linear differential equations of second higher … linear differential...linear differential...

LINEAR DIFFERENTIAL EQUATIONS OF

SECOND AND HIGHER ORDER

9aaaaa

577

9.1 INTRODUCTION

A differential equation in which the dependent variable, y(x) and its derivatives, say, 2

2,

dy d ydx dx

etc. occur in the first degree and are not multiplied together is called linear differentialequation. Then, we classify them as linear differential equation with constant co-efficientsand the other with variable coefficients.

The linear differential equations with constant coefficients generally arises in practical problemsrelated to the study of mechanical, acoustical and electrical vibrations, whereas lineardifferential equations with variable coefficients arise generally in mathematical modeling ofphysical problems. Some of the important linear differential equations with variable coefficientsare Bessel equation, Legendre’s equation, Chebyshev equation etc.

The solution of linear differential equations with constant coefficients are generally foundin terms of known standard functions while there exists no such procedure in case ofdifferential equations with variable coefficients and their solutions many a times results inthe form of an infinite series.

The general form of the nth order linear differential with constant coefficients is

( )−

−−+ + … + + =1

0 1 11

n n

n nn n

d y d y dyk k k k y X x

dx dx dx…(1)

where k0, k1, k2,…, kn are constants and X is a function of ‘x’ only.The general linear differential equation with variable coefficients is written as

( )− −

−− −+ + + … + + =1 2

0 1 2 11 2

n n n

n nn n n

d y d y d y dyP P P P P y X x

dx dx dx dx… (2)

where P0 (≠ 0), P1, P2,…, Pn and X are function of ‘x’ only.If X(x) = 0 in (1) and (2), then they are called linear homogeneous differential equations

with constant coefficients and variable coefficients respectively.

9.2 SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS

In equation (1), x varies on some interval of definition, say I, which may be open, semi open,closed or infinite and the differential equation may be valid for all x ∈ (0, ∞), (–∞, 0), (–∞, ∞).If y1(x) is a solution of the equation (1), then it must satisfy the equation identically and

Engineering Mathematics through Applications578

whence y1(x) must be continuously differentiable (n – 1) times and 1( )n

nd y x

dx must be

continuous in that interval.Further, if coefficients P0(x), P1(x) ,…….., Pn(x), P0(x) ≠ 0, in the linear homogeneous

equation (2) are continuous on some interval of def, say I, then this equation has n linearlyindependent solutions. If y1(x), y2(x) ,…, yn(x) are n linearly independent solutions, then thegeneral solution is their combination.i.e. y(x) = c1y1 + c2y2 + … + cnyn … (3)

Theorem: If y1(x), y2(x) ,…, yn(x) be n linearly independent solutions of

−

−−+ + … + + =1

0 1 110,

n n

n nn n

d y d y dyk k k k y

dx dx dxwhere k0, k1 ,…, kn are all constants, then y = c1y1 + c2y2 + … + cnyn is also a solution of (1). Thisis called the Principle of Superposition or Principle of linearity.

Proof: Putting y = c1y1 + c2y2 + … + cnyn into left hand side of equation (1), we get

1

0 1 1 2 2 1 1 1 2 21

n n

n n n nn nd dk c y c y c y k c y c y c y

dx dx

−

−+ + … + + + + … +

1 1 1 2 2 1 1 2 2n n n n n ndk c y c y c y k c y c y c ydx−+ … + + + … + + + + … +

1

1 1 11 0 1 1 11

n n

n nn n

d y d y dyc k k k k y

dx dx dx

−

−− = + + … + +

1

2 2 22 0 1 1 21

n n

n nn n


dx dx dx

−

−− + + + … + +

1

0 1 11

n nn n n

n n n nn n


dx dx dx

−

−− +…+ + + … + +

[ ] [ ] [ ]1 2 1 20 0 0 0, since ( ), ( ) , , ( )n nc c c y x y x y x= + + … + = …

are solution of the linear equation. This proves the theorem.

Remarks: The above n linearly independent solutions y1(x), y2(x),…,yn(x) are called the fundamental solutionsof equation (1) and the set comprising them forms a basis of the nth order linear homogeneous equations.

The solution (3), y = c1y1 + c2y2 + … + cnyn = u(x) is a combination of n linearly independentsolutions containing n arbitrary constants c1, c2,…, cn. It is called the general solution ofequation (1). It is also known as the complementary function (C.F.).

Further, if y = v(x) be a solution of the non–homogeneous equation containing no arbitraryconstant is called it’s particular solution (P.I.). Therefore, y = u(x) + v(x) = C.F. + P.I. is calledthe complete solution of equation (1). Hence, in order to solve equation (1), first find thegeneral solution (C.F.) and then find the particular solution (P.I.).

Linear Differential Equations of Second and Higher Order 579

Linear Independence and Dependence of Solutions

Functions y1(x), y2(x) ,…, yn(x) are said to linearly independent on some interval of definition,say I, if the relation (3) viz. c1y1 + c2y2 + … + cnyn = 0. Implies c1 = 0 = c2 = … = cn. This system(or set) of linearly independent solutions is called fundamental system (or set) of solution(or integrals).

However, these functions are said to be dependent on the interval of definition, say I, ifrelation (3) holds for c1, c2 ,…, cn not all zero. In this case, one or more functions can beexpressed as a linear combination of the remaining functions.

e.g. if 1 1 2 2 3 31

10, then n nc y c y c y c yc

≠ = − + + … + …(4)

Conversely, if any of yi's can be expressed as the linear combination of the remainingfunctions y1, y2 ,…, yi – 1, yi + 1 ,…, yn then the given set of functions are linearly dependent.

Theorem: The necessary and sufficient condition that n integrals y1, y2, …, yn of the linear

differential equation (2) viz.1

0 1 10

n n

nn n

d y d yP P P y

dx dx

−

−+ + … + = , where P0, P1, P2 ,…, Pn (P0 ≠ 0)

are continuous functions of x on a common interval I or constants, be linearly independent isthat the determinant, W* (Wronskian),

1 2

1 2

1 1 11 2

n

n

n n nn

y y yy' y ' y '

y y y− − −

… … … … …… … … … …

… … … … … … … …… … … … … … … …

… … … … …

does not vanish identically on I.

Proof: Necessary Condition: If y1(x), y2(x) ,…, yn(x) are not linearly independent then thereare constants c1, c2 ,…, cn not all zero such that c1y1 + c2y2 … + cnyn = 0.

Also c1y1(i) + c2y2

(i) + … + cnyn(i) = 0, i = 1, 2 ,…, n – 1.

It follows the determinant

1 2

1 2

1 1 11 2

* ( )

n

n

n n nn

y y yy' y ' y '

W x

y y y− − −

… … … … …… … … … …

= … … … … … … … …… … … … … … … …

… … … … …

1 1 2 2 2

1 1 2 2 2

1 1 1 1 11 1 2 2 2

´ ´ ´ ´ ´n n n

n n n

n n n n nn n n

c y c y c y y yc y c y c y y y

c y c y c y y y− − − − −

+ + ………… + ………+ + ………… + ………

…………………………… … ………=…………………………… …………

+ + … + ………


2

2

1

1 12

00

1 0

0

n

n

n nn

y yy ' y '

c

y y− −

… … … … …… … … … …

= =… … … … … … … …… … … … … … … …

… … … … …

Thus, if y1(x), y2(x) ,…, yn(x) are not linearly independent, then the determinant W*(Wronskian) will be identically zero.

Sufficient Condition: If W*(x) = 0, then there can be found constants c1, c2, …, cn not allzero such that c1y1 + c2y2 + … + cnyn = 0;whence c1y1

(i) + c2y2(i) + … + cnyn

(i) = 0, i = 1, 2, ,…, n – 1.Now W* = 0 implies that the determinant must be reducible to the form wherein all the

elements of one row (or one column) are zero or in other words there must be certainmultiplier λ1, λ2 ,…, λn such that

1 1 2 2

1 1 2 2

1 1 11 1 2 2

0 ( )´ ´ ´ 0 ( )

0 ( )

n n

n n

n n nn n

y y y iy y y ii

y y y n− − −

λ + λ + … … … … … +λ = …λ + λ + … … … … … +λ = …

… … … … … … … … … … … … … … … …λ + λ + … … … … … +λ = …

…(5)

Differentiating each of these equations and subtracting from the next following equations,we get

1 1 2 2

1 1 2 2

1 1 11 1 2 2

´ ´ ´ 0 ( )´ ´ ´ ´ ´ ´ 0 ( )

´ ´ ´ 0 ( )

n n

n n

n n nn n

y y y iy y y ii

y y y n− − −

λ + λ + … … … … … + λ = …λ + λ + … … … … … + λ = …

… … … … … … … … … … … … … … … …λ + λ + … … … … … + λ = …

…(6)

Ist operation is explained as: d(λ1y1 + λ2y2 + … + λnyn) = 0 implies (λ1y'1 + λ2y'2 + … + λny'n) + (λ1' y1 + λ2'y2 + … + λn'yn) = 0

and on subtraction this from (λ1y1 + λ2y2 + … + λnyn) = 0 results in (l1y1 + l2y2 + … + lnyn) = 0 and so on.

Equation (6) in matrix form is equivalent to

1 2

1 2

1 1 11 2

´ ´ ´0

n

n

n n nn

y y yy y y

y y y− − −

… … … … …… … … … …

=… … … … … … … …… … … … … … … …

… … … … …

If one of the determinant, say formed by omitting the rth column, then there will be arelation, a1y1 + a2y2 + – – – – + ar – 1yr – 1 + ar + 1yr + 1 + – – – + anyn = 0


In general, if one of these determinants vanishes, then from (6) and the first (n – 1) equationsin (5), we get

1 2

1 2

´´ ´(say),n

n

λλ λ= = … = = µλ λ λ

implying 1 1

1 1 11 1

´ ( )or or log or dxd

dx c e µ∫λ λ= µ = µ λ = µ λ =λ λ ∫

Similarly, 2 2 1 2, , ; , , ,dx dxn n nc e c e c c cµ µ∫ ∫λ = … λ = … being constants. On substituting these

λi's in the equation (5(i)), we get c1y1 + c2y2 + … + cnyn = 0.Hence, if the condition holds for (n – 1) functions, it also holds for n. Whence the necessary

and sufficient condition that y1, y2, y3, …, yn forms a system of linearly independent integralis that determinant W* does not vanish indentically.

9.3 OPERATOR �D� AND COMPLEMENTARY FUNCTION

To solve the equation

1

1 110

n n

n nn n

d y d y dyK K K y

dx dx dx

−

−−+ + … + + = …(1)

We first define the operator ‘D’

Operator D: If we write so thatdy dDy Ddx dx

= = is an operator which when applied to a

function of ‘x’, differentiates that function with respect to ‘x’. Likewise, n

nn

d yD y

dx= and so

on. Further, orm m nn

m n n m m nn m m n

d y d yd D D D D Ddx dx dx

++

+ = = =

Means the differential operator,

‘D’ fully obeys the laws of algebra.Whence, the linear differential equation (1) in symbolic form may be written as

Dny + k1Dn – 1y + … + kn – 1Dy + kny = X(x) …(2)Or more precisely, f(D)y = X where in f(D) can be treated much the same as an algebraic

expression in D.Further, (Dn + K1D

n – 1 + … + kn – 1D + kn) = 0 or f(D) = 0 … (3)is called auxiliary equation to (2) with m1, m2, --------, mn as its n roots, means furtherdiscussion on solution of (1), depends on the nature of these n roots.

Case1: When all the roots are real and distinct:In this case, equation (2) will become

(D – m1) (D – m2) … (D – mn)y = 0 … (4)It will be satisfied by the solutions of

(D – m1)y = 0; (D – m2)y = 0; ……; (D – mn)y = 0,


where (D – m1)y = 0 means 1 0dy

m ydx

− = which is a Leibnitz linear equation

Here, 1 1 11 1I.F. and . orm x m x m xe y e c y c e− −= = = . Like wise others.

Therefore, general solution becomes

1 21 2( ) nm xm x m xny x c e c e c e= + + … + … (5)

Case 2: When two roots are real and equal:In this case, equation (2) corresponding to the two repeated roots becomes

(D – m1) (D – m1)y = 0Let (D – m1)y = z, so that above equation reduces to

(D – m1)z = 0 or Dz – m1z = 0

Again a Leibnitz Linear, resulting in 11m xz c e−= so that (D – m1)y = z becomes

1 11 1 1 1( ) orm x m xdy

D m y c e m y c edx

− = − = , a Leibnitz Linear with 11 1, m xP m Q c e= − = and its I.F. is

1m xe− and solution becomes ( ) ( )1 1 1 11 1 2 1 2orm x m x m x m xye c e e dx c x c y c x c e− −= = + = +∫

Thus, the general solution of equation (2), corresponding to two repeated roots becomes

( ) 311 2 3

nm x m xm xny c x c e c e c e= + + + … + … (6)

Case 3: (i) When one pair of roots is imaginary (complex):Let m1 = α + iβ and m2 = α – iβ and the remaining roots are real, then the general solution ofequation (1) becomes

( ) ( ) 31 2 3

ni x i x m x m xny c e c e c e c eα+ β α− β= + + + … +

( ) 31 2 3

nm x m xx i x i xne c e c e c e c eα β − β= + + + … +

( ) ( ) 31 2 3cos sin cos sin nm x m xx

ne c x i x c x i x c e c eα = β + β + β − β + + … +

( ) ( ) 31 2 1 2 3cos sin mm x m xxne c c x i c c x c e C eα = + β + − β + + …+

[ ] 31 2 3cos sin nm x m xx

ne C x C x C e C eα= β + β + + … + … (7)

where (c1 + c2) = C1, i(c1 – c2) = C2, c3 = C3 etc.Case 3: (ii) When one pair of roots is equal and imaginary:

(complex roots equal means m1 = α + iβ = m2, m3 = α – iβ = m4)In this case general solution becomes.

( ) ( ){ } 51 2 3 4 5cos sin nm x m xx

ny e c x c x c x c x c e c eα= + β + + β + + … + … (8)

Example 1: Find the solution of the differential equation 4y”’ + 4y” + y’ = 0.

Solution: The given equation in symbolic form is written as (4D3 + 4D2 + D)y = 0 and itsauxiliary equation is 4D3 + 4D2 + D = 0


D(4D2 + 4D + 1) = 0 D(2D + 1)2 = 0 i.e. 1 10, ,2 2

D = − −

Whence ( ) ( )0 21 2 3. .

x

y C F c e c x c e−

= + + .

Example 2: Solve 3

30

d yy

dx+ =

Solution: The given equation in symbolic form is given by (D3 + 1)y = 0 and its auxiliary

equation is D3 + 1 = 0 i.e. (D + 1) (D2 – D + 1) = 0 or 1 31,2 2

D i= − ±

Whence 21 2 3

3 3. . cos sin2 2

xxC F c e e c x c x− = + +

Example 3:4

41 2 3 44

, show that cos sin cosh sinhd x m x x c mt c mt c mt c mtdt

= = + + + .

Solution: The given equation in symbolic form is (D4 – m4)x = 0The auxiliary equation becomes (D2 – m2) (D2 + m2) = 0Which implies either D2 – m2 = 0 or D2 + m2 = 0

i.e. D = ±m and D = ±imHence the required solution is

mt mt imt imtx ae be ce de− − = + + +

( ) ( )cosh sinh cosh sinh

2 2mt mt mt mt

a b+ −

= +

( ) ( )cos sin cos sin

2 2mt i mt mt i mt

c d+ −

+ +

( ) ( ) ( ) ( )

cosh sinh cos sin2 2 2 2

a b a b c d c dmt mt mt i mt+ − + − = + + +

⇒ [ ] [ ]1 2 3 4cosh sinh cos sinx C mt C mt C mt C mt= + + +

where 1 2 3 4( ), , and

2 2 2 2i c da b a b c dC C C C −+ − += = = =

Example 4: Solve 4

44

0d y

a ydx

+ = [NIT Kurukshetra, 2007; NIT Jalandhar, 2007]

Solution: Symbolic form of the given equation is


(D4 + a4)y = 0 …. (1)Therefore, the auxiliary equation becomes

D4 + a4 = 0 or D4 + a4 + 2a2D2 – 2a2D2 = 0

or ( ) ( )222 2 2 0D a aD+ − =

( )( )2 2 2 22 2 0D a aD D a aD+ − + + =

Which implies, either 2 2 2 0D a aD+ − = …. (2)

or 2 2 2 0D a aD+ + = ….(3)

On solving the two quadratic equations (2) and (3), we get

and respectively2 2 2 2a a a aD i D i−= ± = ±

Hence the required solution is

2 21 2 3 42 2 2 2( cos sin ) ( cos sin )

a ax xa a a ay e c x c x e c x c x−= + + + … (4)

Alternately: + π= − = − = =144 4 (2 1)

i.e. ( 1) , 0, 1, 2, 34

nD a D a a cis n

Using, D. Moivre`s Theorem,

11 10, cos sin (1 ),

4 4 2 2 2an D a i a i i π π = = + = + = +

23 3 1 11, cos sin (1 ),4 4 2 2 2

an D a i a i i π π = = + = − + = − −

35 5 1 12, cos sin (1 ),4 4 2 2 2

an D a i a i i π π = = + = − + − = − +

47 7 1 13, cos sin (1 )4 4 2 2 2

an D a i a i i π π = = + = − = −

Now roots are comparable to ( ), ( )i iα ± β −α ± β , so that complementary function becomes,

1 2(cos sin ) (cos sin )x xy e c x i x e c x i xα −α= β + β + β − β

2 21 2 3 4cos sin cos sin

2 2 2 2

a ax xa a a ae c x c x e c x c x− = + + +


ASSIGNMENT 1

Solve (i)2

26 9 0,

d y dyy

dx dx− + = (ii) ( )2 1 0,D D y+ + =

(iii)3

30,

d yy

dx− = (iv)

4 2

4 28 16 0

d y d yy

dx dx− + =

(v) ( ) ( )2

22 10 0, 0 4; 0 1

d y dyy y y y'

dx dx− + = = =

(vi) 4 4 0y y y+ + =′′′ ′′ ′ (vii) ( )4 2 2 42 0D n D n y+ + =

(viii) }2

20, with at 00

dl g tdt

θ = αθ + θ = =θ =

9.4 INVERSE OPERATOR

(a) Defn: ( ) ( )1X x

f D is that function of 'x' independent of arbitrary constants which when

operated on by f(D) gives X(x). Hence ( )1

f D is the inverse operator of f(D).

Further, ( ) ( )1X x

f D is the particular integral of the equation ( ) ( ).f D y X x=

(b)1

,X XdxD

= ∫ no arbitrary constant is being added to.

Let 1 1( ) . ( ) or or

dyX x y D X x Dy X dy Xdx

D D dx= ⇒ = = =

Integrating both sides, .y Xdx= ∫

(c) ( )1 ax axX x e e X dxD a

−=− ∫

Let, ( ) ( ) ( ) ( ) ( )1 1X x y D a X x D a yD a D a

= ⇒ − = −− −

ordy

X Dy ay ay Xdx

= − − =

which is a Leibnitz linear equation whose, integrating factor is e–ax and its solution becomes

. orax ax ax axy e e X dx y e e X dx− − −= =∫ ∫


Problems based on use of Inverse operator in finding particular integral.

Example 5: Solve the differential equation + =2

22

secd y

a y axdx

Solution: Equation in its symbolic form: (D2 + a2)y = secax … (1)Thus auxiliary equation, D2 + a2 = 0 i.e. D = ± ia … (2)

∴ C.F. = eox (c1cosax + c2sin ax … (3)

( ) ( )( )2 2 2 21 1 1. . sec sec secP I ax ax ax

D a D a D ia D ia= = =

+ − − + −

1 1 1 sec ,2

axia D ia D ia

= − − + (By Partial Fractions) … (4)

Now, 1 1sec sec , using ( )iax iax ax axax e ax e dx X x e Xe dx

D ia D a− − = = − −∫ ∫

( ) ( )sec cos sin 1 taniax iaxe ax ax i ax dx e i ax dx= − = −∫ ∫

( )log cos log cosiax iaxi ax axe x e x i

a a

− = − = + … (5)

On the same lines, changing i to –i, we get

log cos1 sec iax ax

ax e x iD ia a

− = − + … (6)

Using (5) and (6), (4) becomes,

{ } { }log cos log cos1. .2

iax iaxax axP I e x i e x i

ia a a− = + − −

∴ ( ) ( )log cos12

iax iax iax iaxaxx e e i e e

ia a− − = − + +

∴ 2

log cos.sin .cos

axx ax axa a

= + … (7)

since sin and cos2 2

iax iax iax iaxe e e eax axi

− − += = Hence the complete solution is

( ) ( )1 2 2

log coscos sin sin cos

axxy c ax c ax ax axa a

= + + +

Note: If a = 1, then differential equation becomes 2

2d y

y secxdx

+ = and its solution is

(c1cosx + c2sin x) + (xsinx + cosx log cosx)


Example 6: Solve the differential equations 2

22

+ = cosec d y

a y axdx

Solution: The given equation in its symbolic form is (D2 + a2)y = cosecaxThus, the auxiliary equation becomes D2 + a2 = 0 i.e. D = ±ia

∴ C.F. = (c1cosax + c2sin ax)

For 2 21. . cosecP I ax

D a=

+

2 21 cosec( )

axD a

=− −

( )( )1 cosec ax

D ia D ia=

+ −

1 1 1 cosec

2ax

ia D ia D ia = − − + (By Partial Fractions)

Now, 1 cosec coseciax iaxax e e ax dx

D ia−=

− ∫ 1sin ax axu g X e e X dxD a

− = − ∫

(cos sin )coseciaxe ax i ax ax dx= −∫ (cot )iaxe ax i dx= −∫ (cot )iaxe ax i dx−∫

log(sin )iax axe ix

a = −

… (1)

On the same lines, changing i to –i, we get

log(sin )1 cosec iax axax e ix

D ia a− = + +

… (2)

Using (1) and (2),

{ } { }log cos log cos1. .2

iax iaxax axP I e x i e x

ia a a− = + − −

log(sin )1 ( ) ( )

2iax iax iax iaxax

e e ix e eia a

− − = − − +

log sin1 sin cos

axax x ax

a a = −

… (3)

since sin and cos2 2

iax iax iax iaxe e e eax axi

− − − += =


Hence the complete solution is

( )1 2log(sin )1cos sin sin cos

axy c ax c ax ax x ax

a a = + + −

Note: If a = 1, then differential equation reduces to + =2

2

d yy sec x

dx and the corresponding solution

becomes y = (c1cosx + c2sinx) + sinx(logsinx) – x cosx. However, these above two problems are easier iffollowed under the method of variation of parameter, discussed in subsequent articles.

9.5 GENERAL PROCEDURE FOR FINDING PARTICULAR INTEGRAL

Consider the equation,

( )1

1 11

n n

n nn n

y y yk k k y X x

x x x

−

−−+ + … + + =d d d

d d d… (1)

Which in symbolic form becomes (Dn + k1D

n – 1 + … + kn – 1D + kn)y = X(x) or f(D)y = X(x) …(2)

Thus the particular integral, ( )1( ) ( )y x X x

f D=

Case 1: when X(x) = eax, then

1. . , provided ( ) 0( ) ( )

axax eP I e f a

f D f a= = ≠ … (3)

Since,

2 2

.....................

.....................

ax ax

ax ax

n ax n ax

De aeD e a e

D e a e

== =

∴ (Dn + k1Dn – 1 + … + kn – 1D + kn)eax = (an + k1a

n – 1 + … + kn – 1a + kn)eax

f(D)eax = f(a)eax

Operating on both sides 1 ,( )f D

1 1 1( ) ( ) or , ( ) 0( ) ( ) ( ) ( )

axax ax ax ef D e f a e e f a

f D f D f D f a⋅ = = ≠

If f(a) = 0, it is a case of failure and then we proceed as

1 1 , provided ( ) 0( ) ( )( )

axax ax ee x e x f adf D f af D

dD

= = ≠′′ … (3a)


If f(a) = 0, then (D – a) must be a factor of f(D)

i.e. ( ) ( ) ( ) and ( ) ( ) ( ) 1. ( ),df D D a D f D D a D DdD

= − φ = − φ + φ′

which means, ( ) ( ) ( ), provided ( ) 0at D a

df a f D a adD =

= = φ φ ≠′

So that1 1 1 1 1( ) ( ) ( ) ( ) ( ) ( )

ax ax ax ax ax axe e e e e e dxf D D a D a D a a

−= = =− φ φ − φ ∫

( ) ( )

ax axe ex xa f a

= =φ ′ , provided f’(a) ≠ 0

Further if f’(a) = 0 then again it is case of failure and

21 , ( ) 0

( ) ( )

axax ee x f a

f D f a= ≠′′

′′ … (3b)

Example7: Solve the differential equation (D2 – 2D – 3)y = 3e2x

Solution: The given equation (D2 – 2D – 3)y = 3 e2x … (1)has auxiliary equation as: D2 – 2D – 3 = 0 or D = 3, – 1 .…(2)Thus the complementary function, C.F. = c1e

3x + c2e–x … (3)

For particular integral,

2

21 1. . 3 , ( ) 02 3 ( ) ( )

axx ax eP I e comparable to e f a

D D f D f a

= = ≠ − −

2 2

23

(2) 2(2) 3x xe e= = −

− − … (4)

Whence the complete solution, y = (c1e3x + c2e– x) – e– 2x

Example 8: Solve the differential equation (D3 – 3D2 + 4D – 2)y = ex

Solution: Auxiliary equation for the given equation becomes D3 – 3D2 + 4D – 2 = 0 or (D – 1)(D2 – 2D + 2) = 0

Implying 2 4 81,2

D + −= i.e. 1, (1 ± i)

Whence yC.F.(x) = c1ex + ex(c2cosx + c3sinx)

And . . 3 21 1( )

3 4 2 1 3 4 2x x

P Iy x e eD D D

= =− + − − + −

A case of failure as, f(a) = 0.

21 ,

3 6 4x

PIy x eD D

=− + ( )

1 , at provide ´( ) 0axx e D a f af D

= ≠ ′

21

3(1) 6(1) 4x xx e x e= =

− +


Example 9: Solve ( )2

24 cosh 2 1 3xd y

y xdx

− = − + [VTU, 2000; KUK, 2003–04]

Solution: Symbolic form of the equation: (D2 – 4)y = cosh (2x – 1) + 3x

Its auxiliary equation becomes,(D2 – 4) = 0 ⇒ D2 = 4 or D = ±2

So that yC.F.(x) = c1e2x + c2e – 2x

And ( ). 2 21 1( ) cos 2 1 3

4 4x

P Iy x xD D

= − +− −

h

implying ( ) ( )2 1 2 1

. 2 21 1 3

4 2 4

x xx

P Ie ey

D D

− − − += + − −

log 31

2 22 2 21 1 ,

2 4 2 4 4

xx xe e ee e

D D D

−−= + +

− − − [3x is comparable to az = ez.loga]

( )(log 3)

2 22

1 1 1 ,2 2 2 2 log 3 4

xx xe ex e x e

e D D−= + +

− [3x = ex.log3 = e(log 3)x]

( )2 2

2. 3 ,

2 4 2. 4 log 3 4

x x xx e x e ee

−= − +

− (Rewrite, e(log 3)x = 3x)

( )

( )2 12 1

23

4 2 log 3 4

xx xx e e− −− −= + −

23sinh(2 1)

4 (log 3) 4

xx x= − +−

∴ Complete Solution, ( )

−= + + − +−

2 21 2 2

3sinh 2 14 3 4

xx x xy c e c e ( x )

log

Case 2: When X(x) = sin(ax + b) or cos(ax + b), then

2

. . 2 21 sin( )sin( ) , ( ) 0

( ) ( )P Iax by ax b f a

f D f a+= + = − ≠

−Since Dsin(ax + b) = acos(ax + b)

D2sin(ax + b) = – a2sin(ax + b)

D3sin(ax + b) = – a3cos(ax + b)

D4sin(ax + b) = a4sin(ax + b)

or (D2)2sin(ax + b) = (– a2)2sin(ax + b)

………………………………

………………………………In general, (D2)rsin(ax + b) = (– a2)rsin(ax + b)Adding all the above expressions,


( )2 2 1 21 1( ) ( ) sin( )n n

n nD k D k D k ax b−−+ + … + + +

( )2 2 1 21 1( ) ( ) ( ) sin( )n n

n na k a k a k ax b−−= − + − + … + − + +

or f(D2) sin (ax + b) = f(–a2) sin (ax + b)

Operating on both sides the inverse operator 21

( )f D

2 2

2 21 1( )sin( ) ( )sin( )

( ) ( )f D ax b f a ax b

f D f D+ = − +

or ( )22 2

sin( )1 sin( ) , 0( ) ( )

ax bax b f af D f a

++ = − ≠−

In case of failure where f(–a2) = 0, we proceed as:

2 2

sin( )1 sin( ) ,( ) ( )

ax bax b xf D f a

++ =−′ provided f ’(–a2) ≠ 0 … (4a)

Similarly, if f ’(–a2) = 0 then

2 22 2

sin( )1 sin( ) , ( ) 0( ) ( )

ax bax b x f af D f a

++ = − ≠′′−′′

and so on. …(4b)

Example 10: Solve the equation (2D2 + D – 1)y = 16cos2x [Jammu Univ. 2002]

Solution: Given equation (2D2 + D – 1)y = 16cos 2x … (1)

Its auxiliary equation as: 2D2 + D – 1 = 0 i.e. 1 , 12

D = − …(2)

Thus complementary function. 21 2( . .)

x xC F c e c e−= + … (3)

2

2 2 2

cos( )1 1. . 16cos2 cos( ) ; ( ) 0(2 1) ( ) ( )

ax bP I x ax b f aD D f D f a

+= + = − ≠ + − − Q

16 cos22( 4) 1

xD

=− + −

(as here a = 2)

( )

( )( )16 916 cos2 cos2

9 9 9D

x xD D D

+= =

− − +

216 ( 9)cos2

81D x

D= +

−

[ ]16 16cos2 9cos2 (2sin 2 9cos2 )4 81 85

D x x x x= + = −− − … (4)


Hence, the complete solution becomes

( )21 2

16 2sin 2 9cos 285

xxy c e c e x x−

= + + −

Example 11: Solve the equation 2

24 sin2 .

d yy x

dx+ =

Solution: Symbolic form of the given equation, (D2 + 4)y = sin2xCorresponding auxiliary equation, D2 + 4 = 0 i.e. D = ±2iThus, y(C.F) = (c1cos2x + c2sin2x)

( ) 21. . sin 2 ;

4y P I x

D=

+ Replace D2 = – a2 = – 4, but here f(–a2) = 0

( ) ( ) + = + −′

2 21 1In case of failure, sin sin

( ) ( )ax b x ax b

f D f a

Implying ( ) 1 cos2 cos2. . sin 2 .2. 2 2 4

x x x xy P I x xD

− = = = −

Hence the complete solution, ( )1 2cos2cos2 sin 2

4x xy c x c x= + − .

Case 3: When X(x) = xm then, [ ]= ,–1m mP.I.

1y = x 1 (D) xf(D)

± φ where m is a positive integer.

From f(D), take the lowest degree term outside so that the remaining expression in f(D)becomes [1 ±φ(D)]. Now take [1 ±φ(D)] to the numerator and expand it by binomial theoremupto Dm so that (m + 1)th and higher order derivatives of xm are zero.

Note: Some useful results:

(i) (1 – D)–1 = 1 + D + D2 + ...…………… ∞(ii) (1 + D)–1 = 1 – D + D2 + ...…………… ∞

(iii) (1 – D)–2 = 1 + 2D + 3D2 + …………… ∞(iv) (1 – D)–3 = 1 + 3D + 6D2 + …………… ∞

Example 12: Solve 3 2

23 2

6 1d y d y dy

xdx dx dx

− − = +

Solution: Symbolic form becomes (D3 – D2 – 6D)y = 1 + x2

Auxiliary Equation, D3 – D2 – 6D = 0

Implying D(D + 2)(D – 3) = 0 or D = 0, 3, – 2

Whence yC.F. = (c1 + c2e3x + c3e– 2x)

And 2 2

. . 23 21 1(1 ) (1 )

6 6 16 6

P Iy x xD DD D D D

= + = + − − − + −


12

21 1 (1 )6 6 6

D D xD

− = − + − +

( )22 2

21 1 16 6 6 6 6

D D D D xD

= − − − + − + − − − + +

( )2 2

21 1 16 6 6 36

D D D xD

= − − − + +

( ) ( ) ( )2 2 2 21 71 1 16 6 36

Dx x D xD

= − + − + + +

( )2 21 7 1 251 (2)6 3 36 6 3 18

x xx xD D

= − + − + = − − +

( )3 23 21 1 25 1 6 3 25

6 3 3 2 18 108x x x x x x = − − + = − − +

Therefore ( ) ( )3 2 3 21 2 3

1( ) 6 3 25108

x xy x c c e c e x x x−= + + − − +

Example 13: Solve the differential equation (D4 + 16D2)y = x2 + 5. [Jammu Univ, 2002; NIT Kurukshetra, 2006]

Solution: Given (D4 + 16D2)y = x2 + 5. …(1)A.E., D4 + 16D2 = 0 i.e. D2(D2 + 16) = 0 or D = 0, 0, ±4i … (2)

Thus, the complementary function, C.F. = (c1 + c2x) + (c3cos 4x + c4sin 4x) … (3)

For particular integral,

( )24 2

1. . 516

P I xD D

= ++

− =

11 ( )( )

m mcomparable to x f D xf D

( ) ( ) ( )2 22 2 2

2

1 1 15 5116 16 1

16

x xD D D D

= + = + + +

( )1

2 22

1 11 516 16

D xD

− = + +

( )2 22

1 11 516 16

D xD

= − +


2 22

1 1 1 1 395 .216 16 16 8

x x dxD D

= + − = + ∫

3 4 21 1 39 1 39

16 3 8 16 12 16x x xx

D = + = +

Hence complete solution becomes,

( ) ( )2 2

1 2 3 439cos 4 sin 4

64 3 4x xy c c x c x c x = + + + + +

Case 4: When X(x) = eaxV(x), then, 1 1( ) ( )( ) ( )

ax axe V x e V xf D f D a

=+

For some u(x), D(eaxu) = eaxDu + aeaxu = eax(D + a)u

D2(eaxu) = eaxD2u + 2aeaxDu + a2eaxu = eax(D + a)2u … … … … … …

Dn(eaxu) = eax(D + a)nuAdding all,

Dn(eaxu) + k1Dn – 1(eaxu) + … + kn – 1D(eaxu) + kn(eaxu)

= eax(D + a)nu + k1eax(D + a)n – 1u + … + kn – 1 (D + a)u + knu

Implying f(D)(eaxu) = eaxf(D + a)u

Operating f(D) on both sides, ( )1 1( ) ( )( ) ( )

ax axf D e u e f D a uf D f D

= +

or 1 ( )( )

ax axe u e f D a uf D

= +

Now put f(D + a)u = V i.e.1

( )u V

f D a=

+

So that ( )=+1 1

( ) ( )ax axe V e V

f D a f D

Example 14: Solve (D4 – 1)y = excosx

Solution: The corresponding Auxiliary Equation is D4 – 1 = 0 i.e. (D2 – 1)(D2 + 1) = 0 or D = ± 1, ±i

Hence the complementary function y(C.F.) = (c1ex + c2e– x) + (c3cosx + c4sinx)

For particular Integral,

( ) ( ) ( )441 1. . cos cos

1 1 1x xy P I e x e x

D D= =

− + −


( )4 3 21 cos

4 6 4 1 1xe x

D D D D=

+ + + + −

( ) ( ) ( )22 2 2

1 cos4 6 4

xe xD D D D D

=+ + +

1 cos

1 4 ( 1) 6( 1) 4xe x

D D=

+ − + − + (replace D2 = – a2 = – 1)

cos

5x xe=

−Hence the complete solution is

( ) ( )1 2 3 41cos sin cos5

x x xy c e c e c x c x e x−= + + + −

Note: For problem (D4 – 1)y = cosx.coshx, rewrite, ( ) ( )cos cos4 1 x xD 1 y e x e x2

−− = +

Here, −

= =− −1 2

cos cosand

5 5

x xe x e xPI PI

Example 15: Solve the differential equation 2

23 2 xed y dy

y edx dx

+ + =

[KUK, 2003; NIT Kurukshetra 2010]

Solution: Symbolic form of the given equation, 2( 3 2) xeD D y e+ + =Auxiliary equation becomes D2 + 3D + 2 = 0 or D = –1, –2.

C.F. = c1e– x + c2e– 2x

And ( )2 21 1( . .)3 2 3 2

x xe x x ey P I e e e eD D D D

−= =+ + + +

1 1Comparable to ( ) ( ) with 1 and ( )( ) ( )

xax ax x ee V x e V x a V x e ef D f D a

= = − = +

( ) ( )21

1 3( 1) 2xx x ee e e

D D−=

− + − +

( )21 xx ee D e

D D−=

+ , ( ) ( ) = = asx x xe e x e xD e e D e e e

( )1 1( 1)

xx ee D eD D

−=+

( )1 . ,

( 1)x xx e x x x ee e e e e e dx

D− − − = = + ∫

1using ( ) ax axX x e e XdxD a

− = − ∫


( ) ( ) 2x x xx x e x x e x ee e d e dx e e e e e− − − − − = = = ∫Alternately:

( ). . 21 13 2 ( 1)( 2)

x xe eP Iy e e

D D D D= =

+ + + +

1 1

1 2x xe ee e

D D= −

+ + (by partial fraction )

2 2x xx x e x x ee e e dx e e e dx− −= −∫ ∫

( ) ( )2 2x x x xx e x x e e x x ee e e e e e e dx e d e dx− − −= − − =∫ ∫

2 .xx ee e−=

Example 16: Solve the differential equation (D3 + 2D2 + D)y = x2e2x + sin2x[PTU, 2003; NIT Kurukshetra, 2008]

Solution: The auxiliary equation given by D3 + 2D2 + D = 0 i.e. D = 0, –1, –1∴ C.F. = c1 + (c2x + c3)e

–x

For ( )2 2 21 23 2

1. . sin . . . .2

xP I x e x P I P ID D D

= + = ++ +

2 21 3 2

1. .2

xP I x eD D D

=+ +

( ) ( ) ( )2 2

3 21

2 2 2 2xe x

D D D=

+ + + + +

2 2

2 3

121 8 118 118 18 18

xe xD D D

= + + +

12

2 27 4118 6 9

xe D D x− = + + + ………

22

2 27 4 7118 6 9 6

xe D D D x = − + + ……… + + ………

2 2

2 2 27 33 7 331 2 .218 6 36 18 6 36

x xe eD D x x x = − + = − +

2

2 7 1118 3 6

xe x x = − +


22 3 2

1. . sin2

P I xD D D

=+ +

3 21 1 cos2

2 2x

D D D−=

+ +

0

2 2 21 1 1 1 cos2

2 1 (2 ) 2 ( ) 2xe x

D D D D D D D= −

+ + + +

1 1 cos2

2 2 ( 4 8 )x

D D D= −

− − +

( )( )( )3 81 1 1 1 cos2

2 2 3 8 3 8D

xD D D

−= +

+ −

( )2

(3 8)cos21 1 3 8 1cos22 2 9 64 2 2 36 64

D xD xx xD

−−= + = +− − −

( )1 3 cos2 8cos22 200x D x x= − −

3sin 2 4cos2

2 100x x x+= +

Hence the complete solution is

( )2

21 2 3

3sin 2 4cos2 7 112 100 18 3 6

xx x x x ey c c x c e x x− + = + + + + + − +

Case 5: When X = xV, V being a function of x, then

1 1 1( )( ) ( ) ( )

dxV x V Vf D f D dD f D

= +

By Leibnitz’s Theorem on successive differentiation, we have Dn(Xx) = nc0(DnX).x + nc1Dn – 1X.1

⇒ Dn(xX) = xDnX + nDn – 1X

⇒ ( ) ,n n ndD xX x D X D XdD

= + 1n nd D nD

dD− = Q … (1)

or ( )( ) ( ) ( )df D xX x f D X f D XdD

= + … (2)

(if Dn represents a polynomial of nth degree in D)


Putting f(D)X = V, we get 1 1 1( ( ) ) or( ) ( ) ( )

f D X V X Vf D f D f D

= = … (3)

With the help of (3), (2) becomes

1 1( ) . . ( )( ) ( )

f D x V xV f D Vf D f D

= + ′

Operating 1( )f D

on both sides of above equation, we get

( )1 1 1. ( . )

( ) ( ) ( ) ( )f D

x V xV Vf D f D f D f D

′= +

or 2( )1 1. ( . )

( ) ( )( )

f Dx V V xV

f D f Df D

′− =

or 1 1 1. ( . )( ) ( ) ( )

dx V V xVf D dD f D f D

− = , ( ) 1

2( )

since ( )( )

f Dd f DdD f D

− ′= −

Example 17: Solve 2

24 sin

d yy x x

dx+ = [Raipur, 2004]

Solution: Symbolic form, (D2 + 4)y = xsinxA.E., D2 + 4 = 0 implying D = ± 2i so that

yC.F.(x) = (c1cos2x + c2sin 2x)

21( . .) ( sin )

4y P I x x

D=

+ (Comparable to case 5.)

( )22 2

1 2sin sin4 4

Dx x xD D

= −+ +

( )2

1 2sin sin( 1) 4 1 4

Dx x x= −− + − +

Replace f(D2) = f(– a2)

( )sin 2 1 2sin sin cos .3 9 3 9

x x D x x x x= − = −

∴ ( ) ( ). . . . 1 21( ) cos 2 sin 2 3 sin 2cos9C F P Iy x y y c x c x x x x= + = + + −


Example 18: Solve 2

23 2 sinxd y dy

y xe xdx dx

+ + = [VTU, 2001; Osmania, 2003]

Solution: Here A.E. is D2 + 3D + 2 = 0 ⇒ D = –1, –2 y(C.F.) = c1e–x + c2e–2x

( ) ( ) ( )( ). . 22

1 1sin sin3 2 1 3 1 2

x xP Iy x e x e x x

D D D D= =

+ + + + + +

( )21 sin5 6

xe x xD D

=+ +

2 21 1sin sin5 6 5 6

x de x x xD D dD D D

= + + + + + ,

( )22

1 2 5sin sin1 5 6 5 6

x De x x xD D D

+ = −− + + + +

( ) ( )21 2 5sin sin

5 1 1 5 6x De x x x

D D

+= − + − + +

( )221 2 5sin sin

5 1 25 1x x D De x x

D D

+ += − − +

( ) ( )( )

2 51sin sin sin10 25 1 2 1

x Dxe D x x xD

+= − + − − + +

( ) 1 2 5sin sin sin10 25 2

x x De D x x xD+ = − + −

( ) 1 5 1cos sin (sin sin )10 25 2

x xe x x x xD

= − + − +

( ) ( )1sin cos 2sin 5cos10 50

x xe x x x x = − + − −

Complete solution, ( ) ( ) ( )2 21 2

1( ) sin cos 2sin 5cos10 50

x x x xy x c e c e e x x x x− − − = + − + + −

Example 19: Solve (D2 – 4D + 4)y = 8x2e2xsin2x

[ Delhi, 2002; JNTU, 2006; NIT Kurukshetra, 2007; UP Tech, 2007]

Solution: A.E. D2 – 4D + 4 = 0 or D = 2, 2


Whence, yC.F.(x) = (c1 + c2x)e2x … (1)

2 21( ) 8 sin 24 4

xPIy x x e x

D D2=

− +

( ) ( )2 2

218 sin 2

2 4 2 4xe x x

D D=

+ − + +

( )2 2 2 22

1 1 18 sin 2 8 sin 2x xe x x e x xD D D

= = … (2)

2 218 sin 2 ,xe x x dxD

= ∫

2

2 1 18 cos 2 sin 2 cos 2 ,2 2 4

x x xe x x xD

= − + + integration by parts

2 21 1 18 cos2 sin 2 cos22 2 4

xe x x dx x x dx x dx = − + + ∫ ∫ ∫

{ }2 21 sin 2 sin 2 1 18 2 sin 2 cos22 2 2 2 4

x x xe x x dx x x dx xdx = − − + + ∫ ∫ ∫

2

2 sin 2 18 sin 2 cos 24 4

x x xe x x dx x dx = − + + ∫ ∫

( ) ( )2

2 cos2 cos 2sin 2 18 1 cos24 2 2 4

x x xx xe x dx x dx− −

= − + − + ∫ ∫

2

2 38 sin 2 cos 2 cos 24 2 4

x x xe x x x dx = − − + ∫

2

2 38 sin 2 cos 2 sin 24 2 8

x x xe x x x = − − + … (3)

Therefore,

( )2

2 21 2

3( ) 8 sin 2 sin 2 cos 24 8 2

x xCF PI

x xy x y y c c x e e x x x = + = + + − + −

= [c1 + c2x – (2x2 – 3)sin2x – 4xcos2x] e2x

Alternately: From equation (2) onwards,

y.P.I. = Imag. Part of ( )

2 2 2 2 2 222

1 18 . . 82

x ix x ixe x e I P e e xD D i

=+


22 2 2 2 2 2

21 18 . 8 1

4 22 1

2

x ix x ix iDimag e e x imag e e xDii

− = = − − +

( )2

2 2 28. 1 2 34 2 4

x ix iDiDimag e e x

= + + + … −

= − + −

2 2 2 3. 2 22

x iximag e e x ix

= + − − 2 2. (cos 2 sin 2 ) (3 2 ) 4ximag e x i x x ix

2

2 38 sin 2 cos2 sin 24 2 8

x x xe x x x = − − +

ASSIGNMENT 2

Solve 1.2

22

cos ,d y

a y x axdx

+ = 2.2

22

cos ,d y

y x xdx

− =

3.2

2cos ,

d yy ec x

dx+ = [MDU, 2002] 4.

4

4sin ,

d yy x x

dx− = [KUK, 2005-06]

5. (D2 + a2)y = tan ax, [MDU, 2002; VTU, 2005]6. (D2 + 4)y = exsin2x 7. (D2 + 4D + 3)y = e– xsinx + x,8. (D2 – 1)y = xsinx + (1 + x2)ex, [KUK, 2010]9. (D2 – 4)y = xsinhx, [Madras, 2002; KUK, 2007]

10. (D4 + 1)y = x2cosx, [Osmania, 2002]

11.4

4cos cosh ,

d yy x x

dx− =

12. (D2 – 4D + 3)y = sin3xcos2x, [Madras, 2000]

13*.2

22 sin ,xd y dy

y x e xdx dx

− + = 14. (D2 + n2)y = x2ex, [*UP Tech; 2009]

[VTU, 2001; Osmania, 2003; UPTech., 2005; JNTU, 2006; SVTU, 2007; PTU, 2009]

21 1 1Hint: ( ) ( ) ( sin ); integration by parts(0) ( 1)

x x xPIy e V x e V x e x x

f f D D

= = = +

9.6 SPECIAL METHODS FOR FINDING PARTICULAR INTEGRAL

I. Method of Variation of Parameters

The method of variation of parameters which is due to J.L Langrange (1736–1813), enablesus to find the general solution of any second order non homogeneous linear differential


equation (including both, with variable coefficients as well as with constant coefficients)whose complementary function is known. Under this method, complementary function ofthe reduced equation with no X(x) is used by replacing the arbitrary constants with functionsso chosen that a particular integral is obtained.

Consider the equation, 2

2( )

d y dyP Qy X x

dx dx+ + = … (1)

If y1 and y2 are two linearly independent solutions of the above equation and 1 2

1 2*

y yW y' y '=

is the Wronskian of y1 and y2 (named after the noted Polish Mathematician and PhilosopherHoene Wronsky (1778–1853) ), then particular integral.

. . 1 1 2 2 1 2 2 1P IX Xy u y u y y y dx y y dx

W W∗ ∗ = + = − + ∫ ∫ … (2)

Proof: Let general solution of the equation (1) be, yC.F. = c1y1 + c2y2, … (3)

where c1 & c2 are arbitrary constants. The idea behind this method is that assume the arbitraryconstants of the general solution c1 & c2 as functions of x i.e. c1 = u1(x) & c2 = u2(x), so that

yP.I. = u1(x)y1 + u2(x)y2 … (4)

Differentiating (4), ( ) ( )1 1 2 2 1 1 2 2dy

u y u y u y u ydx

′ ′ ′ ′= + + +

On assumption that u’1y1 + u’2y2 = 0 …(6)

We get ( )1 1 2 2dy

u y u ydx

′ ′= + … (5)

Further, differentiating (6),

( ) ( )2

1 1 2 2 1 1 2 22

d yu y u y u y u y

dx″ ″ ′ ′ ′ ′= + + + … (7)

On substituting values of 2

2, ,

dy d yy

dx dx from (4), (6), (7) respectively into (1), we see

u1’y1’ + u2’y2’ = X … (8)Now equations (5) and (8) constitute two linear equations in u1’ and u2’, which on solving,

results in

1 2 2 1and ,X Xu y u yW W∗ ∗

′ ′= − = where W* = y1y2’ – y1’y2

From above u1(x) and u2(x) are obtained by integration and, whence the particular integraland complete solution, y = yC.F. + yP.I. .

Observations:1. Here constant of integration is not required, since we find particular integral, and any function u1(x) and

u2(x) satisfying the required conditions will take care of it.2. Though this method is commonly applied for 2nd order equations but it can easily be extended to

equations of any order.


e.g. Consider 3rd order equation3 2

0 1 2 33 2( ) ( ) ( ) ( ) ( )d y d y dyP x P x P x P x y R xdx dx dx

+ + + =

with yC.F.(x) = c1y1(x) + c2y2(x) + c3y3(x), where y1(x), y2(x) and y3(x) are three linearly independentsolutions corresponding to homogeneous equation and c1, c2, c3 are arbitrary constants. In this case,yP.I.(x) = u1(x) y1 + u2(x) y2 + u3(x) y3:Here also we follow the same procedure as discussed earlier and obtain the required correspondingequations for solving u1’, u2’, u3’ as

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 30

00

( ) ( )( )

u y u y u yu y u y u y

R xu y u y u y X xP x

′ + ′ + ′ = ′ ′ + ′ ′ + ′ ′ = ′ ″ + ′ ″ + ′ ″ = =

Here, Wronskian (a non-zero entity) becomes, 1 2 3

1 2 3

1 2 3

*y y y

W y' y ' y 'y " y " y "

=

Example 20: Solve 2

2tan

d yy x

dx+ = by the method of variation of parameters.

[KUK, 2004–05; PU, 2003-04; Raipur, 2004; NIT Kurukshetra, 2008; UP Tech, 2009]

Solution: Rewrite the given equation as (D2 + 1)y = tanxHere A.E D2 + 1 = 0 or D = ±i

Thus yC.F.(x) = c1y1 + c2y2 = c1cosx + c2 sinx

Assume yP.I.(x) = u1(x) y1 + u2(x) y2 = u1cosx + u2 sinx

By method of variation of parameter, we have

1 21 2

1 22 1

( )cos sin

and * 1; tansin cos( )

Xu x y y y x xW W X xx xX y yu x yW

∗

∗

′ = − = = = = −′ ′ ′ =

On integration,

2

1sin cos 1( ) sin tan sincos cos

x xu x x x dx x dx dxx x

−= − = − =∫ ∫ ∫ ( ) ( )cos sec sin log sec tanx x dx x x x= − = − +∫

And 2( ) cos tan sin cosu x x x dx x dx x= = = −∫ ∫Whence yP.I.(x) = cosx[sinx – log(secx + tanx)] + sinx(– cosx)

= – cos x log(secx + tanx)And therefore the complete solution becomes,

y = (c1cosx + c2sinx) – (cosx) log (secx + tanx)


Example 21: Solve 2

2sin

d yy x x

dx+ = , using method of variation of parameter.

[SVTU 2007; JNTU, 2005]

Solution: Symbolic form of the given equation, (D2 + 1) y = xsinxAuxiliary equation, D2 + 1 = 0 i.e. D = ±i

Thus C.F. = c1y1 + c2y2 = c1cosx + c2sinx … (1)

Let P.I. = u1(x)y1 + u2(x)y2;Now by method of variation of parameter,

( )

( )

1 21 2

1 22 1

cos sinand W* = 1;sin cos

Xu x yy y x xWy' y ' x xXu x y

W

∗

∗

′ = − = = − ′ = …(3)

So that ( ) ( )1 2

1 cos2( ) sin sin

2xXu x y dx x x x dx x dx

W∗−

= − = − = −∫ ∫ ∫

2 sin 2 cos24 4 8x x x x = − + + … (4)

2 11( ) cos ( sin ) sin 22

Xu x y dx x x x dx x x dxW∗

= = =∫ ∫ ∫ … (5)

sin 2cos2

4 8x xx = − +

Thus, 2 1 sin 2. . cos sin 2 cos2 sin cos2

4 4 8 4 8x x x xP I x x x x x = − + + + − +

( ) ( )2 1cos sin 2 cos sin cos2 cos cos2 sin sin 2

4 4 8x xx x x x x x x x x= − + − + +

2 1cos sin cos

4 4 8x xx x x= − + + … (6)

Therefore complete solution

( )2

1 21cos sin cos sin cos8 4 4

x xy c x c x x x x= + + + −

Example 22: Solve y” – 2y’ + 2y = ex tanx by variation of parameter. [Burdwan, 2003]

Solution: Rewrite the given equation, (D2 – 2D + 2)y = extanx.Here A.E becomes, D2 – 2D + 2 = 0 D = 1 ± i


Thus, yC.F.(x) = c1y1 + c2y2 = ex[c1cosx + c2sinx] … (1)

Assume: yP.I.(x) = u1(x)y1 + u2(x)y2 … (2)By method of variation of parameter,

1 2 2 1´( ) , ´( ) ;X Xu x y u x yW W∗ ∗

= − =

1 2’ ’1 2

cos sin*

sin cos sin cos

x x

x x x xy y e x e x

W y y e x e x e x e x= =

− + +

cos sin(cos sin ) (cos sin )

x x

x xe x e xe x x e x x

=− +

Thereby on integration of u1 ́& u2´, we get

. . 1 2 2 1( )P IX Xy x y y dx y y dx

W W∗ ∗= − +∫ ∫

1 22 2tan tansin cos

x xx x

x xe x e xy e x dx y e x dx

e e= − +∫ ∫

1 2 1 1 2 2sin tan cos tany x x dx y x x dx y I y I= − + = − +∫ ∫ … (3)

21 tan sin tan cos sec cosI x x dx x x x x dx= = − +∫ ∫ , integration by parts.

= – sin x + log(sec x + tan x) … (4)

2 sin cosI x dx x= = −∫ … (5)

On putting the values of integrals, (4) and (5) again into (3), yPI = – excosx[– sinx + log(secx + tanx)] – exsinxcosx

= – excosx log(sec x + tanx)Hence the complete solution is

y = ex(c1cosx + c2sinx) – excosx log(secx + tanx).

Example 23: Solve (D2 – 1)y = e–2xsin e–x using method of variation of parameter.

Solution: The given equation (D2 – 1)y = e– 2xsin e–x … (1)Auxiliary form D2 – 1 = 0 or D = ±1

So that yC.F.(x) = c1y1 + c2y2 = (c1ex + c2e–x) … (2)

Assume yPI(x) = u1(x)y1 + u2(x)y2 … (3)Now by the method of variation of parameter,

21 2 2 1, ; where sinx xX Xu' y u ' y X e e

W W−

∗ ∗ = − = − =


And 1 2

1 2* 2´ ´

x x

x xy y e e

W y y e e

−

−= = = −−

Thereby giving22

1. sinsin

2 2

x xx t t te e dtu e dx

t

− −−= − = −

− − −∫ ∫ (taking e–x = t)

( ) ( ) ( )2 21 1sin cos 2 cos2 2

t t dt t t t t dt− = − = − − − ∫ ∫ ( ) ( )21 cos 2 sin 2 sin

2t t t t t dt = − − + − ∫

2

cos sin cos2t t t t t= − −

2

cos sin cos2

xx x x xe e e e e

−− − − −= − − … (4)

And 2

2sin 1 sin

2 2

x xx x xe eu e dx e e dx

− −− −= − =

−∫ ∫

1 1 1sin cos cos2 2 2

xt dt t e−− −= = =∫ … (5)

Whence using (4) and (5) in (3), we get

2

. .1( ) cos sin cos cos

2 2

xx x x x x x x

P Iey x e e e e e e e

−− − − − − − − = − − +

= –(sine–x + excose–x) … (6)And therefore complete solution becomes

y(x) = yCF(x) + yPI(x) = (c1ex + c2e–x) – (sine–x + excos e–x) … (7)

Example 24: Solve by the method of variation of parameter, 2

22

1 x

d yy

dx e− =

+.

[VTU 2005; Hisar, 2005]

Solution: Here auxiliary equation becomes D2 – 1 = 0, means D = ±1.Thus yC.F.(x) = c1y1 + c2y2 = c1e

x + c2e–x … (1)

By method of variation of parameter, we assume the particular integral as yPI(x) = u1y1 + u2y2 = u1e

x + u2e–x, … (2)

1 2

1 2* 2´ ´

x x

x xy y e e

W y y e e

−

−= = = −−

and 1 2 2 1( ) , ( ) ,X Xu x y u x yW W∗ ∗

′ ′= − = …(3)


Implying 12 1( )

1 2 1

xx

x xeu x e dx dx

e e

−− = − ⋅ = + − +∫ ∫ (Taking ex = t, exdx = dt)

( )2 21 1 1 1

1 1dt dt dt dt

t t t t t= = − +

+ +∫ ∫ ∫ ∫ (By partial fractions)

( )1 1log log 1 logx

xxet t e

t e− + = − − + + = −

and ( )2( )

log 11 ( )

xx

x

f xeu dx dx ee f x

′= − = − = − +

+∫ ∫

Whence, ( ). .1( ) log log 1

xx x x x

P I xey x e e e e

e− − + = − + − +

And the complete solution becomes

( ) ( ). . . . 1 21( ) log log 1 1.

xx x x x x

C F P I xey x y y c e c e e e e

e− − + = + = + + − + −

Example 25: It is given that 1 21andy x yx

= = are two linear independent solution of

the associated homogeneous equation 2

22

, 0.d y dy

x x y x xdx dx

+ − = ≠ Find particular integral

and the general solution. [NIT Kurukshetra, 2010]

Solution: Rewrite the given equation as 2

2 21 1 , 0.

d y dy yx

dx x dx x x+ − = ≠ … (1)

Further, we are given . . 1 1 2 2 1 21( )C Fy x c y c y c x cx

= + = + … (2)

By method of Variation of Parameter,

. . 1 1 2 2 1 21( ) ( ) ( )P Iy x u x y u x y u x ux

= + = + … (3)

The Wronskian of y1 = x and 21yx

= is

2

12* , 011

xxW x

xx

= = − ≠−

1 2 2 11( ) and ( ) withX Xu x y u x y X

W W x∗ ∗ ′ ′= − = =

Implying 11 1 1( ) log

2 2xu x dx x

x x = − ⋅ = −∫


2

21( )

2 4x xu x x dx

x = ⋅ = − −∫ … (4)

Whence 2

. .1 1( ) log2 4P I

xy x x xx

− = + And complete solution,

{ } 21 2 1 1 1

1 1( ) log log , * .2 4 2 4

cx x xy x c x c x c x x c cx x

∗ = + + − = + + = −

II Method Of Undetermined Coefficients:

Under this method which is applicable only to equations with constant coefficients, we find yP.I.(x)of f(D)y = X(x) by assuming a trial solution containing unknown constants which are determinedby substituting it and its derivatives in the given equation. Here, assumed yP.I.(x) dependson the form of X(x) in each case. In cases, when the right hand side of the equation viz. X(x)is of the form containing exponential, polynomial, cosine and sine functions, sums or product ofthese functions, then particular integral can easily be found by this method. The accompanyingtable illustrates the procedure as:

X(x) Assumed yP.I.(x)eax Aeax

cosβx or sinβx (Asin βx + Bcosβx)xn (A0xn + A1xn – 1 + …… + An – 1x + An)eaxxn eax(A0x

n + A1xn – 1 + …… + An – 1x + An)

eaxcosβx or eaxsinβx eax(Asinβx + Bcosβx)

This method holds so long as no term of the assumed Particular Integral appears inComplementary Function. If any term of the assumed expression (P.I.) is present in thecomplementary function, then multiply the assumed particular integral by x repeatedly untilno terms of product is present in the complementary function. It fails, when X(x) = tanx,sec x, cot x, since the differentiation of X(x) results in infinite number of terms.

Example 26: Using the method of undetermined coefficients, solve the differentialequation: (D2 – 2D + 3)y = x3 + cosx

Solution: The corresponding auxiliary equation is

2 2 4 122 3 0 . . 1 22

D D D i e D i± −− + = ⇒ = = ±

∴ ( )1 2. . cos 2 sin 2xC F e c x c x= + … (1)

Supposing the particular integral

( ) ( )3 21 2 3 4 5 6cos siny a x a x a x a a x a x= + + + + + … (2)

So that ( ) ( )21 2 3 5 63 2 sin cos ,

dya x a x a a x a x

dx= + + + − + … (3)


( ) ( )2

1 2 5 626 2 cos sin ,

d ya x a a x a x

dx= + + − − … (4)

On substituting above values of 2

2, and

dy d yy

dx dx in the given equation, we get

( ) ( ) ( ) ( )21 2 5 6 1 2 3 5 66 2 cos sin 2 3 2 sin cosa x a a x a x a x a x a a x a x + + − − − + + + − +

( ) ( )3 2 31 2 3 4 5 63 cos sin cosa x a x a x a a x a x x x + + + + + + = + … (5)

Comparing the coefficients of x3, x2, x, constant term, cosx and that of sinx on both sides ofequation (5) we get

1133

a = 3a2– 6a1 = 0 6a1 – 4a2 + 3a3 = 0 2a2 – 2a3 + 3a4 = 0 2a5 – 2 a6 = 1,

⇒ 113

a = ⇒ 223

a = ⇒ 36 8 3 03 3

a− + = ⇒ 44 4 3 03 9

a− + = a5 + a6 = 0

i.e. 329

a = i.e. 48

27a = − i.e. 5 6

1 1,4 4

a a= = −

With the above values of a1, a2, a3, a4, a5 and a6, particular integral becomes

( )2

22 2 8 1 1. . cos sin3 3 9 27 4 4xy P I x x x x = + + − + −

( ) ( )2 21 19 18 6 8 cos sin27 4

x x x x x= + − − + − … (6)

Hence the complete solution,

( ) ( )2 21 2

1 1( cos 2 sin 2 ) 9 18 6 8 cos sin27 4

xy e c x c x x x x x x= + + + − − + −

Example 27: Solve by Method of undetermined, the equation (D2 – 2D)y = exsinx[VTU, 2006; NIT Kurukshetra, 2009]

Solution: Here yC.F.(x) = (c1 + c2e2x) … (1)

Assume yPI(x) = ex(Acosx + Bsin x) … (2)

( ) ( )sin cos cos sinx xdyDy e A x B x e A x B x

dx= = − + + + … (3)

( ) ( )cos sinxe A B x B A x= + + −

( ) ( ) ( )2

22

( )sin cos cos sinx xd yD y e A B x B A x e A B x B A x

dx= = − + + − + + + −


= ex(2Bcosx – 2Asin x) … (4)On substituting values D2y, Dy and y in the given equation, we get

( ) ( ) ( )2 cos 2 sin 2 cos sin sinx x xe B x A x e A B x B A x e x− − + + − = … (5)

Comparing the coefficient of excosx, exsinx on both sides,

and ( )( )

2 2 0 012 2 12

B A B A

A B A B

− + = ⇒ = − − − = ⇒ = −

… (6)

Whence, 1( ) sin2

xPIy x e x= − and complete solution is

( )21 2

1( ) sin2

x xy x c c e e x= + −

Example 28: Solve (D2 – 2D + 1)y = ex by method of undetermined coefficients.[NIT Kurukshetra, 2006]

Solution: Here yC.F. = c1ex + c2xex Since the right hand member of the complete equation isX(x) = ex, therefore the first possible assumption for particular integral would be yP.I. = Aex,but this solution has already occurred in yC.F. We, therefore, yP.I.(x) = Ax ex. Still this solutionalso has appeared in yC.F. We, therefore, again multiply by x and try yP.I.(x) = Ax2ex

Thus 2 22 (2 )x x xdyA xe x e A x x e

dx= + = +

And ( ) ( ) ( )2

2 22

2 2 2 4 2x x x x xd yA e xe xe x e A x x e

dx = + + + = + +

On substituting values of D2y, Dy and y in the given equation, we get A(x2 + 4x + 2)ex – 2A(2x + x2)ex + Ax2ex = ex

or 2Aex = ex implying 12

A = (ex ≠ 0)

Whence, 2. .

1( )2

xP Iy x x e=

and complete solution becomes 21 2

1( )2

x x xy x c e c e x e−= + + .

III The Complete Solution in Terms of a known Integral

OR

Solution: By Method of Reducing the Order of the Equation

If an integral included in the complementary function of a second order equation,

2

2( ) ( )

d y dyP x Q x y X

dx dx+ + = … (1)


be known, then the complete solution can be found.Let y = u(x) be a known solution (i.e. complementary function) of the homogeneous

equation corresponding to (1).

∴ 2

2( ) ( ) 0d u duP x Q x u

dx dx+ + = … (2)

Now, let y = u(x)v(x) be the general solution of (1), so that

2 2 2

2 2 2and 2

dy d ydv du d v d u du dvu v u vdx dx dx dx dx dx dx dx

= + = + + … (3)

On substituting these values of 2

2,

dy d ydx dx

into equation (1) and re-arranging the terms,

we get. 2 2

2 22d v du dv d u duu Pu P Qu v X

dx dx dx dx dx + + + + + =

or 2

22d v du dvu Pu X

dx dx dx + + = , using (2)

or 2

22d v du dv XP

dx u dx dx u + + = … (4)

If we take ,dv pdx

= equation (4) reduces to

2dp du XP p

dx u dx u + + = … (5)

Clearly, this equation being a Leibnitz Linear differential equation of 1st order can be

solved for ‘p’, where dvpdx

= which on integration w.r.t. x results in an expression for v.

∴ y = uv gives a solution of equation (1).

Note: Complete solution corresponding to equation 2

2( ) ( ) ( )

d y dyP x Q x y X x

dx dx+ + = can be found by the

following rules:1. y = x is the part of the C.F. if P + xQ = 02. y = x2 is the part of the C.F. if 2 + 2xP + x2Q = 03. y = ex is the part of the C.F. if 1 + P + Q = 04. y = e–x is the part of the C.F. if 1 – P + Q = 0

5. y = eax is the part of the C.F. if2

1 0P Qa a

+ + =

Example 29: Solve y” – 2y’ + y = exsinx given that y = ex is a solution corresponding tohomogeneous equation.


Solution: The given equation is comparable to 2

2( ) ( )

d y dyP x Q x y X

dx dx+ + = with P= – 2, Q = 1

so that 1 + P + Q = 0, means y = ex is a part of the complementary function. Let y = vex be asolution of the given equation.

Thus, 2 2

2 2, 2x x x x xdy d ydv dv d vv e e ve e e

dx dx dx dx dx= + = + +

On using these values of dydx

and 2

2

d ydx

into the given equation,

2

22 2 sinx x x x x x xdv d v dvve e e ve e ve e x

dx dx dx + + − + + =

On simplification,2 2

2 2sin or sin ( 0)x x xd v d ve e x x e

dx dx= = ≠

Which on integration implies, 1cosdv x cdx

= − +

Further, v = – sin x + c1x + c2

∴ The complete solution, y = ex(– sin x + c1x + c2)

Example 30: Apply Method of Variation of Parameter to solve the equation

( ) ( )2

22

1 1d y dy

x x y xdx dx

− + − = −

Solution: Rewrite the given equation as

( ) ( ) ( )2

21 1

1 1d y dyx y xdx x dx x

+ − = +− − … (1)

Here P + xQ = 0, with 1and ;1 1

xP Qx x

= = −− −

means y = x is a part of the

complementary function. Thus for finding complementary function of the equation (1),Put y = vx so that equation (1), reduces to

( )2

22 0

1d v x dvdx x x dx

+ + = − … (2)

or 2 0, where

1dp x dvp pdx x x dx

− − − = = −

or 1 21 0

1dp

pdx x x

− + − = − ⇒1 21

1dp

dxp x x

= + − −(case of variable-separable)

On integration, logp = x + log(x – 1) – 2logx + logc1


implying ( )1

2

1 xc x ep

x−

= …(3)

i.e. 1 2

x xdv e ecdx x x

= − again a case of variable-separable, …(4)

11 2 22

x xxce ev c dx dx c e c

x x x = − + = + ∫ ∫ … (5)

∴ y(C.F) = vx =(c1ex + c2x) … (6)

Now let, y(P.I.) = u1y1 + u2y2,where u1 and u2 are functions of x and y1 = ex, y2 = x.

Now under this method ∗ ′ = − 1 2

Xu yW

and ∗ ′ = 2 1

Xu yW

where 1 2

1 2* (1 ); and (1 )

1

xx

xy y e x

W e x X xy' y ' e= = = − = −

∴( )( ) ( )1 21

11

x xx

xXu y dx x dx xe dx x eW e x

− −∗

− = − = − ⋅ = = + −∫ ∫ ∫ … (7)

And ( )( )2 111

xx

xXu y dx e dx xW e x∗

− = = ⋅ = −∫ ∫ … (8)

∴ y(P.I.) = u1y1 + u2y2 = (x + 1)e–xex + x2 = (1 + x + x2)And complete solution y = (c1e

x + c2x) + (1 + x + x2)


22

1 cot cot sin .d y dy

x y x xdx dx

+ − − =

Solution: The given equation is comparable to

2

22

sind y dy

P Qy xdx dx

+ + = … (1)

Here, P = 1 – cotx, Q = – cotx and 1 – P + Q = 0.Therefore, y = ve – x is complementary solution of the equation,

( )2

21 cot cot 0

d y dyx y x

dx dx+ − − = … (2)

Take, y = ve– x so that

2 2

2 22

x

x

dy dve vdx dxd y d v dve vdx dx dx

−

−

= − = − +


By this substitution, the given equation reduces to

( ) ( )2

21 cot or 1 cot ford v dv dz dvx x z z

dx dx dx dx= + = + = … (3)

Implying ( ) ( )11 cot or log log sindz x dx c z x xz

= + = +∫∫or 1 ( sin )xdvc e x

dx= ... (4)

Integrating, ( ) 21

1 1

sin or sin cos2

xx cec dv e x dx v x x

c c= = − +∫ ∫ … (5)

∴ ( ) 2

1 1

sin cos ,2

x

x

y ce x xe c c−

= − +

as y = ve– x

or yC.F. = a1(sinx – cosx) + a2e – x;2

1 21

1 ,2

ca a

c c= = …(6)

Let yP.I.(x) = u1y1 + u2y2, where u1 and u2 are functions of 'x'(i.e. replace arbitrary constants in C.F by u1(x) and u2(x).)

Then by method of variation of parameter,

1 2 2 1; ,X Xu y u yW W∗ ∗

′ ′= − = ... (7)

sin cos

2 sincos sin

xx

xx x e

W e xx x e

−∗ −

−−= = −+ − and X = sin2x …(8)

Now 2

1 2sin 1 cossin

2 sin 2 2x

xX x xu y dx e dx x dx

W e x−

∗ −= − = − = =−∫ ∫ ∫ … (9)

And, ( )2

2 1sinsin cos

2 sinxX xu y dx x x dx

W e x∗ −= = −−∫ ∫

( )21 sin sin cos2

xe x x x dx= − −∫

( )1 cos21 1 sin 22 2 4

x xxe dx e x dx

−= − +∫ ∫

( ) ( )cos 2 2 sin 2 2 cos 2 sin 220 20 4

x x xe e ex x x x −= + − − − …(10)

On substituting values of u1, u2, and y1, y2 particular integral becomes.

( ) ( ) ( ). .cos

sin cos cos 2 2sin 2 2cos2 sin 22 20 20 4

x x xx

P Ix e e ey x x e x x x x− − = − + + − − −

and, y(complete) = yC.F. + yP.I.


ASSIGNMENT 3

1. Using Method of Variation of Parameter, solve the following equations

(i)2

22

cos ;d y

y ec xdx

+ = (ii) (D2 – 2D + 2)y = excosec x;

(iii) ( )2 24 4 8 sin 2 ;xD D y x e x− + = (iv) y” – 2y’ + 2y = extan x;

(v) (D2 – D – 2)y = e(ex + 3x); (vi)3

26 9 ;

xey" – y' + y =x

[PTU, 2006]

2. Apply Method of variation of parameter to the following equations:(i) x2y” – 4xy’ + 6y = x2logx (ii) x2y2 + xy1 – y = x2ex

[Hint: Either follow example 27 or reduce the given Cauchy equation with variablecoefficient to equations with constant coefficients by putting x = et and then applyMethod of variation].

3. Solve by Method of variation of parameter, the equation

( ) ( )2

2 32

2 1 2 1d y dy

x x x x y xdx dx

− + + + =

[Hint: y = x is a part of C.F., take y = vx. as y (C.F.)]4. Using method of undetermined coefficients, solve the following equations:

(i) (D2 – 3D + 2)y = 10; (ii)2

2;

d yy x

dx+ =

(iii)2

22 4 ;xd y dy

y edx dx

− + = − (iv)2

24 4 sin ;xd y dy

y xe xdx dx

+ + = +

5. Use Method of undetermined coefficient to find solution of(i) (D2 – 3D + 2)y = x2 + ex (ii) (D2 – 1)y = xsin3x + cosx

[KUK, 2008]

6. Solve ( )2

2cot 1 cot sin ;xd y dy

x x y e xdx dx

− − − = by reducing the order of integration .

[Type III, complete solution in terms of known integral.]

9.7 SOLUTION OF EULER�CAUCHY AND LEGENDRE: LINEAR EQUATIONS

These are two special type of homogeneous linear differential equations with variable coefficientswhich on application of certain transformation reduce to homogeneous linear differentialequations with constant coefficients. Cauchy–Euler equation is more commonly known asCauchy’s equation. The details of the two are as follows:

I Cauchy�s Linear Equation

The differential equation of the form

1

10 1 11

n nn n

n nn n

d y d y dyK x K x K x K y X

dx dx dx

−−

−−+ + …… + + = … (1)


where K0, K1, ……, Kn are constants and X is a function of x only, known as Cauchy's linearequation. Such equations can be reduced to linear differential equation with constantcoefficients by putting.

x = et or t = logx so that1dt

dx x=

1 or . if .

dy dy dy dy dydt dx Dy Ddx dt dx dt x dx dt dt

= = = = =

Again, 2 2

2 2 2 2 21 1 1 1 1d y dy dy dy dy dy d yd d d dt

dx dx dx dx x dt x dt x dt dx dt x dt x dt = = = − + = − +

or ( )2

2 22

1d y

x D y Dy D D ydx

= − = − and so on.

By substituting all these values in (1), we obtain linear equation with constant coefficients,which has already been discussed.

Example 32: Solve the differential equation [Raipur, 2004]

( )3 2

3 23 2

3 8 65cos log , 0d y d y dy

x x x y x xdx dx dx

+ + + = >

Solution: The equation ( )3 2

3 23 2

3 8 65cos log , 0d y d y dy

x x x y x xdx dx dx

+ + + = > represents

Cauchy's Homogeneous Linear Equation. In this case, we substitute, x = et

so that ( )2

22

; 1dy dx Dy x y D D ydx dx

= = − and so on. for ;d Ddt

=

The given equation becomes, D(D – 1)(D – 2)y + 3D(D – 1)y + Dy + 8y = 65cos t

⇒ [D3 + 8]y = 65cos t

A.E. D3 + 8 = 0 ⇒ D = – 2, 1 3i±

Hence the complementary function (C.F) ( )21 2 3cos 3 sin 3t tc e e c t c t−= + +

Particular Integral, ( )31 1. . 65 cos 65 cos

8 8P I t t

D D= =

+ − +

21 865 cos 65 cos

8 64Dt t

D D+= =

− −

(8 )cos65 (8cos sin ), log

64 1D t t t t x+= = − =

+



222

3 4 1d y dy

x x y xdx dx

− + = + . [SVTU, 2007]

Solution: Here we substitute x = et so that 1log and dtx tdx x

= =

If ( )2

22

; , 1dy d yd D x Dy x D D y

dt dx dx= = = − and so on.

Therefore the given equation reduces to [D(D – 1) – 3D + 4]y = (1 + et)2

or (D2 – 4D + 4)y = 1 + 2et + e2t. …(1)

A.E. D2 – 4D + 4 = 0 i.e. D = 2, 2 so that

yC.F.(x) = (c1 + c2t)e2t = (c1 + c2logx)x2 …(2)

and ( )21 2 32

1( ) 1 24 4

t tPIy x e e PI PI PI

D D= + + = + +

− +

01 2

1 1 ,4 4 4

tPI eD D

= =− + (replace D by 0)

( ) ( )2 2 21 12 2 2 2 .

2 1 2t t tPI e e e x

D= = = =

− −

( ) ( )( )2 22 2

2 23 2

log1 1 .2 2 2 1 22

tt t x xt ePI e t e

DD= = = =

−−

Whence complete solution becomes,

( ) ( )2 22

. . . . 1 21( ) log 2 log4 2C F P I

xy x y y c c x x x x= + = + + + +


22

log sin(log ), 0d y dy

x x y x x xdx dx

+ + = >

[KUK 2006, Madras 2006; Kerala, 2005; Raipur, 2005]

Solution: This is a Cauchy's linear homogeneous equation, where

,dy

x Dydx

= ( )2

22

1 , withd y dx D D y Ddx dt

= − … =

Given equation reduces to [D(D – 1) + D + 1]y = tsin t or (D2 + 1)y = tsin t …(1)

∴ yC.F.(x) = (c1cost + c2sint) …(2)

And, ( ). . 22 21 1 1( ) sin Imaginary Imaginary

1 1 1it it

P Iy x t t te e tD D D i

= = =+ + + +


11 1Img. Img. 12 22 1

2

itit e De t t

D i D iiDi

− = = + +

1 1 1 1Img. Img.

2 2 2 2

itit i ee tti D i D D i

= − = − −

2 2 2

Img Img2 2 2 2 2 2

it iti e i et i t t iti

= + = + − −

( ) ( )2cos sin

Img4

i t i tt it

+= +

−

( )sin cos4t t t t= − …(3)

Therefore, complete solution

( ) ( )1 2( ) cos sin sin cos , log4CF PIty x y y c t c t t t t t x= + = + + − = …(4)

II Legendre's Linear Differential Equation:

The differential equations of the form

( ) ( ) ( )1

10 1 11

n nn n

n nn n

d y d y dyK ax b K ax b K ax b K y X

dx dx dx

−−

−−+ + + + ……… + + + =

where K0, K1, …, Kn are constants and X is a function of x only, known as Legendre's Equation.Such equation can be reduced to linear differential equation with constant coefficients byputting

(ax + b) = et or t = log(ax + b) so that .dt adx ax b

=+

( )or , ifdy dy dy dy dydt a dax b a aDy Ddx dt dx dt ax b dx dt dt

= = + = = =+

Again, ( )

2 2

22

d y dy dy dy dyd d a a a d dtdx dx dx dx ax b dt dt ax b dt dx dtax b

= = = − + + ++

( ) ( )22 2

2 2 2

dy d ya adt dtax b ax b

= − ++ +

or ( )2 22 2

2 2

d y d y dyax b a

dx dt dt + = − = a2D(D – 1)y and so on

By substituting all these values in (2), we obtain linear equation with constant coefficientswhich has already been discussed.


Example 35: Solve ( ) ( ) ( )( )2

22

1 1 4 cos log 1d y dy

x x y xdx dx

+ + + + = +

[MDU, 2005; NIT Kurukshetra, 2010]

Solution: As the given equation is Legendre's Linear equation. Here we take (1 + x) = et

( )

( ) ( )

( ) ( )( )( )

222

323

1 ,

1 1 ,

1 1 2 , so on

dx y Dydx

dx y D D ydxdx y D D D ydx

+ = + = − + = − − …

where dDdt

=

The given equation reduces to D(D – 1)y + Dy + y = 4cost ⇒ (D2 + 1)y = 4cos t, t = log(1+ x)

A.E. D2 + 1 = 0 ⇒ D = ± i

∴ C.F. = (c1cos t + c2sin t),

2 21 1. . 4cos 4 cos 2 cos 2 cos 2 sin

1 2 1D DP I t t t t t t t t t

D D D= = = = =

+ −

Complete solution, y = (c1cos t + c2sin t) + 2tsin t, t = log(1 + x).

Example 36: Solve [(3x + 2)2D2 + 3(3x + 2)D – 36]y = (3x2 + 4x + 1) [Sambalpur, 2002]

Solution: Legendre’s equation ((3x + 2)2D2 + 3(3x + 2)D – 36)y = (3x2 + 4x + 1) … (1)

Take (3x + 2) = et so that t = log (3x + 2) and3

3 2dtdx x

=+ … (2)

∴ 3 3 , where

3 2 3 2dy dy dydt dydx dt dx x dt x dt

= = = δ δ =+ +

or ( )3 2 3dy

x ydx

+ = δ … (3)

Similarly ( ) ( )2 2

22 23 9 9.

3 2 3 23 2d y dy dy dy d yd ddx dx dx dx x dt dt x dtx

− = = = + + ++

or ( ) ( )2 2

22 2

3 2 9 9 1d y d y dy

x ydx dt dt

+ = − = δ δ − … (4)

Therefore, the given equation reduces to

( ) ( )219 1 3(3 ) 36 9 12 33

y y y x xδ δ − + δ − = + +

Implying ( ) ( )22 3 2 1

9 363

xy

+ −δ − =


( ) ( )2

2 19 4 ; as 3 23

ttey x e−δ − = + = … (5)

A.E. δ2 – 4 = 0 or δ = ± 2 and thus yC.F.(t) = c1y1(t) + c2y2(t) = c1e2t + c2e–2t

And ( )2 2 0

2 02 2

1 1 1 1 1 1 1( ) .( ) 3 3 9 4 4 27 2 4

t t tt t

PIe e ey t e e t

f

− = = − = − δ δ − δ − δ −

2 2

, log(3 2)1 1 1

27 4 4 108

t tt x

t e te = + += + =

Complete solution, ( )2

2 21 2

1( )108

tt t tey t c e c e− += + +

Example 37: Solve ( ) ( )2

22

2 3 2 3 12 6d y dy

x x y xdx dx

+ − + − =

[VTU, 2003, 2007; Kerala, 2005; KUK, 2005; NIT Kurukshetra, 2008]

Solution: Take 2x + 3 = et, t = log(2x + 3) so that the given equation reduced to [4D(D – 1) – 2D – 12]y = 3(et – 3) as 6x = 3(2x) = 3(et – 3)

or 2(2D2 – 3D – 6)y = 3(et – 3)

A.E 21 2

3 572 3 6 0 or ,4

D D D m m±− − = = =

3 57 3 57

1 2 1 2 4 41 2 1 2 1 2( ) ( ) ( ) ( ) ( )m t m t t m t m t t

CFy t c e c e c e c e c e c e+ −

= + = + = +

( )21. . 3 3

4 6 12tP I e

D D= −

− −

02 2

1 1 3 33 94 6 12 4 6 12 14 4

t t te e eD D D D

= − = − +− − − −

Whence ( ) ( ) ( )1 21 2

3 3( ) ( ) ( ) 2 3 2 3 2 3 .14 4

m mCF PIy x y x y x c x c x x= + = + + + − + +

ASSIGNMENT 4

Solve the following equations

1. (i)2

2 22

4 6 ;d y dy

x x y xdx dx

− + = (ii)2

2 42

2 4 ;d y dy

x x y xdx dx

− − =

(iii)2

22 2

2 1 ;d y y

x xdx x x

− = + (iv)2

2 2

12 log1 , 0;xd y dy

xdx x dx x

+ = >

(v)3 2

3 23 2

12 2 10 ;d y d y

x x y xdx dx x

+ + = + [SVTU, 2006; PTU, 03]


(vi)2

2 22

2 4 2log ;d y dy

x x y x x x odx dx

− − = + > [Bhopal, 2003; KUK, 2010]

(vii) The radial displacement u in a rotating disc at a distance r from the axis is given

by 2

2 32

0,d u dur r u krdr dr

+ − + = where k is a constant. Solve the equation under

the conditions u = 0, when r = 0; u = 0 when r = a.

(viii) ( ) ( ) ( )( )2

22

1 1 2sin log 1d y dy

x x y xdx dx

+ + + + = +

(ix) ((1 + 2x)2D2 – 6(1 + 2x)D + 16)y = 8(1 + 2x)2

(x) ((x + 1)2D2 + (x + 1)D)y = (2x + 3)(2x + 4)2. Establish the Euler`s Cauchy equation of IIIrd order whose general solution is

y = Ax + Bx2 + Cx3.3. Establish the Euler-Cauchy equation of IIIrd order whose general solution is

y = Ax + Bx(logx) + Cx(logx)2

9.8 SIMULTANEOUS LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT

COEFFICIENTS

Linear differential equations consisting of two or more dependent variables and singleindependent variable say x(t), y(t) etc. are called simultaneous differential equations. Tosolve these type of equations, eliminate one of the dependent variable and then solve theresulting equation which gives the solution of the dependent variable in terms of theindependent variable. Repeat the process for 2nd dependent variable. In this section, weshall restrict our discussion to solutions of linear equations with constant coefficients only.

Example 38: Solve 2 , 2 .t tdydx y e x edt dt

−+ = − =

Solution: The given equation in symbolic form can be written asDx + 2y = et …(1)

Dy – 2x = e–t …(2)Operate D on (2) and add to it 2 time of (1), we get

(D2 + 4)y = 2et – e–t … (3)Here A.E. is D2 + 4 = 0 i.e. D = ± 2i

∴ yC.F.(t) = (c1cos2t + c2sin2t) … (4)

And ( ). . 2 4 21 1 1( ) 2 2 2

4 4 4 5 5

t tt t t t

P Ie ey t e e e e

D D D

−− −= − = − = −

+ + + … (5)

Hence ( )1 22( ) cos2 sin5 5

tt ey t c t c t e

− = + + − …(6)


Now from (2), ( )1 21 1 2cos2 sin 22 2 5 5

tt t tex Dy e D c t c t D e e

−− = − = + + − −

1 21 22 sin 2 2 cos22 5 5

tt tec t c t e e

−− = − + + + −

1 22sin 2 cos 2

5 5

ttec t c t e−= − + + − … (7)

Example 39: Solve (D – 1)x + Dy = 2t + 1, (2D + 1)x + 2Dy = t

Solution: For elimination of y, take difference of 2 time of 1st from 2nd i.e. ((2D + 1)x + 2Dy) – 2((D – 1)x + Dy) = t – 2(2t + 1)

or 3x = – 3t – 2 or 2( )3

x t t= − − Implying 1.dxdt

= −

Now using above values of x(t) and dxdt in Ist equation, we get

Dx – x + Dy = 2t + 1 or 21 2 13

t Dy t − − − − + = +

Implying 24 4i.e. ( )

3 2 3tDy t y t t c= + = + +

where c is a constant of integration.

Example 40: Solve 7 0, 2 5 0dydx x y x y

dt dt− + = − − =

Solution: The given equations in symbolic form are written as: (D – 7)x + y = 0 …(1)

–2x + (D – 5)y = 0 …(2)To eliminate y, operate (D–5) on (1) and add the two equations to get

(D – 5)(D – 7)x + 2x = 0 or (D2 – 12D + 37)x = 0 … (3)So that A.E. is D2 – 12D + 37 = 0 or D = 6 ± i

∴ xC.F.(t) = e6t(c1cost + c2sin t) … (4)

Implying ( ) ( )6 61 2 1 2sin cos 6 cos sint tdx e c t c t e c t c t

dt= − + + + … (5)

On substituting values of x(t) and Dx from (4) and (5) respectively in equation (1), we get e6t(–c1sin t + c2cos t) + 6e6t(c1cos t + c2sin t) – 7e6t(c1cos t + c2sin t) + y = 0

or y = e6t[(c1 – c2)cos t + (c1 + c2)sin t]

= e6t[Ccost + Dsin t] … (6)

where, C = c1 – c2 and D = c1 + c2


Example 41: Solve the simultaneous differential equations

0, 0dydxt y t x

dt dt+ = + = with conditions: x(1) = 1 and y(–1) = 0

Solution: 0dxt ydt

+ = … (1)

0dy

t xdt

+ = … (2)

Differentiating (1) with recept to ‘t’,

2 2

22 2

0 or 0dy dyd x dx d x dxt t t t

dt dt dt dt dt dt+ + = + + = … (3)

Substituting the value of dy

tdt

from (2), equation (3) reduces to

2

22

0d x dxt t xdt dt

+ − = …(4)

Which is Cauchy’s homogeneous linear differential equation in t.

Put t = ep or log t = p so that 1dpdx dx dx

dt dp dt t dp= = or dx dxt Dx

dt dp= =

Likewise ( )2

22

1d xt D D xdt

= − and so on

Therefore (4) reduces to, D(D – 1)x + Dx – x = 0 or (D2 – 1)x = 0 …(5)

Whence . . 1 2 1 21( ) or ( )p p

C Fx p c e c e x t c t ct

− = + = + … (6)

Implying 21 2

cdx cdt t

= − … (7)

From (1), we get 2 21 12

c cdxy t t c c tdt t t

= − = − − = − + … (8)

On using the conditions, x (1) = 1 and y (–1) = 0 in (6) & (7), we get 1 212

c c= =

Example 42: To small Oscillations of a certain system with two degrees of freedom aregiven by the equations

2

2 23 2 0

, where3 5 0

D x x y dDD x D y x y dt

+ − = =+ − + = If x = 0, y = 0; Dx = 0, Dy = 2 when t = 0, find x and y when t = 1/2.

Solution: Re–write the given set of simultaneous equations as (D2 + 3)x – 2y = 0 …(1)


(D2 – 3)x + (D2 + 5)y = 0 …(2)In order to make above equations separately in x(t) and y(t) i.e. to eliminate x, first operate

these equations by (D2 – 3) and (D2 + 3) respectively and then subtract (1) from (2), we get [(D2 + 3)(D2 + 5) + 2(D2 – 3)]y = 0

On simplification, (D4 + 10D2 + 9)y = 0 … (3)Corresponding auxiliary equation becomes

D4 + 10D2 + 9 = 0 or (D2 + 1)(D2 + 9) = 0 i.e. D = ± i, ± 3iThus, y(C.F.) = (a1cos t + a2sin t) + (a3cos3t + a4sin3t) … (4)

To find x, eliminate y from (1) and (2).Operate (1) by (D2 + 5) and multiply (2) by 2 and then add the two,

(D4 + 10D2 + 9)x = 0 (an equation identical to (3)) i.e. D = ± i, ± 3i ...(5)

Thus, x(C.F.) = (b1cos t + b2sint) + (b3cos3t + b4sin3t) …(6)

To find the relation between constants involved in (4) and (6)Substitute values of x and y in either of the given equations, say in (1), we get

2(b1 – a1) cost + 2(b2 – a2) sin t – 2(3b3 + a3) cos3t – 2(3b4 + a4)sin3t = 0 …(7)Which must holds for all t.On equating co–efficient of cost, sin t, cos3t, sin3t, we get

b1 = a1, b2 = a2,3

3 ,3a

b = − 44 3

ab = − …(8)

Thus on replacing bi‘s by respective ai´s, equation (6) becomes

( ) ( )1 2 3 41cos sin cos3 sin 33

x a t a t a t a t= + − + …(9)

Clearly, relations (4) and (9) are the solutions of the set of given equation.In order to find out these constants, make use of the initial conditions.Using x = y = 0 when t = 0, equation (4) and (9) gives,

1 3

1 31 3

0implying 0.1 0

3

a aa aa a

+ = = =− =

With above values of a1 and a3, equation (9) and (4) reduces to

4

2

2 4

sin sin 33

sin sin 3

ax a t t

y a t a t

= −

= + and }2 4

2 4

cos cos3cos 3 cos3

Dx a t a tDy a t a t

= −= + …(10)

With the conditions Dx = 3 and Dy = 2 when t = 0, equations (10) give

}2 4

2 4

32 3

a aa a

= −= + implying 2 4

11 1,4 4

a a= = − .

Hence from equation (10),


( )

1 111sin sin 34 31 11sin sin 34

x t t

y t t

= + = −

Further when 1 ,2

t = ( ) ( )1 111sin 0.5 sin 1.5 1.40154 3

x = + =

( ) ( )1 11sin 0.5 sin 1.5 1.0694

y = − =

ASSIGNMENT 5

Solve the following simultaneous equations:

(i) 5 2 ,

2 0;

dx x y tdtdy

x ydt

+ − =

+ + = [KUK, 2005]

(ii)

2

2 3 0,

3 2 2 ;t

dx x ydt

dyx y e

dt

+ + =

+ + =[Delhi, 2002]

(iii) ( ) ( )( ) ( )

1 2 1 ,1 1 1;

tD x D y eD x D y

+ + + =− + + =

(iv) 0,

0;

dxt ydtdy

t xdt

+ =

+ =given ( )

( )1 1

1 0xy

= − =

(v)22

2 2sin , cos ;

d yd x y t x tdt dt

+ = + = [NIT Jalandhar, 2006]

(vi) 2 , 2 , 2dydx dzy z x

dt dt dt= = =

(vii) A mechanical system with two degrees of freedom satisfies the equation

2 2

2 22 3 4, 2 3 0

d y dy d y dxdt dt dt dt

+ = − =

Obtain expression for x and y in terms of t, given , ,dy dy

xdx dt

all vanish at t = 0.

ANSWERS

Assignment 1

(i) y = (c1 + c2x)e3x, (ii) 21 2

3 3cos sin2 2

xy e c x c x− = +

(iii) 21 1 2

3 3cos sin2 2

xxy c e e c x c x− = + +

[UP Tech, 2008]


(iv) y = (c1 + c2x) e2x + (c3 + c4x) e–2x

(v) ( ) }11 2

2

4cos3 sin 3 , ,3

x cy e c x c x c

== + = − (vi) y = c1 + (c2 + c3x)e–x/2

(vii) y = (c1 + c2x) cosnx + (c3 + c4x)sinnx, (viii) cos /gθ = α l

Assignment 2

1. ( ) ( )1 2 2cos sin cos sin

4xy c ax c ax ax ax axa

= + + +

2. ( ) ( )21 2

1 1 cos sin2

x xc e c e x x x x−+ − − +

3. (c1cosx + c2sinx) + [sin log(sinx) – xcosx]

4. ( ) ( ) ( )1 2 3 4cos sin cos 3 sin8

x x xy c x c x c e c e x x x x−= + + + + −

5. ( ) ( )1 2 21cos sin cos log sec tany c ax c ax ax ax axa

= + − +

6. ( ) ( )1 21 1cos2 sin 2 4sin 2 cos2

2 5 17

xey c x c x x x = + + − +

7. ( )31 2

1 42cos sin5 3 3

xx x ey c e c e x x x

−− − = + − + + −

8. ( ) ( ) ( )21 2

1 sin cos 32 4

xx x xey c e c e x x x x x−= + − + + − +

9. ( )2 21 2

2sinh cosh3 7

x x xy c e c e x x−= + − −

10. 2 21 2 3 4cos sin cos sin

2 2 2 2

x xx x x xy e c c e c c− = + + +

2 2 cos 2 sin2

x x x x−+ −

11. ( ) ( )1 2 3 41cos sin cos cosh5

x xy c e c e c x c x x x−= + + + −

12. ( ) ( ) ( )31 2

1 110cos 5 11sin 5 sin 2cos884 20

x xy c e c e x x x x= + + − + +

13. y = (c1 + c2x)ex – ex(xsinx + cosx)

14. ( ) ( )2

1 2 22 2 2

4 2 8cos sin .1 1 1

xe xy c nx c nx xn n n

−= + + + + + + +


Assignment 3

1. (i) y = (c1cosx + c2sinx) – xcosx + sinx (logsinx)(ii) y = (c1cosx + c2sinx)ex – xexcosx + exsin (logsinx)(iii) y = (c1 + c2x)e2x – e2x[4xcos2x + (2x2 – 3)sin x](iv) y = ex(c1cosx + c2sin x) – ex(cosx) log(secx + tan x)

(v) ( ) ( )21 2 3 6xx x e x xy c e c e e e e− −= + + − +

(vi) y = (c1 + c2x)e3x – e3xlogx

2. (i) ( ) ( ) ( )2 22 3 2

1 2 log log2xc x c x x x x+ − − (ii) 2

1

xxc ec x e

x x + + −

3. ( )2

21 2 2

x xy c x e c x= + −

4. (i) y = (c1ex + c2e

2x) + 5; (ii) y = (c1cosx + c2sinx) + x;(iii) y = (c1 + c2x)ex – 2x2λex;

(iv) ( ) 21 2

1 2 3 4sin cos9 27 25 25

x xc c x e x e x x− + + − + −

5. (i) ( )2 21 2

1 73 22 2

x x xy c e c e x x xe = + + + + −

(ii)1 3 cossin 3 cos3

10 5 2xy x x x = − + −

6. ( )12

cos 2sin cos .2 5

xxxc ee xy x x c e

−= − − + +

Assignment 4

1. (i) y = c1x2 + c2x3 – x2logx; (ii)

424

1 log5

c xy c x xx

= + +

(iii) 22 21

1 1 log ;3

cy c x x x

x x = + + − (iv) y = 2(logx)3 + c1logx + c2.

(v) ( ) ( )12 3

logcos log sin log 5 10 ;

xcy x c x c x x

x x = + + + +

(vi) ( )2

1 42

1 3log ;6 2 8

c xy c x xx

= + − − +

(vii) ( )2 2

8kru a r= −

(viii) ( ) ( ) ( ) ( )( )1 2cos log(1 ) sin log(1 ) log 1 cos log 1y c x c x x x= + + + − + +


(ix) ( ) ( ) ( )( ){ }221 21 2 log 1 2 log 1 2y x x c x c= + + + + +

(x) ( ) ( )( )2 21 2 log 1 log 1 8y c c x x x x = + + + + + +

2. 3 23 6 6 0x y x y xy y− + − =′′′ ′′ ′

3. 3 ' 0x y xy y+ − =′′′

Assignment 5

(i) ( ) ( ) ( ) ( )3 31 1 2 21 6 1 3 , 2 3 2 327 27 27 27

t tx t e t y t e t− −= − + + + = − + + −

(ii) , sint t t tx e e y e e t− −= + = − +

(iii) 2 21 2 1 22 , 3 2 3t t t t t tx c e c e e y c e c e e− − − −= + + = + +

(iv)1 1 1 1,2 2

x t y tt t

= + = − +

(v) ( ) ( )1 2 3 4cos sin cos sin ,4 4

t t t tx c e c e c t c t t t−= + + + − +

( ) ( ) ( )( )1 2 3 41cos sin 2 sin cos4

t ty c e c e c t c t t t t−= − − + + + + −

(vi) ( )21 2 3cos 3 ,t tx c e c e t c−= + − 2

1 2 32cos( 3 )3

t ty c e c e t c− π= + − +

21 2 3

4cos( 3 )3

t tz c e c e t c− π= + − +

(vii)8 3 4 8 31 cos , sin9 2 3 9 2

x t y t t = − = −

linear differential equations of second higher … linear differential...linear differential...

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