Engineering Mathematics II

1
Engineering Mathematics -II
MAT -21
PARTS
D
A
C
B
Lap. Tran.
Diff. Eqns.
Diff. Cal.
Int. Cal. Vec. Cal
Multiple Int.
Gamma Beta Fun.
Applications
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Differential Equations
Introduction Linear differential equations with
constant coefficients are solved using three
methods in this chapter. The method of inverse
operator, variation of parameters and the method
of undetermined coefficients for solving a linear
differential equation with constant coefficients
are the three methods that are discussed
here. The transformation of two special types of
linear differential equations with variable
coefficients into linear differential equations
with constant coefficients using appropriate
substitutions will also be dealt with.
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Differential Equations
Definition An equation of the form
where p0, p1, p2, ..pn and ?(x) are
functions of x, is called a Linear Differential
equation (L.D.E) of order n.
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Differential Equations

• Note
• If p0, p1, p2, .. pn are all constants,
• then (1) is called a linear differential
  equation or order n
• with constant coefficients.
• If ?(x) 0, then (1) is called a homogenous
  linear
• differential equation.
• Denoting

(1) can be written as
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Differential Equations
Linear Differential Equations with constant
coefficients (L.D.E) A linear differential
equation with constant coefficients is of the
form
Or f(D) y
where
(can be considered as a polynomial in D)
The complete solution of a L.D.E consists of two
parts namely, the Complementary Function (C.F)
and the Particular Integral (P.I).
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Differential Equations
Complementary Function is the most general form
of a function
satisfying the homogeneous L.D.E f
(D)y 0. Particular Integral is a function
satisfying the L.D.E
f (D)y ?(x) . Complete Solution
y C.F P.I is called the complete solution
of f (D)y ?(x).
eg the D.E. y e-2x has c1 c2 x
as its C.F.
and e-2x /4 as its P.I.
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Differential Equations
Note If y1 and y2 are two solutions of the
homogeneous L.D.E., then a1y1 a2y2 is
also its solution, where a1 and a2 are constants.

• To find the C.F.
• i.e., to find the solution of f (D)y 0.
• Form the Auxiliary Equation (A.E.) f (m) 0.
  
• (b) Solve f (m) 0 to obtain n roots.
• (c) Then C.F. can be written using the following
  table

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Differential Equations
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Differential Equations
Solution of the given D.E. is
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Differential Equations
(2) Solve (D5 2D4 2D3 4D2 - 11D - 10)y
0 Here f (m) m5 2m4 2m3 4m2 -
11m 10 The roots are m 2, -1, -1, 1 ?
2i The C.F. (the solution) of the given
D.E. is
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Differential Equations
f (m) 0 (m2 m 1)2 0
The roots are
This is also the solution of the D.E.
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Differential Equations

• Note
• The roots of the quadratic equation ax2 bx
  c 0

are given by

• If the degree of the A.E. is 3, reduce it to a
  second
• degree equation using synthetic division
  after finding one root by trial and error.

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Differential Equations
To find the P.I. We study the following three
methods to find the P.I. of a L.D.E. with
constant coefficients. (1) Method of Inverse
Differential Operator (2) Method of Variation
of Parameters (3) Method of Undetermined
Coefficients
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Differential Equations
Method of Inverse Differential Operator The P.I.
of the D.E. f (D) y ? (x) is given by
Note
are operators and are inverses of each other.
The following table can be used to find the P.I.
for different types of functions ? (x).
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Differential Equations
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Differential Equations
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Differential Equations
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Differential Equations
and proceed as above
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Differential Equations
Note Method of inverse differential operator
can be applied to find the P.I. of the L.D.E.
f(D) y ?(x) only if ?(x) consists of
exponential, sine or cosine, polynomial functions
or a combination of these functions.
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Differential Equations
Problems
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Differential Equations
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Differential Equations
Solution of the given D.E. is y C.F. P.I.
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Differential Equations
A.E. is m3 4m2 4m 0 ? m 0, -2, -2
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Differential Equations
Solution is y C.F. P.I.
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Differential Equations
Method of Variation of Parameters This is
applicable to solve a L.D.E. with any function in
its R.H.S. It makes use of the C.F. to obtain the
general solution. This is done by converting the
arbitrary constants found in the C.F. (called
parameters) into arbitrary functions and then
finding them. Eg. Let the equation be
where P, Q, R are functions of x.
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Differential Equations
Let the C.F. of this L.D.E. be c1u(x)
c2v(x). Assume the general solution (G.S.) of the
L.D.E. as y Au Bv, where A and B are also
functions of x.
Then y ' (Au ' Bv ' ) (A'u B'v) Let
(A' u B' v ) 0 . (1) so that y '
(Au ' Bv ' )
But since y is the general solution, it
satisfies the L.D.E. i.e.,
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Differential Equations
Since u and v are solutions of the homogeneous
equation
y'' P y' Q y 0, we have u'' P u' Q
u 0 and v'' P v' Q v 0. ?
equation (2) reduces to A'u' B'v' R .
(3). Solving equations (1) and (3), we get
Integrating, we get A(x) and B(x) involving 2
arbitrary constants, say c1 and c2 ? y Au
Bv gives the general solution of the L.D.E.
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Differential Equations

• Note
• The general solution includes both C.F. and
  P.I.
• (2) The Jacobian

is called the Wronskian of the L.D.E., denoted
by W
(3) The method of variation of parameters can be
extended to higher order differential
equations also.
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Differential Equations
Problems Use the method of Variation of
Parameters to solve the following
The C.F. is c1 cos x c2 sin x. we shall
choose u cos x and v sin x. Hence W
1 Assume the G.S. in the form y Au Bv so
that
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Differential Equations
- sec x - cos x c1
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Differential Equations
B ?cos x tan2x dx ?sec x dx - ?cos x dx
ln (sec x tan x) - sin x c2
? G.S. is y A cos x B sin x ? y
c1 cos x c2 sin x - 2 sin x ln (sec x tan
x).
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Differential Equations
The A.E. is m2 1 0. The functions u and
v can be chosen as cos x and sin x so that
W 1. If the G.S. is y Au Bv, then
(multiplying and dividing by (1 -
sin x) )
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Differential Equations
A - sec x tan x - x c1, using tan2 x
sec2 x - 1
? y c1cos x c2
sin x - x cos x - 1 sin x ln (1 sin x)
is the solution.
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Differential Equations
Method of Undetermined Coefficients This is
another method to find the P.I. of the L.D.E.
f(D)y ?(x). This is applicable, as in the
method of Inverse differential operator, only for
exponential functions, sine or cosine function,
polynomial in x or a combination of these in
?(x). The method consists of assuming a trial
P.I. of the form y a1 ?1(x) a2
?2(x) a3 ?3(x) .
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Differential Equations
containing the unknown constants a1, a2,
which are determined by substitution in the given
L.D.E. The functions ?1(x), ?2(x), ?3(x) .
are the different terms in ?(x) along with
those that arise from differentiation of these
terms. The ?i(x)s are chosen using the rules
given below
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Differential Equations
Case 1 None of the terms in ?(x) is a term
of the C.F. of the L.D.E. Then the functions
?i(x) are the terms of ?(x) along with those
arising from differentiation of these functions.
eg. (D2 - 1)y e2x 3sinx x2 The trail
P.I. is y c1 e2x c2 sinx c3
cosx c4 x2 c5x c6
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Differential Equations
Case 2 If a term of ?(x) is ?k(x) where
?k(x) is a term of the C.F. corresponding
to an s-fold root m then the function x s
?k(x) and all its derivatives should be
introduced for ?i(x)s. eg. (D 2) (D
3)3 y 2 e-3x
Since e-3x corresponds to 3 fold root m -3
of the A.E, the trial P.I. is
y c1 x3 e-3x c2 x2 e-3x c3 xe-3x c4
e-3x Note To simplify the calculation of the
unknowns, the second, third and fourth terms in
the trial P.I. above can be omitted as they are
already present in the C.F. (as part of the
solution)
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Differential Equations
Case 3. If a term of ?(x) is xs ?k(x)
where ?k(x) is a term of C.F. corresponding to
an r fold root m of the A.E., then the
function xrs ?k(x) along with all its
derivatives constitute the ?i(x)s.
eg. (D 2) (D 3)2 y xe -3x
Since e -3x corresponds to 2-fold root m -3
of the A.E., the trial P.I. is y c1 x3
e-3x c2 x2 e-3x c3x e-3x c4e-3x
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Differential Equations
Problems Use the method of Undetermined
Coefficients to solve the following
problems (1) (D2 - D - 2)y x2 - sin x 2ex
.. (1) the A.E is m2 - m - 2 0,
its C.F. is c1e2x c2e-x
To get the P.I. As the RHS of (1) contains 3
independent functions, we can find the 3 P.I's
separately
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Differential Equations
(i) (D2 - D - 2)y x2 Trial P.I. Is y
a1x2 a2x a3 Since this satisfies the
L.D.E., we have 2a1 - 2a1x - a2 - 2a1x2 -
2a2x - 2a3 x2
? - 2a1 1 ?
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Differential Equations
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Differential Equations
(ii) (D2 - D -2)y - sin x Trial P.I. is
y a1 sin x a2 cos x Therefore -a1 sin
x - a2 cos x - a1 cos x a2 sin x - 2a1 sin x -
2a2 cos x - sin x
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Differential Equations
(iii) (D2 - D - 2)y 2ex Trial P.I.
Is y a1ex
P.I3 is -ex Therefore solution of the given
L.D.E is
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Differential Equations
(2) (D2 - 1)y 7e-x C.F is c1ex
c2e-x and trial P.I. is a1xe-x No
need to take e-x as it is present in C.F
Therefore solution of the given L.D.E is
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Differential Equations
(3) (D2 1)y 4xcosx - sin x The C.F
is c1cosx c2sin x Therefore trial P.I.
can be y a1 x2 cos x a2 x2 sin x a3
x cos x a4 x sin x
cos x and sin x terms are present in C.F and
therefore can be omitted from the trial
P.I. Substituting in the given L.D.E and
simplifying, we get 4 a2 x cos x - 4a1 x sin x
(2a1 2a4) cos x (2a2 2a3) sin x
4x cos x - sin x.
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Differential Equations
Comparing the coefficients of like terms on both
sides, we get
Therefore the solution of the L.D.E is
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Differential Equations
Legendre's linear differential equation
This is a linear Differential Equation with
variable coefficients of the form
where a, b and ki s are constants
This L.D.E. can be transformed to one with
constant coefficients using the substitution
given below
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Differential Equations
ax b et or t ln (ax b) which gives
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Differential Equations
Cauchy's homogeneous linear differential
equation This is a special case of Legendres
linear differential equation when a 1 and b 0
and therefore is of the form
Note Cauchy's homogeneous linear differential
equation can be reduced to one with constant
coefficients using the substitution
x et or t ln x
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Differential Equations
Problems
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Differential Equations
? given differential equation reduces to
A.E. is m2 - 4m 4 0 ? m 2, 2 C.F. is
( c1 c2 t ) e2t
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Differential Equations
? P.I. is t2 e2t, because f ' (2) 0
The solution of the given D.E. is y C.F P.I
. with et 1 2x and t ln (1 2x)
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Differential Equations
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Differential Equations
A.E is m2 1 0 ? m ? i ? C.F is c1 cos t c2
sin t
f (D) 0 when D2 is replaced by -1
P.I is 2t sin t Therefore the solution is y C.F
P.I with t log (1 x)
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Differential Equations
Multiplying by x2, we get
Using x et, the equation reduces to
D2y 12t ? y 2t3 c1t c2, by integrating
w.r.t 't' Therefore the solution is y 2 (log
x)3 c1 log x c2
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Differential Equations
Conclusion We saw how the method of variation of
parameters takes an upper hand over the other two
methods, as it can be applied to a linear
differential equation, with any function on its
right hand side.
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Differential Equations
Multiple choice
1. The degree of the following differential
equation is
(a) 3 (b) 2 (c) 1 (d) 4
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Differential Equations
2. An example of a first degree, second order,
linear differential equation is
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Differential Equations
3. Given the methods of (i) undetermined
coefficients (ii) variation of
parameters and (iii) inverse differential
operators, the differential equation y'' -
3y' 4y ex cos x sec 3x can be solved
using
(a) (i) and (iii) only (b) (iii) only (c)
(i) only (d) (ii) only
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Differential Equations
4. Given (i) ex (ii) e2x (iii) e3x which of
the following
(a) (i) only (b) (i) and (ii) only (c) (ii)
and (iii) only (d) (i), (ii) and (iii)
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Differential Equations
5. Given the differential equation
which of the following functions satisfy it
(a) ex and sin 3x (b) e2x and sin x (c) ex and
sin 2x (d) e2x and cos 3x
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Differential Equations
6. Which of the following functions is a solution
of the differential equation y'' - 5y' 6y
e2x
(a) x e2x (b) -x e2x (c) x2 e2x (d) -x2
e2x
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Differential Equations
7. The Wronskian of the differential equation
(D2 - 2D 1)y ex / x is
(a) e2x (b) x ex (c) e2x (d) x e2x
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Differential Equations
8. A trial solution for the P.I using method of
undetermined coefficients of differential
equation (D-3)2 (D-1)y xe3x x2 is
(a) a1x3e3x a2x2e3x a3x2 a4x a5 (b)
a1xe3x a2x2 (c) a1xe3x a2e3x a3x2 a4x
a5 (d) a1xe3x a2x2 a3ex
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Differential Equations
9. A trial solution for the P.I using method of
undetermined coefficients of differential
equation (D - 1) (D - 2)y ex cosx is
(a) a1ex cosx a2 ex sin x (b) a1 ex a2 e2x
(c) a1x ex cos x a2 ex cos x (d) a1x ex cos
x a2 x ex sin x a3 ex cos x a4 ex sin x
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Differential Equations
10. Given the linear differential equation
(x - 1) y ' ' - xy ' y (x - 1)2, the
following pairs of functions are its
complimentary functions
(a) sinx and ex (b) cosx and ex (c) x and
ex (d) sinx and cosx
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Differential Equations
Questions and Answers
1. Solve (D4 - 2D3 2D2 - 2D 1)y 0
Auxiliary equation is m4 - 2m3 2m2 - 2m 1 0
whose roots are m 1, 1 ? i. Therefore the C.F.
and hence the solution of the given D.E is
y (c1 c2 x) ex c3 cos x c4 sinx
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Differential Equations
2. Solve using the method of inverse
differential operator (D2 4)y x sin
2x. A.E. is m2 4 0 whose roots
are m ? 2i
C.F. is c1 cos2x c2 sin2x
sin 2x is the imaginary part of e2ix
cos 2x i sin 2x
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Differential Equations
P.I. is
Solution of the given D.E. is y C.F P.I