾ + ¾ S )
=
= Å(* + 0 + 1)
...
0
?
?
−
?
?
?
?
j
ž/ = * − 20
+2
?
?
∴The complete solution is = = ¿
...
0
27
Solution: Given eq
...
„ + „ − 6 = 0 ⇒ „ = 2, −3
...
• = +# (/ + 20) + + (/ − 30)
...
0,where* = / − 30
=
1
•(30 + *)sin 0 − µ 3 sin0
...
0,where* S = / + 20
= (* S − 20)(−*¹J0) − µ(−2)(−*¹J0)
...
Practice Questions
Solve the following partial differential equations:
1
...
= = +# (/ + 30) + + (/ + 40) +
...
(¾ + 7¾¾ S + 12¾ )= = JQ…ℎ0
...
= = +# (/ − 0) + + (/ − 20) + !
3
...
Ans
...
•¾ + (L + N)¾¾ S + LN¾ S €= = 0/
...
(¾ + 5¾¾ S + 6¾ S )= =
#
...
=
@$ ?
?$@
Ans
...
Ans
...
Ѿ − 2¾¾ S + ¾ S Ò= =
?$:@
#
˜
6
...
Ans
...
#˜˜×˜
P~T
!
0!
= +# (/ − 0) + + (/ + 20) + 0 log(/ − 20)
...
(¾ + 2¾¾ S + ¾ S )= = 2 *¹J / − 0JQ… /Ans
...
28
29
Title: Mathematical inductions
Description: This is mathematical inductions class notes.
Description: This is mathematical inductions class notes.