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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

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.

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.

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

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).

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.

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

Differential Equations

Differential Equations

Solution of the given D.E. is

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

Differential Equations

f (m) 0 (m2 m 1)2 0

The roots are

This is also the solution of the D.E.

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.

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

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).

Differential Equations

Differential Equations

Differential Equations

Differential Equations

and proceed as above

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.

Differential Equations

Problems

Differential Equations

Differential Equations

Solution of the given D.E. is y C.F. P.I.

Differential Equations

A.E. is m3 4m2 4m 0 m 0, -2, -2

Differential Equations

Solution is y C.F. P.I.

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.

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 ) (Au Bv) 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.,

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 Au Bv 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.

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.

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

Differential Equations

- sec x - cos x c1

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).

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) )

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.

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) .

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

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

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)

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

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.Is

separately

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

Differential Equations

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

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

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

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.

Differential Equations

Comparing the coefficients of like terms on both

sides, we get

Therefore the solution of the L.D.E is

Differential Equations

Legendres 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

Differential Equations

ax b et or t ln (ax b) which gives

Differential Equations

Cauchys 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 Cauchys homogeneous linear differential

equation can be reduced to one with constant

coefficients using the substitution

x et or t ln x

Differential Equations

Problems

Differential Equations

given differential equation reduces to

A.E. is m2 - 4m 4 0 m 2, 2 C.F. is

( c1 c2 t ) e2t

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)

Differential Equations

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)

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

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.

Differential Equations

Multiple choice

1. The degree of the following differential

equation is

(a) 3 (b) 2 (c) 1 (d) 4

Differential Equations

2. An example of a first degree, second order,

linear differential equation is

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

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)

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

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

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

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

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

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

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

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

Differential Equations

P.I. is

Solution of the given D.E. is y C.F P.I

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