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SYLLABUS
ENGINEERING MATHEMATICS – II : NAS 203
UNIT - I : DIFFERENTIAL EQUATION Linear differential equations of nth order with constant coefficients, Complementary function and Particular integral, Simultaneous linear differential equations, Solution of second order differential equations by changing dependent & independent variables, Normal form, Method of variation of parameters, Applications to engineering problems (without derivation).
UNIT – II : SERIES SOLUTION AND SPECIAL FUNCTIONS Series solution of second order ordinary differential equations with variable coefficient (Frobenius method), Bessel and Legendre equations and their series solutions, Properties of Bessel function and Legendre polynomials.
UNIT – III : LAPLACE TRANSFORM Laplace transform, Existence theorem, Laplace transforms of derivatives and integrals, Initial and final value theorems, Unit step function, Dirac- delta function, Laplace transform of periodic function, Inverse Laplace transform, Convolution theorem, Application to solve simple linear and simultaneous differential equations.
UNIT – IV : FOURIER SERIES AND PARTIAL DIFFERENTIAL EQUATIONS Periodic functions, Fourier series of period 2 , Euler’s Formulae, Functions having arbitrary periods, Change of interval, Even and odd functions, Half range sine and cosine series, Harmonic analysis. Solution of first order partial differential equations by Lagrange’s method, Solution of second order linear partial differential equations with constant coefficients.
UNIT – V : APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS Classification of second order partial differential equations, Method of separation of variables for solving partial differential equations, Solution of one and two dimensional wave and heat conduction equations, Laplace equation in two dimension, Equation of transmission lines.
TEST BOOKS 1. E. Kreyszig, : Advanced Enginnering Mathematics,Volume-II, John Wiley&Sons 2. B. V. Ramana, Higher Engineering Mathematics, Tata Mc Graw- Hill Publishing Company Ltd. 3. R.K.Jain & S.R.K. Iyenger, Advance Engineering Mathematics, Narosa Publishing House.
REFERENCE BOOKS 1. Chandrika Prasad, Advanced Mathematics for Engineers, Prasad Mudranalaya 2. Peter V. O’ Neil, Advanced Engineering Mathematics, Thomas (Cengage) Learning. 3. A. C. Srivastava & P. K. Srivastava, Engineering Mathematics, Vol. – II, PHI Learning Pvt. Ltd.
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UNIT 1 : DIFFERENTIAL EQUATIONS
CONTENTS 1.01 Differential Equation ....................................................................................................................................2
order and degree of a differential equation.........................................................................................................2
1.02 Linear Differential Equations ........................................................................................................................2
1.03 Linear Differential Equations Of nth Order ....................................................................................................3
linear differential equation...................................................................................................................................3
1.04 Linear Differential Equations of Second Order with Constant Coefficients .................................................4
LINEAR INDEPENDENCE AND DEPENDENCE .........................................................................................................4
METHOD FOR FINDING THE COMPLEMENTARY FUNCTION ................................................................................4
TO FIND THE VALUE OF 1
sinnx ax.f(D)
.......................................................................................................... 11
GENERAL METHOD OF FINDING THE PARTICULAR INTEGRAL OF ANY FUNCTION (x) ................................ 12
1.05 Cauchy Eular Homogeneous Linear Differential Equations ....................................................................... 12
1.06 Legendre’s Homogeneous differential Equations ..................................................................................... 14
1.07 Simultaneous Linear Differential Equations .............................................................................................. 15
simultaneous differential equations ................................................................................................................. 15
METHOD OF REDUCTION .................................................................................................................................. 19
RULE TO FIND OUT PART OF THE COMPLEMENTARY FUNCTION ..................................................................... 23
1.08 Method of reduction of order ................................................................................................................... 24
Method 1. To Find the Complete Solution of RQyPyy '" when part of Complementary Function
is known (Method of reduction of order) ......................................................................................................... 24
Certain rules to find one part of CF(Remember !) ............................................................................................ 26
1.09 Reduced to Normal Form (Removal of first derivative) ............................................................................ 27
Method 2. To Find the Complete Solution of RQyPyy '" when it is Reduced to Normal Form
(Removal of first derivative) .............................................................................................................................. 27
1.10 Changing the Independent Variable .......................................................................................................... 29
Method 3. To Find the Complete Solution of " 'y Py Qy R Changing the Independent Variable ....... 29
1.11 Method of Variation of Parameters .......................................................................................................... 31
Method 4 : To Find the Complete Solution of " 'y Py Qy R by the Method of Variation of Parameters
........................................................................................................................................................................... 31
1.12 Applications of differential equations to engineering problems .............................................................. 35
Application to Electric Circuits ........................................................................................................................... 37
Assignment-I .......................................................................................................................................................... 38
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UNIT 1 : DIFFERENTIAL EQUATIONS
1.01 DIFFERENTIAL EQUATION
An equation which involves differential co-efficient is called a differential equation.
For Example
1. 2
2
1
1
y
x
dx
dy
2. 082
2
2
ydx
dy
dx
yd 3.
2
22/3
2
1dx
ydk
dx
dy
ORDER AND DEGREE OF A DIFFERENTIAL EQUATION
The order of a differential equation is the order of the highest differential co-efficient present in the equation. Consider
1. .sin2
2
wtEc
q
dt
dqR
dt
qdL 2. .tan8sincos
2
2
2
xydx
dyx
dx
ydx
3.
2
2
23
2
1
dx
yd
dx
dy
The order of the above equation is 2.
The degree of a differential equation is the degree of the highest derivative after removing the radical sign and fraction.
The degree of the equation (1) and (2) is 1. The degree of the equation (3) is 2.
Q.1 The order and degree of the differential equation 06
8
2
4
3
3
dx
dyx
dx
yd are …..and …..
Sol. The highest order of the derivative in the given differential equation is 3. So, the order is 3.
The power of the highest order derivative is 4. So the degree is 4.
1.02 LINEAR DIFFERENTIAL EQUATIONS
A differential equation of the form
Qpydx
dy
(1)
is called a linear differential equation, where P and Q are functions of x (but not of y ) or
constants.
In such case, multiply both sides of (1) by Pdx
e
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.
Pdx
Pdxdye Py Q e
dx
(2)
The left hand side of (2) is
Pdx
eydx
d. , but
PdxPdx
eQeydx
d
..
Integrating both sides, we get
CdxeQeyPdx
Pdx
..
This is the required solution.
Note : Pdx
e is called the integrating factor.
Solution is CdxFIQFIy .].[.].[
Q.2 Solve : 2( 1) ( 1)xdyx y e x
dx
Sol. We have, )1(1
xex
y
dx
dy x
Integrating factor 1log( 1) log( 1)1
1
1
dx
x xxIF e e ex
The required solution is
1 1. .( 1).
1 1
x xy e x dx c e dx cx x
( 1)( )xy x e c
1.03 LINEAR DIFFERENTIAL EQUATIONS OF NTH
ORDER
LINEAR DIFFERENTIAL EQUATION
If the degree of the dependent variable and all derivatives is one, such differential equations are called linear differential equation e.g.
22
25 6 1
d y dyy x x
dx dx ,
222
22 xd y dy
y edx dx
, )1(
2
2
fxdt
dx
dt
xd
The order of a differential equation is the highest order of the derivative involved. All the above differential equations are of second order.
Fourier and Laplace transforms are mathematical tools to solve the differential equations.
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1.04 LINEAR DIFFERENTIAL EQUATIONS OF SECOND ORDER WITH CONSTANT COEFFICIENTS
The general form of the linear differential equation of second order is
RQydx
dyP
dx
yd
2
2
Where P and Q are constants are R is a function of x or constant.
Differential Operator . Symbol D stands for the operation of differential i.e.,
22
2,
dy d yDy D y
dx dx
D
1 stands for the operation of integration.
2
1
D stands for the operation of integration twice.
RQydx
dyP
dx
yd
2
2
can be written in the operator form.
2 2( )D y P Dy Qy R D PD Q y R
LINEAR INDEPENDENCE AND DEPENDENCE
Two solutions )(1 xy and )(2 xy are said to be linearly independent if
0)()( 21 xByxAy
A and B are not equal to zero.
COMPLETE SOLUTION = COMPLEMENTARY FUNCTION + PARTICULAR INTEGRAL
METHOD FOR FINDING THE COMPLEMENTARY FUNCTION
(1) In finding the complementary function, R.H.S. of the given equation is replaced by zero.
(2) Let mxeCy 1 be the C.F. of
2
20
d y dyP Qy
dx dx (1)
Putting the values of dx
dyy, and
2
2
dx
yd in (1)
0)( 2
1 QPmmeC mx
.02 QPmm [This equation is called Auxiliary Equation.]
(3) Solve the auxiliary equation :
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Case I : Roots, Real and Different. If 1m and 2m are the roots, then the C.F. is
xmxmeCeCy 21
21
Case II : Roots, Real and Equal. If both the roots are 1m , 1m then the C.F. is
xmexCCy 1)( 21
Q.3 Solve : 2
26 9 0
d y dyy
dx dx
Sol. Here, we have
0962
2
ydx
dy
dx
yd 0)96( 2 yDD
A.E. : 2 26 9 0 ( 3) 0 3, 3.m m m m
xexCCFC 3
21 )(.. .
Q.4 Find the general solution of the differential equation 5 3
5 30
d y d y
dx dx
Sol. 03
3
5
5
dx
yd
dx
yd
5 3 0D y D y
5 3( ) 0D D y
3 2( 1) 0D D y
A.E.: is 3 2( 1) 0m m
0, 0, 0, 1, 1m
Hence, the solution is
1 2 3 4 5( ) x xy C C x C x C e C e .
Case III : Roots, Imaginary. If the roots are , i then the solution will be
].[ 21
)(
2
)(
1
xixixxixi eCeCeeCeCy
)]sin(cos)sin(cos[ 21 xixCxixCe x
]sin)(cos)[( 2121 xCCixCCe x
]sincos[ xBxAe x
Q.5 Solve : 2
24 5 0
d y dyy
dx dx
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Sol. Here the auxiliary equation is
2 4 5 0m m 4 16 20
22
m i
The complementary function is
)sincos(2 xBxAey x
Rules to find particular integral
(i) xaxa eaf
eDf )(
1
)(
1
If 0)( af then xaxa eaf
xeDf
)(
1
)(
1
If 0)(' af then xaxa eaf
xeDf
)("
1
)(
1 2
(ii) 11[ ( )]
( )
n nx f D xf D
expand 1)]([ Df and then operate.
(iii) axaf
axDf
sin)(
1sin
)(
122
and axaf
axDf
cos)(
1cos
)(
122
If 0)( 2 af then axaf
xaxDf
sin)(
1sin
)(
122
and axaf
xaxDf
cos)('
1cos
)(
12
(iv) )()(
1)(
)(
1x
aDfexe
Df
axxa
(v)
dxxeexaD
axxa )()(1
Q.6 Solve the following differential equation:
2
26 sin 3 cos 2 .
d y dyy x x
dx dx
Sol. Here, we have
xxydx
dy
dx
yd2cos3sin6
2
2
xxyDD 2cos3sin)65( 2
A.E.: 0652 mm
0662 mmm
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0)6(1)6( mmm
0)6()1( mm
6,1 m
xx eCeCFC 6
21..
)2cos3(sin65
1..
2xx
DDIP
xDD
xDD
2cos65
13sin
65
122
xD
xD
2cos652
13sin
653
122
xD
xD
2cos105
13sin
155
1
xDD
Dx
DD
D2cos
)2)(2(5
)2(13sin
)3)(3(5
)3(
xD
Dx
D
D2cos
)4(5
23sin
)9(5
322
xD
xD
2cos)42(5
23sin
)93(5
322
xD
xD
2cos)8(5
23sin
)18(5
3
)2)(cos2(40
1)3(sin
90
)3(xDx
D
]2cos22sin2[40
1]3sin33cos3[
90
1xxxx
)2sin2(cos20
1)3sin3(cos
30
1xxxx
Complete solution : .... IPFCy
6
1 2
1 1(cos3 sin 3 ) (cos 2 sin 2 )
30 20
x xy C e C e x x x x
Q.7 Solve : 3 2
3 23 4 2 cosxd y d y dy
y e xdx dx dx
Sol. We have, xeyDDD x cos)243( 23
A.E. is 0243 23 mmm
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imeimmm 1,1.,.,0)22)(1( 2
xx exCxCeCFC )sincos(.. 321
xDDD
eDDD
IP x cos243
1
)22)(1(
1..
232
xDD
eD
x cos24)1(3)1(
1
)221)(1(
1
xD
eD
x cos13
1
)1(
1
xD
D
Dex cos
19
133.
11
12
19
)cossin3(1.
1
xx
Dex
)cossin3(10
1. xxxex
Hence, complete solution is
y )cossin3(10
1)sincos( 321 xxexexCxCeC xxx
Q.8 Solve : 2
2sin sin 2
d yy x x
dx
Sol. We have,
xxydx
yd2sinsin
2
2
xxyD 2sinsin)1( 2
A.E. is 012 m im
C.F. xCxC sincos 21
P.I. ]3cos[cos2
1
1
12sinsin
1
122
xxD
xxD
x
Dx
D3cos
1
1cos
1
1
2
122
xx
xxx
Dx 3cos
8
1sin
22
13cos
19
1cos
2
1
2
1
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]3cossin4[16
1xxx
Complete Solution is y C.F.+P.I.
)3cossin4(16
1sincos 21 xxxxCxCy
Q.9 Obtain the general solution of the differential equation
" 2 ' 2 cosxy y y x e x
Sol. We have, xexyyy x cos2'2"
A.E. : 0222 mm im 1
)sincos(.. xBxAeFC x
xeDD
xDD
IP x cos22
1
22
1..
22
where,
xD
D
xD
D
xDD
I
212
1
212
1
22
12221
xD
DxD
D
...2
12
1
21
2
1 21
2
12
1...
22
1 2
xx
DDxx
And
xeDD
I x cos22
122
xD
xexD
exDD
e xxx cos2
1.cos
1
1cos
2)1(2)1(
122
)(
)('
1)(
)(
1,0)(sin
2
12
2thenIf x
Dfxx
Dfafxxex
.... IPFCy
.sin2
1)1(
2
1)sincos( xxexxBxAe xx
Ans.
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Q.10 Solve : 2 3 2( 4 4) xD D y x e
Sol. We have, xexyDD 232 )44(
A.E. is 0442 mm 0)2( 2 m 2,2m
C.F. xexCC 2
21 )(
P.I. 3
2
223
2 4)2(4)2(
1.
44
1x
DDeex
DD
xx
20
.4
1.
1 52
423
2
2 xe
x
Dex
De xxx
The complete solution is 20
.)(5
22
21
xeexCCy xx Ans
Q.11 Solve 4( 1) cosxD y e x
Sol. Here, we have
xeyD x cos)1( 4
A.E. is 014 m 0)1)(1)(1( 2 mmm
iim ,,1,1
C.F. )sincos( 4321 xCxCeCeC xx
P.I. xeD
x cos1
14
xDDDD
exD
e xx cos646
1cos
1)1(
12344
xDD
ex cos6)1(4)1(6)1(
12
3
coscos
6461
1 xex
DDe
xx
Complete solution is .... IPFCy
3
cos)sincos( 4321
xexCxCeCeCy
xxx Ans.
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TO FIND THE VALUE OF 1
sinnx ax.f(D)
Now niaxiaxnn xiaDf
eexDf
axiaxxf(D) )(
1
)(
1)sin(cos
1
axxf(D)
n sin1
Imaginary part of iaxenx
iaDf
)(
1
axxf(D)
n cos1
Real part of niax xiaDf
e )(
1
Q.12 Solve the differential equation 2( 2 1) cosD D y x x
Sol. xxyDD cos)12( 2
Auxiliary equation is
0122 mm 0)1( 2 m 1,1 m
C.F. xexCC )( 21
P.I.
xxD
cos)1(
12
Real part of ]sin[cos)1(
12
xixxD
Real part of
ixexD 2)1(
1Real part of x
iDe xi
2)1(
1
Real part of
xiDiD
e xi
22 )1()1(2
1Real part of x
iDiDe xi
2)1(2
12
Real part of
x
i
DD
i
ii
e xi
2
11
1
22
Real part of xi
DD
i
i
i
e xi1
2
2
11
2
Real part of
xD
i
i
i
e xi
...1
12
Real part of
i
ixxix
i
1)sin(cos
2
1
Real part of )1)(sincos(2
1 ixxxi
Real part of xxxixxxi cos2
1)1(sin
2
1)1)(sincos(
2
1
Complete solution is .... IPFCy
xxxexCCy x cos2
1sin)1(
2
1)( 21
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Q.13 Solve 2
22 sin
d y dyy x x
dx dx
Sol. Auxiliary equation is 0122 mm
xexCCFC )(.. 21
P.I. xxDD
sin12
12
)sincos( xixeix
Imaginary part of
)sin(cos12
12
xixxDD
Imaginary part of xiexDD
12
12
Imaginary part of xiDiD
eix 1)(2)(
12
Imaginary part of xiDiD
eix 2)1(2
12
Imaginary part of xDi
Dii
eix
1
2
2
1)1(1
2
1
Imaginary part of xDii
xx ])1(1[2
)sin(cos
Imaginary part of ]1)[sincos(2
1ixxxi
P.I. xxxx sin2
1cos
2
1cos
2
1
Complete solution is )sincoscos(2
1)( 21 xxxxexCCy x
GENERAL METHOD OF FINDING THE PARTICULAR INTEGRAL OF ANY FUNCTION (x)
1 ax ax(x) e e (x)dx
D a
1.05 CAUCHY EULAR HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS
1
1
1 01... ( )
n nn n
n nn n
d y d ya x a x a y x
d x d x
where ...,,, 210 aaa are constants, is called a homogeneous equation.
Put ,zex ,log xz e Ddz
d
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Dydx
dyx
dz
dy
dx
dyx
dz
dy
xdx
dz
dz
dy
dx
dy
1.
Again, dx
dz
dz
yd
xdz
dy
xdz
dy
xdx
d
dx
dy
dx
d
dx
yd2
2
22
2 111
yDDxdz
dy
dz
yd
xxdz
yd
xdz
dy
x)(
11111 2
22
2
22
2
2
yDDdx
ydx )( 2
2
22
yDDdx
ydx )1(
2
22
Similarly. yDDDdx
ydx )2)(1(
3
33
Q.14 Solve : 2
2
24 cos(log ) sin(log ).
d y dyx x y x x x
dx dx
Sol. We have,
).sin(log)cos(log42
22 xxxy
dx
dyx
dx
ydx
(1)
Putting dz
dDxzex z ,log and Dy
dx
dyxyDD
dx
ydx ,)1(
2
22 in (1), we get
zezyDDD z sincos]4)1([
i.e. zezyDD z sincos)42( 2
A.E. is 0422 mm2
164)2( m
im 31
C.F. ]3sin3cos[ 21 zCzCe z (2)
P.I. )sin(cos42
12
zezDD
z
zeDD
zDD
z sin42
1cos
42
122
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zDD
ezD
z sin4)1(2)1(
1cos
421
12
zDDD
ezD
z sin42212
1cos
23
12
zezD
zD
ezD
D zz sin31
1cos
49
23sin
3
1cos
49
2322
3 2 1 1 1 1cos sin sin (3cos 2sin ) sin
13 2 2 13 2
z zDz e z z z z e z
(3)
Complete solution is
.... IPFCy
zezzzCzCey zz sin2
1)sin2cos3(
13
1]3sin3cos[ 21 (4)
Replacing xz log and xez in (4), we get
1 2[ cos 3(log ) sin 3(log )]y x C x C x 3 2 1
cos(log ) sin(log ) sin(log )13 13 2
x x x x
1.06 LEGENDRE’S HOMOGENEOUS DIFFERENTIAL EQUATIONS
A linear differential equation of the form
11
1 1( ) ( ) ...
n nn n
nn n
d y d ya bx a a bx a y X
dx dx
(1)
Where naaaba ...,,,, 21 are constants and X is a function of ,x is called Legendre’s linear equation.
Put
zexba )log( bxaz (2)
Dybdx
dybxa )( (3)
)1()( 2
2
22 DDb
dx
ydbxa (4)
yDDDbdx
ydbxa )2)(1()( 3
3
33 (5)
Q.15 Solve 2
2
2(2 1) 2(2 1) 12 6
d y dyx x y x
dx dx
Sol. Here, we have
22
2(2 1) 2(2 1) 12 6
d y dyx x y x
dx dx (1)
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Put zex 12 )12(log xz
Dydx
dyx 2)12( and yDD
dx
ydx )1(2)12( 2
2
22 in (1), we get
)1(
2
161222)1(4 zeyDyyDD
)1(312444 2 zeyDyDyyD
A.E. is 01284 2 mm 0322 mm
0)1)(3( mm 1,3 mm
zz eCeCFC 2
3
1..
)33(1284
1..
2
ze
DDIP
zz eDD
eDD
)0(
22 1284
133
1284
1
)1(1200
13
12)1(8)1(4
13
2
ze
4
1
16
3
ze
Complete Solution is .... IPFCy
4
1
16
32
3
1 zzz eeCeCy
4
1)12(
16
3)12()12( 1
2
3
1 xxCxCy .
1.07 SIMULTANEOUS LINEAR DIFFERENTIAL EQUATIONS
SIMULTANEOUS DIFFERENTIAL EQUATIONS
If two or more dependent variables are functions of a single independent variable, the equations involving their derivatives are called simultaneous equations, e.g.,
tydt
dx 4 , tex
dt
dy 2
The method of solving these equations is based on the process of elimination, as we solve algebraic simultaneous equations.
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Q.16 The equations of motions of a particle are given by
0dx
w ydt
, 0dx
w xdt
Find the path of the particle and show that it is a circle.
Sol. On putting Ddt
d in the equations, we have
0wyDx
0 Dywx
On multiplying (1) by w and (2) by D, we get
02 ywwDx
02 yDwDx
On adding (3) and (4), we obtain
022 yDyw 0)( 22 ywD
Now, we have to solve (5) to get the value of .y
A.E. is 022 wD 22 wD iwD
wtBwtAy sincos
wtBtADy cossin
On putting the value of Dy in (2), we get
0cossin wtBwwtAwwx
wtBwwtAwwx cossin
wtBwtAx cossin
On squaring (6) and (7) and adding, we get
)sin(cos)sin(cos 22222222 wtwtBwtwtAyx
2222 BAyx
This is the equation of circle.
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Q.17 Solve the following
3 8dx
x ydt
3dx
x ydt
With (0) 6x and (0) 2.y
Sol. Here we have
yxdt
dx83
yxdt
dx3
On putting Ddt
d in (1) and (2), we get
08)3(083 yxDyxDx
and 0)3(03 xyDyxDy
Multiplying (3) by )3( D and (4) by 8 adding them we get
0)1( 2 xD
A.E. is 012 m 12 m 1m
tt eCeCFC 21..
0.. IP
tt eCeCx 21
From (3) we get yeCeCD xx 8])[3( 21
tttt eCeCeCeCy 2121 338
tt eCeCy 21 428
)2(2
121
tt eCeCy
Initially when 0t then .2x
From (5); 0
2
0
12 eCeC 221 CC
Also when 0t , 2y .
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18
From (6), )2(4
12 0
2
0
1 eCeC
21 28 CC
Solving (7) and (8), we get
41 C and 62 C
Hence, the required solution is
tt eex 64
and )124(4
1 tt eey
Q.18 Solve : 2
2sin ,
d xy t
dt
2
2cos
d yx t
dt
Sol. Here, we have
2 sinD x y t (1)
tyDx cos2 (2)
Operating equation (1) by ,2D we get
tyDxD sin24
Subtracting (2) from (3), we get
ttxD cossin)1( 4
Auxiliary equation is
014 m 0)1)(1( 22 xm
im ,1,1
tctct
ect
ecFC sin4cos321..
)cos(sin14
1.. tt
DIP
)cos(sin34
1. tt
Dt
)sincos(4
ttt
1
D
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19
)cos(sin4
1sin
4cos
321tttctctectecx
From first equation (1), 2
2sin
dt
xdty
)cos(sin
4
1)sin
4cos
3()
21(
2
2sin tttctctectec
dt
dt
)cos(sin
4
1)sin(cos
4
1cos
4sin
321sin tttttctctectec
dt
dt
)sin(cos
4
1)sin(cos
4
1)cossin(
4
1sin
4cos
321sin tttttttctctectect
)cos(sin2
1)cos(sin
4
1sincos 4321 tttttctcececy tt
and )cos(sin4
1sincos 4321 tttctcececx tt .
METHOD OF REDUCTION
(WHOSE ONE SOLUTION OF COMPLEMENTARY FUNCTION IS KNOWN)
If uy is given solution be longing to the complementary function of the differential equation.
Let the other solution be .vy
Then vuy . is complete solution of the differential equation. Let
2
2
d y dyp Qy R
dx dx (1)
be the differential equation and u is the complementary function of (1)
02
2
Qudx
dup
dx
ud (2)
vuy . so that dx
dvu
dx
duv
dx
dy
2
2
2
2
2
2
2dx
ydu
dx
du
dx
dv
dx
udv
dx
yd
Substituting the values of 2
2
,,dx
yd
dx
dyy in (1), we get
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20
RvQudx
dvu
dx
duvP
dx
vdu
dx
du
dx
dv
dx
udv
.2
2
2
2
2
On arranging, we get
Rdx
dv
dx
du
dx
dvP
dx
vduQu
dx
duP
dx
udv
2
2
2
2
2
The first bracket is zero by virtue of relation (2), and the remaing is divided by .u
u
R
dx
dv
dx
du
udx
dvP
dx
vd
22
2
u
R
dx
dv
dx
du
uP
dx
vd
22
2
Let ,zdx
dv so that
dx
dz
dx
vd
2
2
Equation (3) becomes
u
Rz
dx
du
uP
dx
dz
2
This is the linear differential equation of first order and can be solved ( z can be found), which will contain one constant.
On integration ,dx
dvz we can get v .
Having found v , the solution is .uvy
Q.19 Solve 2" 4 ' (4 2) 0y xy x y given that 2xy e is an integral included in the
complementary function.
Sol. Here, we have 0)24('4" 2 yxxyy
On putting 2
. xevy in (1), the reduced equations as in the article 4.7.
02
2
2
dx
dv
dx
du
uP
dx
vd ]0,24,4[ 2 RxQxP
0)2(2
42
22
2
dx
dvxe
ex
dx
vd x
x
0]44[2
2
dx
dvxx
dx
vd 0
2
2
dx
vd 1c
dx
dv 21 cxcv
DR. U. P. SINGH REC SONBHADRA http://upsingh.in/Students
21
uvy ][2xeu
)( 21
2
cxcey x Ans.
Q.20 Solve 2 2 3" ( 2 ) ' ( 2) xx y x x y x y x e given that y x is a solution.
Sol. Here, we have xexyxyxxyx 3)2(')22("2
xexyx
xy
x
xxy
2
2'
2
22" (1)
On putting vxy in (1), the reduced equation as in the article 4.7
u
R
dx
dv
dx
du
uP
dx
vd
2
2
2
x
xex
dx
dv
xx
xx
dx
vd
)1(
2
2
22
2
2
xezdx
dzxedx
dv
dx
vd
2
2
dx
dvz
Which is a linear differential equation
xdx
eeFI
..
Its solution is cdxeeez xxx .
xxx cexezcxze .
xx ecxedx
dv .
1. ceceexv xxx
1)1( cecexv xx
xcecxxxvxy x
1
2 )( Ans.
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Q.21 Solve 2
2(2 1) ( 1) 0
d y dyx x x y
dx dx
Given that xey is an integral included in the complementary function.
Sol. Here, we have 0)1()12(2
2
yxdx
dyx
dx
ydx
0112
2
2
yx
x
dx
dy
x
x
dx
yd (1)
By putting xvey in (1), we get the reduced equation as in the article 4.7.
02
2
2
dx
dv
dx
du
uP
dx
vd (2)
Putting xeu and zdx
dv in (2), we get
0212
ze
ex
x
dx
dz x
x
0212
zx
xx
dx
dz 0
x
z
dx
dz
cxzx
dx
z
dzlogloglog
21111 log cxc
x
dxcvd
x
c
dx
dv
x
cz
)log(. 21 cxcevuy x .
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RULE TO FIND OUT PART OF THE COMPLEMENTARY FUNCTION
Rule Condition Part of Complementary Function u
1 01 QP xe
2 01 QP xe
3 01
2
a
Q
a
P
xae
4 0QxP x
5 022 2 QxPx 2x
6 0)1( 2 QxPnxnn nx
Q.22. Solve 2
2 3
22 [1 ] 2(1 )
d y dyx x x x y x
dx dx
Sol. 3
2
22 )1(2)1(2 xyx
dx
dyxx
dx
ydx
xx
yx
dx
dy
x
xx
dx
yd
222
2 )1(2)1(2
Here, 0)1(2)1(2
22
x
x
x
x
xxQxP
Hence xy is a solution of the C.F. and the other solution is .v
Putting xvy in (1), we get the reduced equation as in article 4.5
u
x
dx
dv
dx
du
uP
dx
vd
2
2
2
x
x
dx
dv
xx
xx
dx
vd
)1(
2)1(222
2
12122
2
zdx
dz
dx
dv
dx
vd
z
dx
dv
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Which is a linear differential equation of first order and xdx
eeFI 22
..
Its solution is
1
22 cdxeez xx
12
22 c
xexez
xecz 212
1
dxxecdvxecdx
dv
2
12
1212
1
2
2
2
1
2cxe
cxv
22
2
1
2cxe
cxxuvy Ans.
1.08 METHOD OF REDUCTION OF ORDER
METHOD 1. TO FIND THE COMPLETE SOLUTION OF RQyPyy '" WHEN PART OF
COMPLEMENTARY FUNCTION IS KNOWN (METHOD OF REDUCTION OF ORDER)
Let uy be a part of the complementary function of the given differential equation
2
2
d y dyP Qy R
dx dx (1)
Where u is a function of .x Then, we have
2
20
d u duP Qu
dx dx (2)
Let uvy be the complete solution of equation (1), where v is a function of x .
Differentiating y w.r.t. x ,
vdx
du
dx
dvu
dx
dy.
Again, 2
2
2
2
2
2
.2dx
udv
dx
du
dx
du
dx
vdu
dx
yd
Substituting the values of dx
dyy, and
2
2
dx
yd in equation (1) we get
RuvQdx
duv
dx
dvuP
dx
udv
dx
dv
dx
du
dx
vdu
)(2
2
2
2
2
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25
RvQudx
duP
dx
ud
dx
dvPu
dx
du
dx
vdu
2
2
2
2
2
Rdx
dvPu
dx
du
dx
vdu
2
2
2
, using (2)
u
R
dx
dvP
dx
du
udx
vd
22
2
(3)
Put Pdx
dv then,
dx
dp
dx
vd
2
2
Now (3) becomes, u
RPP
dx
du
udx
dp
2 (4)
Equation (4) is a linear differential equation of I order in p and x .
dxPdxPdu
udxp
dx
du
u eueeFI 2
22
..
Solution of (4) is given by
1
22 cdxeuu
Repu
dxPdxP
Where 1c is an arbitrary constant of integration.
12
1cdxeRue
up
dxPdxP
12
1cdxeRue
udx
dv dxPdxP
Integration yields, 212
1cdxcdxeRue
uv
dxPdxP
Where 2c is an arbitrary constant of integration.
Hence the complete solution of (1) is given by,
uvy
ucdxcdx
dxpeRu
dxpe
uuy
212
1
To find out the part of C.F. of the linear differential equation of II order given by
.2
2RQy
dx
dyP
dx
yd
DR. U. P. SINGH REC SONBHADRA http://upsingh.in/Students
26
CERTAIN RULES TO FIND ONE PART OF CF(REMEMBER !)
CONDITION PART OF C.F.
02
1 a
Q
a
P axe
01 QP xe
01 QP xe
02)1( QxmxPmm mx
0QxP x
0222 QxPx 2x
Q.23 Solve : 2
cot (1 cot ) sin .2
d y dy xx x y e xdxdx
Sol. Comparing with the standard form, we get
xxeRxQxP sin),cot1(,cot
0cotcot111 xxQP
A part of xeFC ..
Let xevy be the complete solution of given equation, then
dx
dvxexevdx
dy
2
22
2
2
dx
vdxedx
dvxexevdx
yd
Substituting for dx
dyy, and
2
2
dx
yd in given equation, we get
xdx
dvx
dx
vdsin)cot2(
2
2
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27
xpxdx
dpsin)cot2( …(1) where .
dx
dvp
This is a linear differential equation of I order in p and x .
x
eeFI
xdxx
sin..
2)cot2(
Solution of (1) is, 1
2
1
22
2sin.sin
sinc
ecdx
x
ex
x
ep
xxx
Where 1c is an arbitrary constant of integration.
xecxp x sinsin2
1 2
1
xecxdx
dv x sinsin2
1 2
1
Integrating, we get 2
2
1 )sin2(cos5
1cos
2
1cxxecxv x
Hence the complete solution is given by,
xxx ecxxecxevy
2
2
1 )sin2(cos5
1cos
2
1.
1.09 REDUCED TO NORMAL FORM (REMOVAL OF FIRST DERIVATIVE)
METHOD 2. TO FIND THE COMPLETE SOLUTION OF RQyPyy '" WHEN IT IS REDUCED TO NORMAL
FORM (REMOVAL OF FIRST DERIVATIVE)
When the part of C.F. can not be determined by the previous method, we reduce the given differential equation in normal form by eliminating the term in which there exists first derivative of the dependent variable.
2
2
d y dyP Qy R
dx dx (1)
Let vuy be the complete solution of eqn. (1), where u and v are the function of x .
dx
dvu
dx
duv
dx
dy
and 2
22
2
2
2
2
dx
vdu
dx
dv
dx
du
dx
udv
dx
yd
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Substituting the value of 2
2,,
dx
yd
dx
dyy in eqn. (1), we get
2 22 1
2 2
d v du du d u P du RP v Q
u dx dx u u dx udx dx
(2)
Let us choose u such that 02
Pdx
du
u (3)
Which on solving gives,
2
Pdx
u e
(4)
From (3), 2
Pu
dx
du
Differentiating, we get
)(
2
12
2
udx
dP
dx
duP
dx
ud
dx
dPuuP
dx
dPu
PuP
2222
1 2
Coefficient of QPu
u
P
dx
dPuuP
uQ
dx
du
u
P
dx
ud
uv
224
21
2
21
IP
dx
dPQ
4
2
2
1 (say)
Then (2) becomes, SIvdx
vd
2
2 (5)
This is known as the normal form of equation (1).
Solving (5), we get v in terms of x . Ultimately, uy is the complete solution.
Q.24 Solve : 22
24 (4 1) 3 sin 2 .2 xd y dy
x x y e xdxdx
Sol. Here, xeRxQxP x 2sin3,14,422
Let uy be the complete solution.
Now, 2)4(
2
1
xdxx
eeu
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29
.1)16(4
1)4(
2
114
42
1 222
xxP
dx
dPQI
Also, xe
xe
u
RS
x
x
2sin32sin3
2
2
Hence normal form is,
xdx
d2sin3
2
2
Auxiliary equation is
imm 012
xcxcFC sin2
cos1
..
where 1c and 2c are arbitrary constants of integration.
xxxD
IP 2sin2sin)14(
3)2sin3(
1
1..
2
Solution is, xxcxc 2sinsincos 21
Hence the complete solution of given differential equation is
)2sinsincos( 21
2xxcxceuy x .
1.10 CHANGING THE INDEPENDENT VARIABLE
Method 3. To Find the Complete Solution of " 'y Py Qy R Changing the Independent Variable
2
2
d y dyP Qy R
dx dx (1)
Let us relate x and z by the relation,
( )z f x (2)
dy dy dz
dx dz dx (3)
2
2
d y d dy d dy dz
dx dx dx dx dz dx
22 2 2
2 2 2
dy d z dz d dy dz dy d z dz d y
dz dx dx dz dz dx dz dx dz dx
(4)
Substituting in (1), we get
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30
RQydx
dz
dz
dyP
dx
yd
dx
dz
dx
zd
dz
dy
2
22
2
2
1112
2
RyQdz
dyP
dz
yd (5)
where 21212
2
2
1 ,,
dx
dz
RR
dx
dz
dx
dz
dx
dzP
dx
zd
P
Here 111 ,, RQP are functions of x which can be transformed into functions of z using the
relation )(xfz .
Choose z such that 1Q constant 2a (say)
2
2a
dx
dz
Q
Q
dx
dza
dxa
Qdz
Integration yields, dxa
Qz
If this value of z makes 1P as constant then equation (5) can be solved.
Q.25 Solve : )1(logcos4)1()1(2
22 xy
dx
dyx
dx
ydx .
Sol. )1log(cos)1(
4
)1(1
1222
2
xxx
y
dx
dy
xdx
yd
(1)
Choose z such that,
2
2
)1(
1
xdx
dz
xdx
dz
1
1 (2)
Integration yields, )1(log xz (3)
From(2), 22
2
1
1
xdx
zd
DR. U. P. SINGH REC SONBHADRA http://upsingh.in/Students
31
0
)1(
1
1
1.
1
1
1
1
2
2
1
x
xxxP
121
dx
dz
zx
dx
dz
RR cos4)1log(cos4
21
From (3)
Reduced equation is
zydz
ydcos4
2
2
Auxiliary equation is 012 m im
zczcFC sincos.. 21
zzzz
zD
IP sin2sin2
.4)cos4(1
1..
2
Complete solution is
zzzczcy sin2sincos 21
)1log(sin)1(log2)1log(sin)1(logcos 21 xxxcxcy .
1.11 METHOD OF VARIATION OF PARAMETERS
METHOD 4 : TO FIND THE COMPLETE SOLUTION OF " 'y Py Qy R BY THE METHOD OF VARIATION
OF PARAMETERS
RQydx
dyP
dx
yd
2
2
(1)
Let the complementary function of (1) be
BvAuy (2)
u and are part of C.F.
02
2
Qudx
duP
dx
ud (3)
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32
and 02
2
Qvdx
dvP
dx
vd (4)
Let the complete solution of (1) be
BvAuy (5)
where A and B are not constants but suitable functions of x to be so chosen that (5) satisfies (1). Now,
vBuABvAuy 11111
)( 11111 vBuABvAuy (6)
Let us choose A and B such that
011 vBuA (7)
Now (6) becomes, 111 BvAuy (8)
2112112 BvvBAuuAy (9)
Substituting the values of 21,, yyy from (5), (8) and (9) in (1) respectively, we get
RBvAuQBvAuPBvvBAuuA )()()( 11211211
RQvPvvBQuPuuAvBuA )()( 12221111
RvBuA 1111 (10)
|Using (3) and (4)
Solving (7) and (10) for 1A and 1B , we get
011 vBuA
01111 RvBuA
vuuvRu
B
Rv
A
11
11 1
)(11
1 xvuuv
RvA
| say (11)
)(11
1 xvuuv
RuB
| say (12)
Integrating (11), we get adxxA )( (13)
Where a is an arbitrary constant of integration.
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33
Integrating (12), we get bdxxB )( (14)
where b is also an arbitrary constant of integration.
Putting the above values in (5), we get
vbdxxuadxxy )()(
bvaudxxvdxxuy )()(
This gives the complete solution of (1).
Q.26 Solve by the method of variation of parameters :
2
2
2sec .
d ya y ax
dx
Sol. Here, ,cos axu axv sin are two parts of C.F.
Also, .sec axR
Let the complete solution be
axBaxAy sincos
where A and B are suitable functions of x .
To determine the values of A and B, we have
1
11
cdxvuuv
RvA
1
}sin)sin(cos.{cos
sin.seccdx
axaxaaxaax
axax
1
tancdx
a
ax
12coslog
1cax
aA
where 1c is an arbitrary constant of integration.
2
11
cdxvuuv
RuB
2}sin)sin(cos.{cos
cos.seccdx
axaxaaxaax
axax
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34
22
1c
a
xcdx
a
where 2c is an arbitrary constant of integration.
Hence the complete solution is given by
axBaxAy sincos
axca
xaxc
a
axsincos
coslog212
Q.27 Solve by method of variation of parameters:
2
22 logxd y dy
y e xdx dx
.
Sol. Parts of C.F. are xx xeveu , and xeR x log
Let xx BxeAey be the complete solution where A and B are some suitable functions of x . To
determine A and B, we have
121
11 )(
.logcdx
xexeee
xexecdx
vuuv
RvA
xxxx
xx
1
22
14
log2
log cx
xx
cdxxx
222
11
.logcdx
e
execdx
vuuv
RuB
x
xx
22 loglog cxxxcdxx
Hence the complete solution is
xx BxeAey
xx xecxxxecx
xx
)log(
4log
221
22
Q.28 Using variation of parameters method, solve :
2
2 3
22 12 log .
d y dyx x y x x
dx dx
Sol. Consider the equation
01222
22 y
dx
dyx
dx
ydx for finding parts of C.F.
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35
Put zex so that xx log and Let dz
dD then the given equation reduces to
0]122)1([ yDDD
0)12( 2 yDD
Auxiliary equation is
4,30122 mmm
4
2
3
1
4
2
3
1.. xcxcececFC zz
Hence, parts of C.F. are 3x and 4x
Let BvAuy be the complete solution, where A and B are some suitable functions of x . A and B
are determined as follows :
14253
4
1
11 )(3)4(.
.logcdx
xxxx
xxxcdx
vuuv
RvA
1
2
112
3
)(log14
1log
7
1
7
logcxcdx
x
xcdx
x
xx
and
22
3
2
11 7
.logcdx
x
xxxcdx
vuuv
RuB
2
77
2
6
7.
1
7.log
7
1log
7
1cdx
x
x
xxcdxxx
2
7
2
77
log7
1
4977
1
7
log
7
1cx
xc
xxx
Hence the complete solution is given by
4
2
73
1
243 log7
1
49)(log
14
1
xcx
xxcxBxAxy
1.12 APPLICATIONS OF DIFFERENTIAL EQUATIONS TO ENGINEERING PROBLEMS
Q.29 A particle of unit mass falls under gravity in a resisting medium whose resistance varies with velocity. Find the relation between distance and velocity if initially the particle starts from rest.
Sol. Let v be the velocity when the particle has fallen distance s in time t from rest so that the resistance due to velocity is vk per unit of mass. Hence the force of resistance on the particle of mass
m is mkv against the direction of motion. Also, the force due to gravity, mg is acting vertically
downwards.
DR. U. P. SINGH REC SONBHADRA http://upsingh.in/Students
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Equation of motion of the particle is
mkvmgdt
dvm
kvgdt
dv
dtkvg
dv
Integration gives,
1)log(1
ctkvgk
(1)
Since 0v when 0t
gk
c log1
1
tkvg
g
k
log
1
)1( ktek
g
dt
dsv (2)
Integrating (2), we get
22ce
k
gt
k
gs kt
Since 0s when 0t ,
2
2
k
gc
)1(2
ktek
gt
k
gs (3)
Eliminating t between (2) and (3), we get
k
v
kvg
g
k
gs
log
2
which gives a relation between distance and velocity.
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APPLICATION TO ELECTRIC CIRCUITS
If Q is the electrical charge on a condenser of capacity C and i is the current, then
(a) dt
dQi or dtiQ
(b) the potential drop across the resistance R is Ri.
(c) the potential drop across the inductance L is dt
diL .
(d) the potential drop across the capacitance C is C
Q.
Also, by Kirchhoff’s Law, the total potential drop (voltage drop) in the circuit is equal to the applied voltage (E.M.F.).
Q.30 An inductance L of 2.0 H and a resistance R of 20 are connected in series with an e.m.f. E volt. If the current I is zero when t=0, find the current i at the end of 0.01 second if E=100V, using the following differential equation.
diL Ri E
dt
Sol. The given equation is
L
Ei
L
R
dt
di (1)
LRtdtLR
eeFI //
..
Solution of (1) is
1
/// ceR
Edte
L
Eie LRtLRtLRt
(2)
at ,0t 0i From (2), R
Ec 1
So, from (2), )1( / LRteR
Ei
Putting the values of E, R, L and t , we get
4758.0]1[20
100 1.0 ei ampere.
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ASSIGNMENT-I
LINEAR DIFFERENTIAL EQUATION WITH CONSTANT COEFFIECIENTS
Solve the following differential equations:
1. D2y – 3 Dy + 2y = Cosh x + Cos3x, 2. (D2 + 4)y = ex + Sin2x, 3. (D2 + 4D + 3)y = e-3x. 4. D3y + 3D2y + 3Dy + y = e-x 5. (D3 – 2D – 4)y = x4 + 3x2 6. (D2-a2)y = eax – e-ax 7. (D+2)(D-1)2y = e-2x+2 sinhx
8. (D-1)2(D2+1)2y = sin2xe
2
x
9. (D2 – 4 D + 2)y = Sin2x, given that y = 1/8 and Dy = 4 when x = 0. also the find y when x = 4
10. (D2+4D+8)y=sin(2x+3) 11.(D2+5D-6)y=sin4x.sinx 12.(D3+8)y= x4+2x+1 11. (D-2)2y=8(e2x+sin2x+x2), 14.D2- 4y =xsinhx 15.(D2 – 4D + 1)y = e2x Sinx
HOMOGENOUS LINEAR DIFFERENTIAL EQUATION:
Solve the following differential equations:
1. 2 3 xx y xy y x e
2. 2 5 4 log .x y xy y x x
3. 3
4
3
d yx
dx
23 2
22 1
d y dyx x xy
dx dx
4. 3 2
3 2
3 2
12 2 10 .
d y d yx x y x
dx dx x
5. 2
2
2
(log )sin(log ) 13
d y dy x xx x y
dx dx x
6. 2
2
2(2 3) 2(2 3) 12 6
d y dyx x y x
dx dx
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SIMULTANEOUS LINEAR DIFFERENTIAL EQUATION:
Solve the following differential equations:
1. 2
( ) 1dx
x ydt t
, 1
( 5 )dy
x y tdt t
2. 2 , 2x t x dt tdy tx ty x t dt
3. 0t x y and 0,ty x given x(1) = 1, y(-1) = 0.
4. sin , y x cost.x y t
5. 2
2
24 5
d xx y t
dt ,
2
25 4 1
d yy y t
dt
6. , 55 , 5t tdx dyx y e x y e
dt dt
METHOD OF VARIAION OF PARAMETERS:
Solve the following differential equations:
1. .nxsecyny 2
2. .x2yy2y
3. .ey6yy x
4. .x2tany4y
5. .xe64y3y2y x
6. .x2secey5y2y x
7. .xlogeydx
dy2
dx
yd x
2
2
8. .e)x1(y2dx
dy)5x2(
dx
yd)2x( x
2
2
NORMAL FORM OR BY REMOVING THE FIRST DERIVATIVE:
1. .0y)2x2x(y)xx(2yx 222
2. .esecx 5yy xtan2y x
3. .ey)2x(yx2y)x2x(
2
1
22
4. xsecxtanydx
dy2xcoty
dx
yd2
2
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5. 0)xx8(x4
y
dx
dy
x
1
dx
yd22
2
BY CHANGING THE INDEPENDENT VARIABLE:
1. ).x1log(cos4yy)x1(y)x1( 2
2. .0xcosy y xtany 2
3. .0y4y)x1(x2y)x1( 222
4. 332
2
2
x2yx4dx
dy)1x4(
dx
yd
5. .xyx4yx
1y 42
SECOND ORDER DIFFERENTIAL EQUATION:
1. y2yxSin 2 , given that y = cot x is a solution of it.
2. 0xcosyyxcosxy)xcosxsinx( of which y = x is a solution.
3. x322 exy)2x(y)x2x(yx , given that y = x is a solution
4. 0y)2x4(yx4y 2 given that 2xey is a solution
Mathematics (NAS-203) PracticeTest Unit-I(Diff. equations.)
Time- 60 min M.M. 30
Note: Attempt any Five Questions. All questions carry equal marks
1. Solve the differential equation D2y – 3 Dy + 2y = Cosh x + Cos3x,
2. Solve the differential equation3 2
4 3 2
3 22 1
d y d y dyx x x xy
dx dx dx
3. Normal form or by Removing the first Derivative solve 2 2 22( ) ( 2 2) 0.x y x x y x x y
4. By Changing the Independent Variable solve 2(1 ) (1 ) 4coslog(1 ).x y x y y x
5. By method of Variaion of parameters solve 2 sec .y n y nx
6. Solve the differential equation 2
2
24 5
d xx y t
dt ,
2
25 4 1
d yy y t
dt
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