M II QuestionBank

I B.
Tech II semester
MATHEMATICS-II
FRESHMAN ENGINEERING
1|Page
INDEX PAGE
S.NO TOPIC PAGE NUMBER

UNIT-I UNIT-II
1 COURSE OBJECTIVES,OUTCOMES 3-4
2 SESSION PLANNER 5-8
3 LECTURE NOTES 9-51 67-85
4 TUTORIAL BANK 52 86
5 DESCRIPTIVE BANK 53 87
6 OBJECTIVE BANK 54-55 88
7 UNIT TEST PAPERS 56-59 89-92
8 SEMINAR TOPICS 60 93
9 ASSIGNMENT BANK 61 94
10 REAL TIME APPLICATIONS 62 95-96

11 NPTEL 63 97
12 BLOOM’S TAXONOMY 64-66 98-101
2|Page
Course Objectives: To learn
• Methods of solving the differential equations of first and higher order.

• Evaluation of multiple integrals and their applications
• The physical quantities involved in engineering field related to vector valued
functions
• The basic properties of vector valued functions and their applications to
line, surface and volume integrals
Course Outcomes: After learning the contents of this paper the student must be able to
• Identify whether the given differential equation of first order is exact or not
• Solve higher differential equation and apply the concept of differential
equation to real world problems
• Evaluate the multiple integrals and apply the concept to find areas,
volumes, centre of mass and Gravity for cubes, sphere and rectangular
parallelopiped
• Evaluate the line, surface and volume integrals and converting them from
one to another
3|Page
MA201BS: MATHEMATICS – II
UNIT-I: First Order ODE

Exact, linear and Bernoulli’s equations; Applications : Newton’s law of cooling, Law of
natural growth and decay; Equations not of first degree: equations solvable for p,
equations solvable for y, equations solvable for x and Clairaut’s type.
UNIT-II: Ordinary Differential Equations of Higher Order
Second order linear differential equations with constant coefficients: Non-
Homogeneous terms of the type eas, sin ax , cos ax, polynomials in x, easV(x) and x
V(x); method of variation of parameters; Equations reducible to linear ODE with
constant coefficients: Legendre’s equation, Cauchy-Euler equation.
UNIT-III: Multivariable Calculus (Integration)

Evaluation of Double Integrals (Cartesian and polar coordinates); change of order of
integration (only Cartesian form); Evaluation of Triple Integrals: Change of variables
(Cartesian to polar) for double and (Cartesian to Spherical and Cylindrical polar
coordinates) for triple integrals.
Applications: Areas (by double integrals) and volumes (by double integrals and triple integrals), Centre of
mass and Gravity (constant and variable densities) by double and triple integrals (applications involving
cubes, sphere and rectangular parallelopiped).
UNIT-IV: Vector Differentiation

Vector point functions and scalar point functions. Gradient, Divergence and Curl.
Directional derivatives, Tangent plane and normal line. Vector Identities. Scalar potential
functions. Solenoidal and Irrotational vectors.
UNIT-V: Vector Integration

Line, Surface and Volume Integrals. Theorems of Green, Gauss and Stokes (without
proofs) and their applications.
TEXT BOOKS:
1. B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010
2. Erwin kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley
& Sons,2006
REFERENCES:
1. Paras Ram, Engineering Mathematics, 2nd Edition, CBS Publishes
2. S. L. Ross, Differential Equations, 3rd Ed., Wiley India, 1984.
4|Page
SESSION PLANER
Name of the faculty: Subject: MATHEMATICS-II
Designation: Asst. Professor Branch:
S.N UNIT CLASS TOPIC T/R DATE DATE Remarks

O PLANNED CONDUCTED
1. LH1 Overview of differential T1/R1
equations
2. LH 2 Overview of differential T1/R1
equations
3. LH3 Exact differential equations T1/R1
4. LH4 non-exact diff. equations T1/R1

8. LH8 Linear differential equations T1/R1
9. LH9 Linear differential equations T1/R1
10. LH10 Bernoulli’s differential equations T1/R1
11. LH11 Equations solvable for P T1/R1
12. LH12 Equations solvable for Y T1/R1
13. LH13 Equations solvable for X T1/R1
14. LH14 Equations solvable for X T1/R2
15. LH15 Clairaut’s form T1/R2
16. LH16 Newton’s Law of cooling T1/R2
17. LH17 Newton’s Law of cooling T1/R2
18. LH18 Law of natural growth and T1/R2
19. LH19 Decay

Law of natural growth and T1/R2
20. Decay
PPT T1/R2
LH20
21. LH21 Active Learning(Collaborative T1/R2
learning)
22. LH22 Test T1/R2
23.
23 LH23 Eax v(x)Differential
Linear -method equations T1/R2
with constant coefficients

24. LH24 Xekaxv(x) -method
method T1/R2
25. LH25 Sin(bx) or cos(bx) method T1/R2
26. LH26 Xk-method T1/R2
27. LH27 eax x v(x) -method T1/R2
28. LH28 Xk v(x) -method T1/R2
I
5|Page
29. LH29 Xk v(x) -method T1/R2
30. LH30 Inverse Operators method T1/R2
31. LH31 Method of variation of T1/R2
parameters
32. LH32 Method of variation of T1/R2
parameters
33 LH33 Cauchy-Euler equations T1/R2
34 LH34 Cauchy-Euler equations T1/R2
35 LH35 Legendre’s equations T1/R2
36 LH36 Eigen values and Eigen vectors

PPT T11/R22
37 II LH37 Active Learning(Stump your
partner)
T1/R2
38 LH38 Test T1/R2

39 LH39 Evaluation of double integral in T1/R2
Cartesian form
Cartesian form
Polar form
42 LH42 Change of variables T1/R2

45 LH45 Change of order of integration T1/R1
48 LH48 Evaluation of Triple integral T1/R1
49 LH49 Evaluation of Triple integral T1/R1
50 LH50 Change of variables in Triple T1/R1
integral
51 LH51 Change of variables in Triple T1/R1
integral
52 LH52 Papers distribution T1/R1
T1/R1
53
54
III LH53
LH54
Areas in Double integral
Areas in Double integral T1/R2
55 LH55 Volumes in double integral T1/R2
56 LH56 Volumes in Triple integral T1/R2
57 LH57 Centre of mass T1/R2
56
58 LH57
LH58 Centre of mass T1/R2
59 LH59 Centre of gravity T1/R2
6|Page
60 LH60 ppt T1/R2
61 LH61 Active Learning(Flipped Class T1/R2
room)
62 LH62 Test T1/R2
6
63 LH62
LH63 Introduction T11/R22
64 LH64 Problem on Gradient T1/R2
65 LH65 Problem on Directional T1/R2
Derivative
66 LH66 Problem on Directional T1/R2
Derivative
67 LH67 Problems on Divergence of T1/R2
vectors
68 LH68 Problems on Solenoidal vectors T1/R2
69 LH69 Problems on Irrotational vectors T1/R2
70 LH70 Problems T1/R2
71 LH71 Vector operators T1/R2
72 LH72 Vector operators T1/R2
73 LH73 Vector Identities T1/R2

74 LH74 Vector Identities T1/R2
75 LH75 PPT T1/R2

76 LH76 Active Learning(TAPPS) T1/R2
77 LH77 Test T1/R2

78 LH78 Line integral T1/R2
79 LH79 Line integral T1/R2
80 IV LH80 Line integral T1/R2
81 LH81 Surface integral T1/R2
82
83
84
v LH82
LH83
LH84
Volume integral
Green’s theorem
Green’s theorem
T1/R2
T1/R2
T1/R2
85 LH85 Green’s theorem T1/R2

86 LH86 Gauss divergence theorem T1/R2
89 LH89 Stoke’s theorem T1/R2
92 LH92 PPT T1/R2
93 LH93 Active learning(Muddiest point) T1/R2
7|Page
94 LH94 Test
95 LH95 Revision
96 LH96 Revision
97 LH97 Revision
98 LH98 Revision
Course: I- B.Tech II SEM
TEXT BOOKS:
1. Higher Engineering Mathematics by B.S. Grewal, Khanna Publishers.
2. Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley& Sons,
REFERENCES: 1.Paras Ram, Engineering Mathematics, CBS publishes
FACULTY H.O.D PRINCIPAL/DIRECTOR
8|Page
UNIT-I
DIFFERENTIAL
EQUATIONS OF FIRST ORDER AND
THEIR APPLICATIONS
9|Page
ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER
& FIRST DEGREE

Definition: An equation which involves differentials is called a Differential equation.
Ordinary differential equation: An equation is said to be ordinary if the derivatives have reference to
only one independent variable.
dy d2y dy
Ex . (1) + 7xy = x 2 (2) 2
+ 3 + 2y = e x
dx dx dx
Partial Differential equation: A Differential equation is said to be partial if the derivatives in the
equation have reference to two or more independent variables.
2
 z   z 
2
E.g: 1.   +   = 4z
 x   y 
z z
2. x + y = 2z
x y
Order of a Differential equation: A Differential equation is said to be of order ‘n’ if the derivative
is the highest derivative in that equation.
E.g : (1). (x2+1) . + 2xy = 4x2
Order of this Differential equation is 1.
d2y
− (2 x − 1) + (x − 1) y = e x
dy
(2) x 2
dx dx
Order of this Differential equation is 2.
2
 dy 
(3). + 5   +2y =0 .
 dx 
10 | P a g e
Order=2 , degree=1.
(4). + =0. Order is 2.
Degree of a Differential equation: Degree of a differential Equation is the highest degree of the
highest derivative in the equation, after the equation is made free from radicals and fractions in its
derivations.
E.g : 1 ) y = x . + on solving we get
(1- ) 2xy . = 0 . Degree = 2
2) a. = on solving . we get
. Degree = 2
Formation of Differential Equation :
In general an O.D Equation is Obtained by eliminating the arbitrary constants c1,c2,c3--------cn from a
relation like (x, y, c1 , c2 ,......cn ) = 0. ------(1).
Where c1,c2,c3,---------cn are arbitrary constants.
Differentiating (1) successively w.r.t x, n- times and eliminating the n-arbitrary

constants c1,c 2,----cn from the above (n+1) equations, we obtain the differential equation F(x , y,
=0.
11 | P a g e
PROBLEMS
1.Obtain the Differential Equationy= A +B by Eliminating the arbitrary Constants:
Sol. y= A +B --------------------(1).
+ B(5) ----------(2).
= A (4) . + B(25) ----------(3).
Eliminating A and B from (1), (2) & (3).
e −2 x e5 x −y
 (− 2 )e −2 x 5e 5 x − y1 = 0
(4).e −2 x 25e 5 x − y2
1 1 y
 (− 2) 5 y1 = 0
4 25 y2
 -10y =0.
The required D. Equation obtained by eliminating A & B is
y2- 3y1 -10y = 0
2) Log  y  = cx
 x
Sol: Log  y  = cx -------------(1).

 x
=> log y –log x= cx
=> =c ---------------(2).
(2) in (1) => Log  y  =x[ ].

 x
3) = c.
Sol: Given equation ) =c
12 | P a g e
+ =0
dy − 1 − y
2
 =
dx 1 − x2
4) y = [Acosx +B sinx]
Sol: Given equation is y = [Acosx +B sinx]
= [Acosx +B sinx] + [-Asinx +B cosx]
=y+ -Asinx +B cosx).
d 2 y dy
 2
= + e x (− A sin x + B cos x ) + e x (− A cos x − B sin x )
dx dx
dy dy
+ − y− y
dx dx
d2y dy
= 2
− 2 + 2 y = 0 is required equation
dx dx
5) y= a + b.
Sol: =
=> ( ). + 2x. =0
=> ( ). + 2x. =0 is the required equation.
6) y=a +b
Sol: -2y =0
13 | P a g e
7) Find the differential equation of all the circle of radius
Sol. The equation of circles of radius a is where (h ,k) are the
co-ordinates of the centre of circle and h,k are arbitrary constants.
Sol: =
8) Find the differential equation of the family of circle passing through the origin and having their
centre on x-axis.
Ans: Let the general equation of the circle is x2+y2+2gx+2fy+c=0 .
Since the circle passes through origin, so c=0 also the centre (-g,-f) lies on x-axis. So the y-
coordinate of the centre i.e, f=0. Hence the system of circle passing through the origin and
having their centres on x-axis is x2+y2+2gx=0.
Ans. 2xy . + =0.
9) =c .
Ans: x . +y+ 4. =0
10) y=
Sol: ( + 2xy -1=0.
11) r=a(1+cos
Sol: r=a(1+cos
= - asin
Put a value from (1) in (2).
14 | P a g e
= -r tan
Hence .
Differential Equations of first order and first degree:
The general form of first order ,first degree differential equation is f(x,y) or [Mdx + Ndy =0
Where M and N are functions of x and y]. There is no general method to solve any first order
differential equation The equation which belong to one of the following types can be easily
solved.
In general the first order differential equation can be classified as:
(1). Variable separable type

(2). (a) Homogeneous equation and
(b)Non-Homogeneous equations which to exact equations.
(3) (a) exact equations and
(b) equations reducible to exact equations.
4) (a) Linear equation &
(b) Bernoulli’s equation.
15 | P a g e
Type –I : VARIABLE SEPARABLE:
If the differential equation =f(x,y) can be expressed of the form or f(x) dx –g(y)dy =0
where f and g are continuous functions of a single variable, then it is said to be of the form variable
separable.
General solution of variable separable is =c
Where c is any arbitrary constant.
PROBLEMS:
1 ) tan y = sin(x+y) + sin(x-y).
Sol: Given that sin(x+y) + sin(x-y) = tan y
 2sinx.cosx = tan y [Note: sinC+sinD =2sin( .cos( ]
 2sinx = tany secy
General solution is 2
=> -2cosx = secy +c
=> sec y + 2 cos x +c =0 .//
2) Solve ( ). +( ) =0, y(0) =1.
Sol: Given ( ). +( ) =0
 + =0
On Integrations
1 1
 dx +  dy = 0
( 1+ x 2
) (
1 + y2 )
16 | P a g e
=> + =c ---------------(1)
Given y(0)=1 => At x=0 ,y=1 ---------(2)
(2) in (1) => 0+ 1 =c.
=> 0+ =c
=> c=
Hence the required solution is + =
Exact Differential Equations:

Def: Let M(x,y)dx +N(x,y) dy =0 be a first order and first degree Differential Equation where
M & N are real valued functions of x,y . Then the equation Mdx + Ndy =0 is said to be an
exact Differential equation if a function f .
f f
d[f (x, y)] = dx + dy
x y
Condition for Exactness: If M(x,y) & N (x,y) are two real functions which have continuous
partial derivatives then the necessary and sufficient condition for the Differential equation
Mdx+ Ndy =0 is to be exact is =
Hence solution of the exact equation M(x,y)dx +N(x,y) dy =0. Is

+ = c.
(y constant) (terms free from x).
-------------------********---------------
PROBLEMS
 x
 x
 x
1 ) Solve 1 + e y dx + e y 1 − dy = 0
   y
 
x x
x
Sol: Hence M = 1 + e & N = e (1 −
y y
)
y
 −1 
x x x
1 y
= e (y
& = e   +
y
)e ( )
 y  y
17 | P a g e
x x
= ey ( & = ey (
equation is exact
General solution is
+ = c.
(y constant) (terms free from x)
+ = c.
x
y
e
=> =c
1
y
=> =C
2. Solve ( +1) .cosx dx + =0.
Ans: ( +1) . sinx =c = cosx
3. Solve (r+sin cas + r (sin d
Ans: r 2 + 2r (sin  − cos  ) = 2c
M N
= = sin .
r 
4. Solve [y( ) +cos y] dx+ [ x +logx –xsiny]dy =0.
Sol: hence M = y( ) +cos y, N = x +logx –xsiny.
= 1+ -siny = 1 + -siny
so the equation is exact
18 | P a g e
General sol + = c.
+ = c.
 y(x+ logx) +x cosy = c.
5. Solve ysin2xdx – ( +cosx) .dy =0.

6. Solve (cosx-xcosy)dy – (siny+(ysinx))dx =0
Sol: N = cosx-x cosy & M = -siny-ysinx
= -sinx - cosy = -cosy - sinx
 the equation is exact.
General sol + = c.
=> + =c
=> -xsiny+ ycos x =c

=> ycosx – xsiny =c.
7. Solve ( sinx . siny - x ) dy = ( cosx-cosy) dx
Ans: x sinx.cosy =c.
8. Solve (x2+y2-a2) x dx +(x2-y2-b2) . y .dy =0
Ans: x4+2x2y2-2a2x2-2b2y2 =c .
19 | P a g e
REDUCTION OF NON-EXACT DIFFERENTIAL EQUATIONS TO
EXACT USING INTEGRATING FACTORS
Definition: If the Differential Equation M(x,y) dx + N (x,y ) dy = 0 be not an exact differential
equation. It Mdx+Ndy=0 can be made exact by multiplying with a suitable function u (x,y)  0.
Then this function is called an Integrating factor(I.F).
Note: There may exits several integrating factors.
Some methods to find an I.F to a non-exact Differential Equation Mdx+N dy =0

Case -1: Integrating factor by inspection/ (Grouping of terms).
Some useful exact differentials

1. d (xy) = xdy +y dx
2. d( =
3. d( =
4. d( ) = x dx + y dy
5. d(log( ) =
6. d(log( ) =
7. d( ( ) =
8. d( ( ) =
9. d(log(xy)) =
10. d(log( )) =
11. d( =
20 | P a g e
PROBLEMS:
1 . Solve xdx +y dy + = 0.
Sol: Given equation x dx + y dy + =0
1
d( ) + d( (
on Integrating
1
+ ( = c.
2 . Solve y(x3. ) dx + x (y + x3. dy = 0.
Sol: Given equation is on Regrouping
We get yx3 - dx+ xy dy +x4 dy =0.
x3 (ydx+ xdy)+ y (x dy – ydx ) = 0

Dividing by x3
(ydx + xdy) +( .( ) =0
d( ) +( .d +( =0
on Integrating
2
 y
e + ½   = C is required G.S.
xy
x
3. Solve (1+xy) x dy + (1- yx ) y dx = 0
Sol: Given equation is (1+xy) x dy +(1-yx ) y dx =0.
(xdy + y dx ) + xy ( xdy – y dx ) = 0.
Divided by x2y2 => ( )+( =0
( ) + dy - dx =0.
−1
On integrating => + log y – log x =log c
xy
- - logx +log y =log c.
21 | P a g e
4. Solve ydx –x dy = a ( dx
Sol: Given equation is ydx –x dy = a ( dx
 = a dx
 x 
 d  tan −1 = a dx 
 y 
−1 x
Integrating on tan = ax +c where c is an arbitrary constant.
y
Method -2: If M(x,y) dx + N (x,y) dy =0 is a homogeneous differential equation and

1
Mx +Ny 0 then is an integrating factor of Mdx+ Ndy =0.
Mx + Ny
1 . Solve x2y dx – ( x3+ y3 ) dy = 0

Sol : Given equation is x2y dx – ( x3+ y3 ) dy = 0-----------------(1)
Where M = x2y & N = (-x3- y3 )
Consider = x2 & = -3x2
equation is not exact .
But given equation(1) is homogeneous differential equation then

So Mx+ Ny = x(x2y) – y (x3+ y3) = - y4 0.
1 −1
I.F = = 4
Mx + Ny y
Multiplying equation (1) by
x2 y x3 + y 3
= > 4 dx - dy = 0----------------------(2)
−y − y4
= >- dx -
This is of the form M1dx + N1dy = 0
22 | P a g e
x3 + y3
For M1 = & N1 =
y4
3x 2 3x 2
=> = & =
y4 y4
=> = equation (2) is an exact D.equation.
General sol  M 1dx +  N1dy = c
(y constant) (terms free from x in N1)
=> + = c.
=> + log |y| = c
2. Solve dx + (
Ans: (x-y) . = (x+y ).
3. Solve y( + x (2 dy =0
Sol:Given equation is y( + x (2 dy =0 -----------(1)
It is the form Mdx +Ndy =0
Where M = y( , N= x (2
Consider = 3y2-2x2& = 2y2-3x2
equation is not exact .
Since equation(1) is Homogeneous differential equation then
Consider Mx+ N y= x[y( ] +y [x (2
= 3xy ( 0.
=> I.F. =
23 | P a g e
Multiplying equation (1) by we get

(
y y 2 − 2x2
+
)
x 2 y2 − x2 (
dy = 0
)
(
3xy y 2 − x 2
dx
)
3xy y 2 − x 2 ( )
Now it is exact
(y − x2 − x2
2
) +
y2 + y2 − x2
dy = 0
( )
(
3x y 2 − x 2
dx
)
3y y2 − x2 ( )
+ =0.
 dx dy  2 ydy 2 xdx
 +  + − =0
 x y  2 y −x
2 2
(
2 y2 − x2 ) ( )
log x +log y + log ( - log (y2-x2)=logc  xy = c
4. Solve r ( + ) d – ( + ) dr =0
Ans: + log - log =c.
Method- 3: If the equation Mdx + N dy =0 is of the form y. f(x, y) .dx + x . g ( x, y) dy = 0 & Mx- Ny
then is an integrating factor of Mdx+ Ndy =0.
Problems:
1 . Solve (xy sinxy +cosxy) ydx + ( xy sinxy –cosxy )x dy =0.
Sol: Given equation (xy sinxy +cosxy) ydx + ( xy sinxy –cosxy )x dy =0 -------(1).
Equation (1) is of the form y. f(xy) .dx + x . g ( xy) dy =0.
Where M =(xy sinxy + cos xy ) y
N= (xy sinxy- cos xy) x
M N

y x
 equation (1) is not an exact
24 | P a g e
Now consider Mx-Ny
Here M =(xy sinxy + cos xy ) y
N= (xy sinxy- cos xy) x
Consider Mx-Ny =2xycosxy
Integrating factor =
So equation (1) x I.F

(xy sin xy + cos xy)y dx + (xy sin xy − cos xy)x dy = 0
2 xy cos xy 2 xy cos xy
 ( y tan xy + ) dx + ( y tan xy - ) dy =0
 M1 dx + N1 dx =0
Now the equation is exact.

General sol  M1 dx +  N1 dy = c.
-1
=> +  y dy =c.
=> +logx + (-logy) =log c
=> log|sec(xy)| +log =log c.
=> . secxy =c.
2. Solve (1+xy) y dx + (1-xy) x dy =0

Sol : I.F =
=> + =c
=> + logx - log y =c.
25 | P a g e
=> +log( ) = where c1 = 2c.
3. Solve ( 2xy+1) y dx + ( 1+ 2xy-x3y3) x dy =0
Ans: log y + + =c.
4. solve (x2y2 +xy +1 ) ydx +( x2y2- xy+1 ) xdy =0
Ans: xy - + log( ) =c .
Method -4: If there exists a continuous single variable function f (x ) such that
 f ( x )dx
=f(x),then I.F. of Mdx + N dy =0 is e
PROBLEMS
1 . Solve ( 3xy – 2a ) dx + ( dy =0
Sol: Given equation is ( 3xy – 2a ) dx + ( dy =0
This is of the form Mdx+ Ndy = 0
=> M = 3xy – 2a &N=
= 3x-4ay & = 2x-2ay
equation not exact .
=
(3x − 4ay ) − (2 x − 2ay )
Now consider
x(x − 2ay )
=> = = f(x).
26 | P a g e
1
 dx
=> e x
= x is an Integrating factor of (1)
equation (1) Multiplying with I.F then
=> x dx + x dy = 0
=> (3x2y -2ay2x) dx + (x3-2ax2y) dy =0
It is the form M1dx + N1dy =0
M 1 = 3x 2 y − 2ay 2 x, N1 = x 3 − 2ax 2 y
M 1 N
= 3x 2 − 4axy, 1 = 3x 2 − 4axy
y x
M 1 N1
=
y x
 equation is an exact
General sol  M1dx +  N dy = c.

1
( )
  3x 2 y − 2ay 2 x dx +  0dy = c
= > x3y –ax2y2 =c .
2 . Solve ydx-xdy+(1+
Sol : Given equation is (y+1+ ( dy =0.

M= y+1+ & N=
= 1 = 2x sin y -1
27 | P a g e
M N
 = > the equation is not exact.
y x
M N
−
y x (1 − 2 x sin y + 1) − 2 x sin y + 2 − 2(x sin y − 1) − 2
So consider = = 2 = =
N x 2 sin y − x x sin y − x x(x sin y − 1) x
1
− 2  dx 1
 f ( x ) dx
I.F = e =e x
= e −2 log x =
x2
Equation (1) X I.F => dy =0
It is the form of M1dx+ N1 dy =0.
Gen soln => + =0
=> - +x- cosy =c.
=>
3. Solve 2xy dy – (x2+y2+1)dx =0
Ans: -x + +
4. Solve (x2+y2) dx -2xy dy =0

Ans: x2-y2=cx.
Method -5: For the equation Mdx + N dy = 0 if = g(y) (is a function of y alone) then
is an integrating factor of M dx + N dy =0.
Problems:
1 .Solve (3x2y4+2xy)dx +(2x3y3-x2) dy =0
Sol: Given equation (3x2y4+2xy)dx +(2x3y3-x2) dy =0 -----------------(1).
Equation of the form M dx + N dy =0.
28 | P a g e
Where M =3x2y4+2xy & N = 2x3y3-x2
M N
= 12 x 2 y 3 + 2 x, = 6x2 y3 − 2x
y x
equation (1) not exact.
So consider = = g(y)
I.F = = = = .
 3x 2 y 4 + 2 xy   2x3 y3 − x 2 
Equation (1) x I.F =>   
 dx +  dy = 0
 y2   y2 
 2 2 2x   3 x2 

  3x y + dx +  2 x y − 2 dy = 0
 y   y 
It is the form M1dx + N1 dy =0
General sol  M 1dx +  N1dy = c

=> + =c.
=> + =c.
=> =c.
2 . Solve (xy3+y) dx + 2(x2y2+x+y4) dy =0
Sol: = = = g(y).
I.F = = =y.
29 | P a g e
( )
Gen sol:  xy4 + y 2 dx +  2 y 5 dy = c ( )
+ x+ =c .
3 . solve (y4+2y)dx + ( xy3 +2 y4 – 4x) dy =0
Sol: = = = g(y).
I.F = = =
 2 
Gen soln :   y + y 2 

dx +  2 ydy = c .
4. Solve (y+ y2)dx + xy dy =0
Ans: x + xy =c .
5. Solve (xy3+y) dx + 2(x2y2+x+y4)dy =0.
Ans: (x2+y4-1) =c.

LINEAR DIFFERENTIAL EQUATIONS OF FIRST ORDER:
dy
Def: An equation of the form + P( x). y = Q( x) is called a linear differential equation of
dx
first order in y.
dy
Working Rule: To solve the liner equation + P( x). y = Q( x)
dx
First find the integrating factor I.F =
General solution is y x I.F =  Q( x )  I.F.dx + c
Note: An equation of the form Q(y)is called a linear Differential equation of first
order in x.
30 | P a g e
Then integrating factor =
General solutionis = x X I.F =  Q( y )  I.F.dy + c

PROBLEMS:
1 . Solve (1+ y2) dx=( y –x ) dy
Sol: Given equation is (1+ y2) =( y –x )
tan −1 y
+ ).x=
1 + y2
It is the form of + p(y).x = Q(y)
I.F = e  p ( y ) dy = =
tan −1 y tan−1 y
=> General solution is x. = .e dy + c
1 + y2
= > x. =
[ put tan-1 y = t
1
 dy = dt ]
1 + y2
 x. =t. - +c
=> x. = - +c
=>x = + c/ is the required solution
2. Solve (x+y+1) = 1.
Sol: Given equation is (x+y+1) = 1.
=> = y+1.
31 | P a g e
It is of the form + p(y).x = Q(y)
Where p(y) = -1 ; Q(y) = 1+y
= > I.F = = =
General solution is x X I.F =  Q( y )  I.F.dy + c

=>x. =
=>x. =
=> x =- yx - +c
=> x =- +c .//
3. Solve
Sol: Given equation is
It is of the form
Where p(x) =1 Q(x) =
=> I.F = = =
General solution is y x I.F =  Q( x )  I.F.dx + c
=> y. =
=> y. =  et dt + c put
=> y. = et+c
=> y. = +c
4. Solve x. + y =log x
32 | P a g e
Sol : Given equation is x. + y =log x
It is of the form +p(x)y = Q(x)
Where p(x) = & Q(x)=
i.e , + .y=
=> I.F = = = =x.
General solution is y x I.F =  Q( x)  I .F . dx + c
=> y.x =
=> y .x = x (logx-1) +c.
5 . Solve (1+y2) + (x- = 0.
Sol : Given equation is
It is of the form + p(y) .x = Q(y)
Where p (y) = , Q(x)=
I.F = = =
General solution is x x I.F =  Q( y )  I.F.dy + c .
= > x. = dy +c
=> x. = dt +c
[Note: put =t
=> = dt ]
=>x. = dt +c
=> x. = +c
33 | P a g e
=> x. = +c
y
6. solve + =
xlog x
−cos 2 x
Ans: ylogx = + c.
2
7. + (y-1). Cosx = e-sinxcos2x
Ans: y.esinx = + + esinx +c
dy 2x 1
8. + . y= given y = 0 , when x= 1.
dx 1 + x 2
( 1 + x 2 )2

Ans : y(1+x2) = −
4
9. Solve = (1+x) . sec y
Sol : The above equation can be written as

Divided by sec y => cos y - = (1+x) ------------(1)
Put sin y = u
= > cos y =
Differential Equation (1) is – . u = (1+x)
this is of the form Q(x)
Where p(x) = Q(x) =(1+x)
=> I.F = = = =
General solution is u x I.F =  Q( y )  I.F.dy + c
=> u. =
=> u. =
34 | P a g e
=> (sin y ) = c
( Or)
= > sin y = (1+x) c . (1+x) is required solution.
10. Solve - ytanx =
Ans : = +c .
11 .Solve – yx =
Ans: = cosx + c.
12. . = 2xy2 + y
Ans : = + c.
13. + ycos x = sin x
Ans : = (1+ 2 sinx ) + c (or) = -(1+ 2 sinx ) +c.
14. + y cot x =
Ans: ysinx (c + =3.

BERNOULLI’S EQUATION :
(EQUATIONS REDUCIBLE TO LINEAR EQUATION)
Q(x)
Def: An equation of the form + p(x) .y = ----- --------(1)
is called Bernoulli’s Equation, where P&Q are function of x and n is a real constant.
Working Rule:
Case -1 : If n=1 then the above equation becomes + p. y = Q.
dy
=> General solution of + ( P − Q) y = 0 is
dx
35 | P a g e
dy
 + ( P − Q)dx = c by variable separation method.
y
Case -2: If 1 then divide the given equation (1) by
 . + p(x) . =Q ---------(2)
Then take =u
(1-n)
 =
Then equation (2) becomes
+ p(x) . u =Q
+ (1-n) p.u = (1-n)Q which is linear and hence we can solve it.
Problems:
1 . Solve x + y = x3
Sol: Given equation is x + y = x3
Given equation can be written as =x2y6
Which is of the form + p(x).y =Q
Where p(x) = Q(x) = & n=6
Divided by y6 => . + = ---------------(2)
36 | P a g e
Take =u
 = }----------------(3)
 = }----------------(3)
(3) in (2) => - u = -5x2
Which is a Linear differential equation in u
I.F = = = =
General solution is u .I.F =  Q( x)  I .F .dx + c
1 1
u. 5
=  −5 x 2 . 5 dx + c
x x
= +c (or) +c
2. Solve ( xy ) =1
Sol: Given equation is ( xy ) =1
dx
This can be written as -x .y= => . - .y = ---------------------(1)
dy
Put =u
 . = ---------------(2).
(2) in (1)  u .y =
(Or) + u .y = - .
37 | P a g e
This is a Linear Differential Equation in ‘ u’
e
P(y)dy
I.F = = =
General solution  u .I.F =  Q( y )  I.F.dy + c
 u. = +c
y2
−
2
e
 = -2( . +c
x
(or)
x(2- )+ cx =1.
3. Solve +y tanx = sec x
Ans: I.F = = e  logcos x = cos x
General solution cos x= -x +c .
4. (1- ) +xy = x
Sol: Given equation can be written as
+ y = x
Which is a Bernoulli’s equation in ‘ y ‘
Divided by  . + = -----------(1).
Let =u
 = => = -------------(2)
38 | P a g e
du 2x − 2 sin −1 x
(2) in (1)  + =  − .u =
dx 1 − x 2 1 − x2
Which is a Linear differential equation in u
 I.F = = = =
General solution  u .I.F =  Q( x )  I.F.dx + c
 = -
=> = -2 [ x + ] +c
5. = 2xy2 + y .
Ans: =- .
NEWTON’S LAW OF COOLING
STATEMENT:The rate of change of the temperature of a body is proportional to the difference of the
temperature of the body and that of the surrounding medium.
Let be the temperature of the body at time ‘t’ and be the temperature of its surrounding
medium(usually air). By the Newton’s law of cooling , we have
(  - k( k is +ve constant
 log = -kt +c.
If initially = 1 is the temperature of the body at time t=0 then
c = log (1 −  0 )  log =-kt+ log (1 −  0 )
39 | P a g e
( −  0 )
 log ( = -kt.
(1 −  0 )
( −  0 )
 =
(1 −  0 )
= + (1 −  0 ) .
Which gives the temperature of the body at time ‘t’ .
Problems:
1 A body is originally at 800C and cools down to C in 20 min . If the temperature of the air is
Cfind the temperature of body after 40 min.
Sol: By Newton’s law of cooling we have

= -k( where is the temperature of the air.
d
 = −k  dt  log ( −  0 ) = −kt + log c
( −  0 )
Here = c
 log( = -kt + log c
 log( ) =-kt
 =
 --------------(1)
When t=0 , = C  80 = 40 +c  c = 40 ------------(2).
When t=20 , = C  60 = 40 +c ------------(3).
Solving (2) & (3)  c
 40 =20
=> k= log2
When t= C => equation (1) is
40 | P a g e
= 40 +40
= 40 + ( 40 x )
 = C
2 . An object whose temperature is 750C cools in an atmosphere of constant temperature 250C,
at the rate of k being the excess temperature of the body over that of the temperature. If
after 10min , the temperature of the object falls to 650C , find its temperature after 20 min. Also
find the time required to cool down to 550C.
Sol : We will take one minute as unit of time.
It is given that =-k
 c ------------(1).
Initially when t=0  =
 c=
Hence C = 50   = 50.e
− kt
-----------------(2)
When t= 10 min  =
 40= 50
 = ---------------------(3).
The value of when t=20  c
50
50
50
When t=20  = C.
41 | P a g e
Hence the temperature after 20min =320+250=570C
When the temperature of the object = 550C
 = 550 − 250 = 30 0 C
Let t, be the corresponding time from equ. (2)
30 = 50.e − kt1
----------------(4)
1
(e )
−k 10 4  4 10
= i.e., e −k =  
From equation (3) 5 5
t1
 4 10 t 4 3
30 = 50   1 log = log
From Equ(4) we get 5 10 5 5
 log 3 
 t1 = 10
( )
5  = 22.9 min
 log

4
5
( )
3. A body kept in air with temperature250C cools from 1400C to 800C in 20 min. Find when the body
cools down in 350C.
d d
Sol : By Newton’s law of cooling = −k ( −  0 )  = − kdt
dt  − 0
 log ( −  0 ) = −kt + c Here o=
 log = -kt +c -------------(1).

When t=0 , = c  log (115) =c
 c =log (115).
 kt = - log + log 115--------(2)
When t=20 , = c
 log (80-25)= -20k + log 115
 20 k =log (115) - log(55) ---------(3)
(2)/ (3) => =
42 | P a g e
When = C 
 = 3. 31
 temperature = 20  3.31 = 66.2

The temp will be C after 66.2 min.
4 . If the temperature of the air is C and the temperature of the body drops from C to
C in 10 min. What will be its temperature after 20min. When will be the temperature C.
Sol: log = -kt + log c
c= C and = .
t= =4.82min
5. The temperature of the body drops from C to C in 10 min. When the surrounding air is at
C temperature. What will be its temp after half an hour.When will the temperature be C.
Sol : = -k(
log = -kt + log c
when t=0 , = => c=80
when t=10 , = => = .
when t =30min => = 20 +80 ( ) = 46OC
log 5 − log 80
when = c = > t = 10 = 74.86 min
(log 11 − log 16)
LAW OF NATURAL GROWTH OR DECAY
43 | P a g e
Statement : Let x(t) or x be the amount of a substance at time ‘ t’ and let the substance be getting
converted chemically . A law of chemical conversion states that the rate of change of amount x(t) of a
chemically changed substance is proportional to the amount of the substance available at that time
(or) = - kx ; ( k >0)
Where k is a constant of proportionality
Note: Incase of Natural growth we take
= k .x (k > 0)
PROBLEMS
1 The number N of bacteria in a culture grew at a rate proportional to N . The value of N was initially
1
100 and increased to 332 in one hour. What was the value of N after 1 hrs
2
Sol: The differentialequation to be solved is = kN
 = k dt
 log N = kt + log c
N = c ------------(1).
When t= 0sec , N =100  100 =c  c =100
When t =3600 sec , N =332  332 =100
 =
Now when t = hors = 5400 sec then N=?
 N =100
44 | P a g e
 N =100
 N = 100 = 605.
 N = 605.
2 . In a chemical reaction a given substance is being converted into another at a rate proportional to the
th
1
amount of substance unconverted. If   of the original amount has been transformed in 4 min, how
5
much time will be required to transform one half.
Ans: t= 13 mins.
3. The temperature of a cup of coffee is C, when freshly poured the room temperature being C.
In one min it was cooled to C. How long a period must elapse, before the temperature of the
cup becomes C.
Sol: : By Newton’s Law of cooling,
= -k( ; k>0
= C  log ( ) = -kt + log c--------------(1).
When t=0 ; =92  c =68
−k 68
When t =1 ; C e =
56
56
 k = log .
68
When C , t =?
65  41
Ans: t = = 0.576 min
682
RATE OF DECAY OR RADIO ACTIVE MATERIALS
45 | P a g e
Statement : The disintegration at any instant is proportional to the amount of material present in
it.
If u is the amount of the material at any time ‘t’ , then = - ku , where k is any constant (k
>0).
Problems:
1) If 30% of a radioactive substance disappears in 10days,how long will it take for 90% of it to
disappear.
Ans: 64.5 days
2) The radioactive material disintegrator at a rate proportional to its mass. When mass is 10
mgm , the rate of disintegration is 0.051 mgm per day . how long will it take for the mass to be
reduced from 10 mgm to 5 mgm.
Ans: 136 days.
3. Uranium disintegrates at a rate proportional to the amount present at any instant. If M1 and M2
are grams of uranium that are present at times T1 and T2 respectively, find the half-life of uranium.
Ans: T = .
4. The rate at which bacteria multiply is proportional to the instantaneous number present. If the
original number double in 2 hrs, in how many hours will it be triple.
Ans: hrs.
5. a) If the air is maintained at C and the temperature of the body cools from C to
C in 12 min, find the temperature of the body after 24 min.
Ans: C
b) If the air is maintained at C and the temperature of the body cools from C
to C in 10 min, find the temperature after 30 min.
Equation not of first degree
Equation solvable for p
46 | P a g e
A differential equation of the first order but of the n th degree is of the form
p n + P1 p n−1 + P2 p n−2 + ............... + Pn = 0

where P1 , P2 , P3 ......... Pn
Splitting up the left hand side of (1) into n linear factors, we have
 p − f1 ( x, y )   p − f 2 ( x, y )  ..............  p − f n ( x, y )  = 0
Equating each ofthe factors to zero
p = f1 ( x, y ) , f 2 ( x, y ) ......... f n ( x, y )
Solving each of these equations of the first order and first degree, we get the solutions
F1 ( x, y, c ) = 0, F2 ( x, y, c ) = 0, F3 ( x, y, c ) = 0,........ Fn ( x, y, c ) = 0
These n solutions constitute the general solution of (1).
Problems:
dy dx x y
1) Solve − = −
dx dy y x
dy dx x y
Solution: The given D E is − = −
dx dy y x
1 x y
p−
= −
p y x
dy
It can be written as where p =
dx
 x y
p2 + p  −  −1 = 0
 y x
 y  x
Factorizing  p −  p −  = 0
 x  y
 y  x
 p−  = 0 .........(i ) and  p −  = 0..........(ii )
 x  y
47 | P a g e
dy y dy x
= and =
dx x dx y
dy dx
= , ydy = xdx
y x
Integration on both side
log( xy) = c , x 2 − y 2 = c
Equations solvable for y
If the given equation, on solving for y , taken the form y = f ( x, p ) (2)
then differentiation with respects to x gives an equation of the form
dy  dp 
p= =   x, p , 
dx  dx 
Now it may be possible to solve this new differential equation in x and p .
Let its solution be F ( x, p, c ) = 0 (2)
The elimination of p from (1) and (2) gives the required solution.
In case of elimination of p is not possible, then we may solve (1) and (2) for x and y andobtained
x = F1 ( x, c ) , y = F2 ( p, c ) , As the required solution, where p is the parameter.
Problems:
1) Solve y − 2 px = tan −1 ( xp )
Solution: The given equation is y − 2 px = tan −1 ( xp ) (1)
Differentiation on both sides w. r. t ‘ x ’
dp
p 2 + 2 xp
dy  dp  dx
= p = 2p + x  +
dx  dx  1+ x p2 4
Substituting this values of ‘ x ’ in (1)
48 | P a g e
2c 1 − n n
y= + p
p 1+ n
dp  dp  p
We get p + 2 x +  p + 2x  =0
dx  dx  1 + x 2 p 4
 dp   p 
 p + 2 x  1 + =0
 dx   1 + x 2 p 4 
 dp  dp
 p + 2 x  = 0  p = −2 x
 dx  dx
dx −2
= dp
x p
This gives Integration on both side (2)
log x + 2log p = log c
log xp 2 = log c
c c
xp 2 = c  p 2 =  p=
x x
Eliminates p from (1) and (2) , we get
c
y=2 + tan −1 (c)
x
Equations solvable for x

If the given equation, on solving for y , taken the form x = f ( y, p ) (2)
then differentiation with respects to x gives an equation of the form
1 dx  dp 
= =   y , p, 
p dy  dy 
Now it may be possible to solve this new differential equation in y and p .
Let its solution be F ( y, p, c ) = 0 (2)
The elimination of p from (1) and (2) gives the required solution.
In case of elimination of p is not possible, then we may solve (1) and (2) for x and y and obtained
y = F1 ( y, c ) , y = F2 ( p, c )
49 | P a g e
As the required solution, where p is the parameter.
Problems:
1) Solve y = 2 px + y 2 p3
Solution : the given D E is y = 2 px + y 2 p3 (1)
y − y 2 p3
Solving (1) for x , takes the form x =
2p
Diff w.r. t ‘ y ’
  dp  dp 
 p 1 − 2 yp 3 − y 3 3 p 2  − ( y − y 2 p3 ) 
= =  
dx 1 1 dx  dx

dy p 2  p 2

 
dp dp dp
2 p = p − 2 yp 4 − 3 yp 3 −y + y 2 p3
dy dy dy
dp dp
p + 2 yp 4 + 2 yp 3 +y =0
dy dy
p (1 + 2 yp 3 ) + y (1 + 2 p3 y ) = 0
dp
dy
(1 + 2 yp3 ) ( p + y ) dp
dy
=0
d
( py ) = 0
dy
Integration on both side py = c (2)

Thus eliminating from the given equations(1) and (2), we get
c c3
y=2 x + 3 y2
y y
y 2 = 2cx + c 3
Clairauits Equation
An equation of the form y = px + f ( p ) (1) is know as clairauts equation.
Diff w. r. t ‘ x ’ , we have
50 | P a g e
dp dp
p = p+x + f l ( p)
dx dx
dp
  x + f l ( p )  =0
dx
dp
= 0 or  x + f l ( p )  = 0
dx
dp
 = 0 gives p = c (2)
dx
Thus eliminating p from (1) and (2) ,we get y = cx + f ( c )
Which is the general solution of (1)
Hence the solution of the clairauts equation is obtained on replacing p by c .
Problems:
1) Solve p = sin ( y − xp ) also find its singular solution .
−1
Solution: The given equation can be written as sin p = y − xp
y = px + sin −1 p (1)
Which is the clairauts equation.
−1
Its solution is y = cx + sin c (2)
To find the singular solution , Differ (2) w. r. t c
1
0= x+ (3)
1 − c2
To eliminate ‘ c ’ from (2) and(3) , we get (3) as
N ( x 2 − 1)
c=
x
Substituting the value of c in (2), we get
 N ( x 2 − 1) 
−1
y = sin   + N ( x 2 − 1)
 x 
Which is the required singular solution.
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TUTORIAL QUESTIONS
1. In a chemical reaction a given substance is being converted into another at a rate proportional to
th
1
the amount of substance unconverted. If   of the original amount has been transformed in 4
5
min, how much time will be required to transform one half.
2. . Solve +y tanx = sec x
 x
 x
 x
 y 
3. Solve: 1 + e dx + e y 1 − dy = 0
   y
 
4. If the air is maintained at C and the temperature of the body cools from C to
5. Sovle: (1- ) +xy = x
6. Solve (xy3+y) dx + 2(x2y2+x+y4) dy =0

7. Solve (1+y2) + (x- = 0.
8. Solve (x+y+1) = 1.
9. Solve y(x3. ) dx + x (y + x3. dy = 0.
10. Solve x2y dx – ( x3+ y3 ) dy = 0

12. A body kept in air with temperature 250C cools from 1400C to 800C in 20 min. Find when the body
cools down in 350C.
13. Solve 2xy dy – (x2+y2+1)dx =0
14. Solve (3x2y4+2xy)dx +(2x3y3-x2) dy =0
15. Solve ( 3xy – 2a ) dx + ( dy =0
52 | P a g e
DESCRIPTIVE QUESTIONS
1. Solve p = sin ( y − xp ) also find its singular solution.
2. State Newton’s law of cooling.

3. Define exact differential equations with an example.
4. Define linear differential equation.
5. Solve y = 2 px + y p
2 3
6. Solve y − 2 px = tan ( xp )
−1
dy dx x y
− = −
7. Solve dx dy y x
8. If the air is maintained at C and the temperature of the body cools from C to
9. If 30% of a radioactive substance disappears in 10days,how long will it take for 90% of it
to disappear.
10. The number N of bacteria in a culture grew at a rate proportional to N . The value of N was
1
initially 100 and increased to 332 in one hour. What was the value of N after 1 hrs
2
53 | P a g e
OBJECTIVE QUESTIONS
1) The equation (ax+hy+g) dx+(hx+by+f)=0 is__________

A)Homogeneous B) variable separiqable c) Exact D) none
xdx + ydy
2) The general solution of = 0 ___________
y2
A) log(x+y)=c B) log( x 2 + y 2 ) C) log(x,y)=c

D) log(x-y)=c
3) The general solution of (1 + x 2 )dy − (1 + y 2 )dy = 0 is _______________
A) tan −1 y − tan −1 x = c B) tan −1 x + tan −1 y = c C) tan −1 x + tan −1 y = cy

D) tan −1 x − tan −1 y = cy
dy
4) The general solution of + xy = x is_______________________
dx
− x2 − x2 − x2 − x2
A) y = 1 + ce 2
B) y = 1 − ce 2
C) y = 1 − 3ce 2
D) y = 1 + 3ce 2
5) The general solution of p 2 − 5 p − 6 = 0 is ____________________
A)(y-2x-c)(y-3x+c) =0 B)(y-2x-c)(y-4x+c) =0 C) )(y-2x-c)(y-5x=c) =0
D))(y-2x-c)(y-3x-c) =0
6) The integrating factor of (1 − x 2 _) y + xy = ax is ___________
1 1 1 1
A) B C) D)
x −1
2
xx 2 − 1 x −1
2
x2 + 1
7) The I,F to the differential equation ydx-x dy+logx dx=0 is _______
1 1 1 y
A) B) − C) − D) −
x2 x2 x3 x3
54 | P a g e
8) The necessary and sufficient condition for exactness of the differential equation M( x,y)
dx+N( x,y) dy=0 is _______
M N M N M N M N
A) =− B) = C) = D) =−
y x y x x y x y
9) The function x or y both which on multiplied to non-exact differential equation converted
into exact is known as_________
A) I.F (Integratinting Factor) B) Division facfor C)M.F( multiplication factor) D) none
dy
10) The integrating factor of x − y = 2 x 2 cos ce2 x is_______________________
dx
1 1 3
A) x B) C) 2 D)
x x x
Fill in the blanks.

1) The integrating factor of x 2 ydx − ( x3 + y 3 )dy = 0 is-
_________________________________
2) The integrating factor of y( x 2 y 2 + 2)dx + x(2 − 2 x 2 y 2 )dy = 0 is
________________________
3) The integrating factor of (3xy − 2ay 2 )dx + ( x 2 − 2axy)dy = 0 is
_________________________
dy
4) The general solution of x + y = log x is _____________________________
dx
dy
5) The general solution of x + y = x3 y 6 is_________________________
dx
6) The general solutionof xp3 = a + bp is____________________
7) The general solutionof y = 2 px − p 2 is ___________________
8) The general solution of p=log(px-y) is _______________
9) The general solution of p= tan(xp-y) is______________
10) The general solution of xdy − ydx = xy 2 dx is _____________________________
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UNIT TEST PAPERS
Name of the student: Reg No: Branch:
Course: I B.TECH I Sem Subject: M-II Marks: 10

TEST- I
SET NO-I
Answer the following question. 1*5=5M
1. If the air is maintained at C and the temperature of the body cools from C
to C in 12 min, find the temperature of the body after 24 min.
(OR)
2. A) Solve
y − 2 px = tan −1
( xp )
dy dx x y
− = −
B) Solve dx dy y x
Fill in the blanks. 10*0.5=5M

1) 1. The integrating factor of x 2 ydx − ( x3 + y 3 )dy = 0 is-
_________________________________
________________________
_________________________
dy
dx
dy
dx
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TEST- I
SET NO-II
1. If 30% of a radioactive substance disappears in 10days,how long will it take for 90%
of it to disappear. (OR)
2.A) Solve p = sin ( y − xp ) also find its singular solution.
B)Solve y = 2 px + y p
2 3

1) The integrating factor of x ydx − ( x + y )dy = 0 is-
2 3 3
_________________________________
________________________
_________________________
dy
dx
dy
dx
57 | P a g e

TEST- I
SET NO-III
y
1. A) solve + =
xlog x
B) Solve ( xy ) =1
(OR)
2. An object whose temperature is 750C cools in an atmosphere of constant temperature 250C, at

the rate of k being the excess temperature of the body over that of the temperature. If
after 10min , the temperature of the object falls to 650C , find its temperature after 20 min. Also
find the time required to cool down to 550C.
2 3 3
_________________________________
________________________
_________________________
dy
dx
dy
dx
58 | P a g e

TEST- I
SET NO-IV
1. A) Solve x2y dx – ( x3+ y3 ) dy = 0
B) Solve 2xy dy – (x2+y2+1)dx =0 (OR)
2. A body kept in air with temperature250C cools from 1400C to 800C in 20 min. Find when the
body cools down in 350C.
2 3 3
_________________________________
________________________
_________________________
dy
dx
dy
dx
10) The general solution of xdy − ydx = xy dx is _____________________________
2
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SEMINAR TOPICS
TOPIC 1:
Exact and Non Exact differential equations
TOPIC 2:
Linear and Bernoulli’s diferential equations
TOPIC 3:
Applications of first order ode
TOPIC 4:
Solvable equations for p,y
TOPIC 5:
Solvable equations for x and clairaut’s equation.
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Assignment Problems
1. Solve ( sinx . siny - x ) dy = ( cosx-cosy) dx
2. Solve x. + y =log x
4. Solve (y+ y2)dx + xy dy =0
5. Solve (x2+y2) dx -2xy dy =0
6. Solve ( 2xy+1) y dx + ( 1+ 2xy-x3y3) x dy =0
7. Solve (xy sinxy +cosxy) ydx + ( xy sinxy –cosxy )x dy =0.
8. = 2xy2 + y .
9. Solve ( xy ) =1
10. The temperature of a cup of coffee is C, when freshly poured the room temperature being
C. In one min it was cooled to C. How long a period must elapse, before the
temperature of the cup becomes C.
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APPLICATIONS
Differential equations have a remarkable ability to predict the
world around us. They are used in a wide variety of disciplines,
from biology, economics, physics, chemistry and engineering.
They can describe exponential growth and decay, the population
growth of species or the change in investment return over time.
Differential equations have wide applications in
various engineering and science disciplines. ... It is practically
important for engineers to be able to model physical problems
using mathematical equations, and then solve
these equations so that the behavior of the systems concerned
can be studied.
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NPTEL
https://nptel.ac.in/courses/111106100/
https://nptel.ac.in/courses/111/108/111108081/
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BLOOMS TAXONOMY
UNIT-1
TOPIC: 1.Exact Differential Equation
ANALYSIS:
Define Exact Differential Equation ?
The first order and first degree differential equation M ( x, y ) dx + N ( x, y ) dy = 0 is said to be
exact if there exisists a function U ( x, y ) such that du ( x, y ) = M ( x, y ) dx + N ( x, y ) dy
SYNTHESIS:
Working rule:
Re write the given differential equation into standard form M ( x, y ) dx + N ( x, y ) dy = 0
M N
* Find ,
y x
M N
Check the condition for exact. i.e =
y x
The general solution of exact equation is  M dx +  Ndy = c

( y −cons tan t ) ( which do not containing x )
EVALUATION
 x
 x
 x
 y 
Solve: 1 + e dx + e y 1 − dy = 0
   y
 
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x x
x
Sol: Hence M = 1 + e & N = e (1 −
y y
)
y
 −1 
x x x
1
= ey ( & = ey   + )ey( )
 y  y
x x
= ey ( & = ey (
equation is exact
General solution is
+ = c.
+ = c.
x
ey
=> =c
1
y
=> =c
2) Topic: LINEAR DIFFERENTIAL EQUATIONS OF FIRST ORDER

ANALYSIS:
Define Linear Differential equation.
dy
An equation of the form + P( x). y = Q( x) is called a linear differential equation of first
dx
order in y.
SYNTHESIS:
dy
The liner equation + P( x). y = Q( x) of first order and first degree in Y
dx
1) Find the integrating factor I.F =
2) General solution is Y( I.F) =  Q( x )  I.F.dx + c
Note: An equation of the form Q(y)is called a linear Differential equation of first
order in x.
65 | P a g e
1) Then integrating factor =
2) General solution is Y( I.F) =  Q( x )  I.F.dx + c
EVALUATION
Solve: (x+y+1) = 1.
Sol: Given equation is (x+y+1) = 1.
=> = y+1.
It is of the form + p(y).x = Q(y)
Where p(y) = -1 ; Q(y) = 1+y
= > I.F = = =
General solution is X( I.F) =  Q( y )  I.F.dy + c

=>x. =
=>x. =
=> x =- yx - +c
=> x =- +c .
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UNIT - II
HIGHER ORDER DIFFERENTIAL
EQUATIONS AND THEIR APPLICATIONS
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LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER
Definition: An equation of the form + P1(x) + P2(x) + --------+
Pn(x) .y = Q(x) Where P1(x), P2(x), P3(x)… …..Pn(x) and Q(x) (functions of x) continuous is
called a linear differential equation of order n.
LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
Def: An equation of the form + P1 + P2 + --------+ Pn .y = Q(x) where
P1, P2, P3,…..Pn, are real constants and Q(x) is a continuous function of x is called an linear
differential equation of order ‘ n’ with constant coefficients.
Note:
1. Operator D = ; D2 = ; …………………… Dn =
Dy = ; D2 y= ; …………………… Dn y=
2. Operator Q= i e D-1Q is called the integral of Q.
To find the general solution of f(D).y = 0 :

Where f(D) = Dn + P1 Dn-1 + P2 Dn-2 +-----------+Pn is a polynomial in D.
Now consider the auxiliary equation : f(m) = 0
i.e f(m) = mn + P1 mn-1 + P2 mn-2 +-----------+Pn = 0
where p1,p2,p3 ……………pn are real constants.
Let the roots of f(m) =0 be m1, m2, m3,…..mn.
Depending on the nature of the roots we write the complementary function
as follows:
Consider the following table
S.No Roots of A.E f(m) =0 Complementary function(C.F)
1. m1, m2, ..mn are real and distinct. yc = c1em1x+ c2e m2x +…+ cnemnx
2. m1, m2, ..mn are and two roots are
equal i.e., m1, m2 are equal and yc = (c1+c2x)em1x+ c3em3x +…+ cnemnx
68 | P a g e
real(i.e repeated twice) &the rest
are real and different.
3. m1, m2, ..mn are real and three yc = (c1+c2x+c3x2)em1x + c4em4x+…+ cnemnx
roots are equal i.e., m1, m2 , m3 are
equal and real(i.e repeated thrice)
&the rest are real and different.
4. Two roots of A.E are complex say yc = (c1 cos x + c2sin x)+ c3em3x +…+ cnemnx
+i -i and rest are real and
distinct.
5. If ±i are repeated twice & rest yc = [(c1+c2x)cos x + (c3+c4x) sin x)]+ c5em5x
are real and distinct +…+ cnemnx
6. If ±i are repeated thrice & rest yc = [(c1+c2x+ c3x2)cos x + (c4+c5x+ c6x2) sin
are real and distinct x)]+ c7em7x +……… + cnemnx
7. If roots of A.E. irrational say  
yc = ex c1 cosh  x + c2 sinh  x + c3em3x + ....... + cnemn x
   and rest are real and
distinct.
Solve the following Differential equations :
1. Solve -3 + 2y = 0
: Given equation is of the form f(D).y = 0

Where f(D) = (D3 -3D +2) y = 0
Now consider the auxiliary equation f(m) = 0
f(m) = m3 -3m +2 = 0  (m-1)(m-1)(m+2) = 0
 m = 1 , 1 ,-2
Since m1 and m2 are equal and m3 is -2
We have yc = (c1+c2x)ex + c3e-2x
2. Solve (D4 -2 D3 - 3 D2 + 4D +4)y = 0

Sol: Given f(D) = (D4 -2 D3 - 3 D2 + 4D +4) y = 0
 A.equation f(m) = (m4 -2 m3 - 3 m2 + 4m +4) = 0
 (m + 1)2 (m – 2)2 = 0
69 | P a g e
 m= -1 , -1 , 2 , 2
 yc = (c1+c2x)e-x +(c3+c4x)e2x
3. Solve (D4 +8D2 + 16) y = 0
Sol: Given f(D) = (D4 +8D2 + 16) y = 0
Auxiliary equation f(m) = (m4 +8 m2 + 16) = 0
 (m2 + 4)2 = 0
 (m+2i)2 (m+2i)2 = 0
 m= 2i ,2i , -2i , -2i
Yc = [(c1+c2x)cos x + (c3+c4x) sin x)]
4. Solve y11+6y1+9y = 0 ; y(0) = -4 , y1(0) = 14

Sol: Given equation is y11+6y1+9y = 0
Auxiliary equationf(D) y = 0  (D2 +6D +9) y = 0
A.equation f(m) = 0  (m2 +6m +9) = 0
 m = -3 ,-3
yc = (c1+c2x)e-3x -------------------> (1)
Differentiate of (1) w.r.to x  y1 =(c1+c2x)(-3e-3x ) + c2(e-3x )
Given y1 (0) =14  c1 = -4 & c2 =2
Hence we get y =(-4 + 2x) (e-3x )
5. Solve 4y111 + 4y11 +y1 = 0
Sol: Given equation is 4y111 + 4y11 +y1 = 0
That is (4D3+4D2+D)y=0
Auxiliary equation f(m) = 0
4m3 +4m2 + m = 0
m(4m2 +4m + 1) = 0
m( =0
m = 0 , -1/2 ,-1/2
y =c1+ (c2+ c3x) e-x/2
6. Solve (D2 - 3D +4) y = 0
70 | P a g e
Sol: Given equation (D2 - 3D +4) y = 0
A.E. f(m) = 0
m2-3m + 4 = 0
m= =
3 7
  i = = i
2 2
y= (c1 cos x + c2sin x)

General solution of f(D) y = Q(x)
Is given by y = yc + yp
i.e. y = C.F+P.I
Where the P.I consists of no arbitrary constants and P.I of f (D) y = Q(x)
Is evaluated as P.I = . Q(x)
Depending on the type of function of Q(x).

P.I is evaluated as follows:
1. P.I of f (D) y = Q(x) where Q(x) =eax for (a) ≠ 0
Case1: P.I = . Q(x) = eax = eax
Provided f(a) ≠ 0
Case 2: If f(a) = 0 then the above method fails. Then
if f(D) = (D-a)k (D)
(i.e ‘ a’ is a repeated root k times).
Then P.I = eax . xk provided (a) ≠ 0
2. P.I of f(D) y =Q(x) where Q(x) = sin ax or Q(x) = cos ax where ‘ a ‘ is constant then
P.I = . Q(x).
sin ax
Case 1: In f(D) put D2 = - a2 f(-a2) ≠ 0 then P.I =
(
f − a2 )
Case 2: If f(-a2) = 0 then D2 + a2 is a factor of (D2) and hence it is a factor of f(D).
Then let f(D) = (D2 + a2) .Ф(D2).
71 | P a g e
sin ax sin ax 1 sin ax 1 − x cos ax
= 2 2 = =
Then
f ( D ) ( D + a ) ( D )  ( − a ) D + a
2 2 2 2
 − a2 2a ( )
cos ax cos ax 1 cos ax 1 x sin ax
= 2 2 = =
f ( D ) ( D + a ) ( D )  ( − a ) D + a
2 2 2 2
 −a 2
2a ( )
3. P.I for f(D) y = Q(x) where Q(x) = xk where k is a positive integer f(D) can be
express as f(D) =[1± ]
Express = = [1± ] -1
Hence P.I = Q(x).
= [1± ] -1 .xk
4. P.I of f(D) y = Q(x) when Q(x) = eax V where ‘a’ is a constant and V is function of x.
where V =sin ax or cos ax or xk
Then P.I = Q(x)
= eax V
= eax [ (V)]
& V is evaluated depending on V.
5. P.I of f(D) y = Q(x) when Q(x) = x V where V is a function of x.
Then P.I = Q(x)
= xV
= [x - f1(D)] V
6. i. P.I. of f(D)y=Q(x) where Q(x)=xmv where v is a function of x.

1 1 1
Then P.I. =  Q( x ) = x mv = I .P. of x m (cos ax + i sin ax)
f ( D) f ( D) f ( D)
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1
= I .P. of x meiax
f ( D)
1 1
ii. P.I. = x m cos ax = R.P. of x meiax
f ( D) f ( D)
Formulae
1. = (1 – D)-1 = 1 + D + D2 + D3 + ------------------
2. = (1 + D)-1 = 1 - D + D2 - D3 + ------------------
3. = (1 – D)-2 = 1 + 2D + 3D2 + 4D3 + ------------------
4. = (1 + D)-2 = 1 - 2D + 3D2 - 4D3 + ------------------
5. = (1 – D)-3 = 1 + 3D + 6D2 + 10D3 + ------------------
6. = (1 + D)-3 = 1 - 3D + 6D2 - 10D3 + ------------------
I. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS:

1. Find the Particular integral of f(D) y = when f(a) ≠0
2. Solve the D.E (D2 + 5D +6) y = ex
3. Solve y11+4y1+4y = 4 e3x ; y(0) = -1 , y1(0) = 3
4. Solve y11 + 4y1 +4y= 4cosx+3sinx , y(0) = 1 , y1(0) = 0
5. Solve (D2+9) y = cos3x
6. Solve y111 + 2y11 - y1 – 2y = 1-4x3
7. Solve the D.E (D3 - 7 D2 + 14D - 8) y = ex cos2x
8. Solve the D.E (D3 - 4 D2 -D + 4) y = e3x cos2x
9. Solve (D2 - 4D +4) y =x2sinx + e2x + 3
10. Apply the method of variation parameters to solve + y = cosecx
11. Solve = 3x + 2y , + 5x + 3y =0
12. Solve (D2 + D - 3) y =x2e-3x

13. Solve (D2 - D - 2) y =3e2x ,y(0) = 0 , y1 (0) = -2
73 | P a g e
SOLUTIONS:
1) Particular integral of f(D) y = when f(a) ≠0
Working rule:
Case (i):
In f(D), put D=a and Particular integral will be calculated.
Particular integral= = provided f(a) ≠0
Case (ii) :
If f(a)= 0 , then above method fails. Now proceed as below.
If f(D)= (D-a)K (D)
i.e. ‘a’ is a repeated root k times, then
Particular integral= . provided (a) ≠0
2. Solve the Differential equation(D2+5D+6)y=ex
Sol : Given equation is (D2+5D+6)y=ex
Here Q( x) =e x
Auxiliary equation is f(m) = m2+5m+6=0
m2+3m+2m+6=0
m(m+3)+2(m+3)=0
m=-2 or m=-3
The roots are real and distinct
C.F = yc= c1e-2x +c2 e-3x
Particular Integral = yp= . Q(x)
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= ex = ex
Put D = 1 in f(D)
P.I. = ex
Particular Integral = yp= . ex
General solution is y=yc+yp
y=c1e-2x+c2 e-3x +
3) Solve y11-4y1+3y=4e3x, y(0) = -1, y1(0) = 3

Sol : Given equation is y11-4y1+3y=4e3x
i.e. -4 +3y=4e3x
it can be expressed as
D2y-4Dy+3y=4e3x
(D2-4D+3)y=4e3x
Here Q(x)=4e3x; f(D)= D2-4D+3
Auxiliary equation is f(m)=m2-4m+3 = 0
m2-3m-m+3 = 0
m(m-3) -1(m-3)=0 => m=3 or 1
The roots are real and distinct.
C.F= yc=c1e3x+c2ex ----→ (2)
P.I.= yp= . Q(x)
= yp= . 4e3x
= yp= . 4e3x
Put D=3
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4e3 x 4 e3 x x1
yp = = = 2 e3 x = 2 xe3 x
(3 − 1)(D − 3) 2 (D − 3) 1!
y=c1e3x+c2 ex+2xe3x -------------------→ (3)
Equation (3) differentiating with respect to ‘x’
y1=3c1e3x+c2ex+2e3x+6xe3x -----------→ (4)

By data, y(0) = -1 , y1(0)=3
From (3), -1=c1+c2 -------------------→ (5)
From (4), 3=3c1+c2+2
3c1+c2=1 -------------------→ (6)
Solving (5) and (6) we get c1=1 and c2 = -2
y=-2e x +(1+2x)e3x
(4). Solve y11+4y1+4y= 4cosx + 3sinx, y(0) = 0, y1(0) = 0
Sol: Given differential equation in operator form
( )y= 4cosx +3sinx
A.E is m2+4m+4 = 0
(m+2)2=0 then m=-2, -2
C.F is yc= (c1 + c2x)
P.I is = yp= put = -1
yp = =
Put = -1
76 | P a g e
yp =
= = = sinx
General equation is y = yc+ yp
y = (c1 + c2x) + sinx ------------ (1)
By given data, y(0) = 0 c1 = 0 and
Diff (1) w.r.. t. y1 = (c1 + c2x) + (c2) +cosx ------------ (2)
given y1(0) = 0
(2)  -2c1 + c2+1=0 c2 = -1
Required solution is y = +sinx
5. Solve (D2+9)y = cos3x
Sol:Given equation is (D2+9)y = cos3x
A.E is m2+9 = 0
m= 3i
yc = C.F = c1 cos3x+ c2sin3x
yc =P.I = =
= sin3x = sin3x
General equation is y = yc+ yp
y = c1cos3x + c2cos3x + sin3x
77 | P a g e
6. Solve y111+2y11 - y1-2y= 1-4x3
Sol:Given equation can be written as
= 1-4x3
A.E is =0
( (m+2) = 0
m=- 2
m = 1, -1, -2
C.F =c1 + c2 + c3
P.I = (1 − 4 x )
3
= 1 − 4 x3 )
= 1 − 4 x3 )
= [1+ + + (
+ …..] 1 − 4 x
3
)
−1  1 3
= 
2  2
( 1
) (
1
1 + D + 2D 2 − D + D 2 − 4D3 + − D3 ) ( )(1 − 4 x )
3
4 8 
= [1- + - D] 1-4 )
= [(1-4 )- + - (-12
= [-4x3+6x2 -30x +16] =
78 | P a g e
= [2x3-3x2 +15x -8]
The general solution is
y= C.F + P.I
y= c1 + c2 + c3 + [2x3-3x2 +15x -8]
7. Solve -8)y = cos2x
Given equation is
-8)y = cos2x
A.E is =0
(m-1) (m-2)(m-4) = 0
Then m = 1,2,4
C.F = c1 + c2 + c3
P.I =
= . . Cos2x
 1 ax 1 
 P.I = f ( D) e v = e f (D + a ) v 
ax
 
= . .cos2x
= . .cos2x (Replacing D2 with -22)
= . .cos2x
79 | P a g e
= . .cos2x
= . .cos2x
= . .cos2x
= (16cos2x – 2sin2x)
2e x
= (8 cos 2 x − sin 2 x )
260
ex
= (8 cos 2 x − sin 2 x )
130
General solution is y = yc + yp
ex
y = c1e x + c2e 2 x + c3e 4 x + (8 cos 2 x − sin 2 x )
130
8. Solve +4)y = +3
Sol:Given +4)y = +3
A.E is =0
( = 0 then m=2,2
C.F. = (c1 + c2x)
P.I = = + (3)
Now )= ) (I.P of )
= I.P of ) )
80 | P a g e
= I.P of . )
On simplification, we get
= [(220x+244)cosx+(40x+33)sinx]
and )= ),
)=
P.I = [(220x+244)cosx+(40x+33)sinx] + )+
y = yc+ yp
y= (c1 + c2x) + [(220x+244)cosx+(40x+33)sinx] + )+
Variation of Parameters :
Working Rule :
d2y
1. Reduce the given equation of the form 2 + P(x ) + Q( x) y = R
dy
dx dx
2. Find C.F.
vRdx uRdx
3. Take P.I. yp=Au+Bv where A= −  and B =  1
uv − vu
1 1
uv − vu1
4. Write the G.S. of the given equation y = yc + y p
9. Apply the method of variation of parameters to solve + y = cosecx
Sol: Given equation in the operator form is ( ----------------(1)
A.E is =0
 m = i
The roots are complex conjugate numbers.
C.F. is yc=c1cosx + c2sinx
81 | P a g e
Let yp = Acosx + Bsinx be P.I. of (1)
u -v = =1
A and B are given by
vRdx
A= −  =- dx = - =-x
uv1 − vu1
uRdx
B=  = = log(sinx)
uv1 − vu1
yp= -xcosx +sinx. log(sinx)
General solution is y = yc+ yp.
y = c1cosx + c2sinx-xcosx +sinx. log(sinx)
10. Solve ( +1)y = 100
Sol:A.E is =0
( then m = .
C.F = (c1+c2x)
P.I = = = = 100
Hence the general solution is y = C.F +P.I
y= (c1+c2x) + 100
HOMOGENEOUS LINEAR EQUATIONS
dny x n−1
d n −1 y
Equations of the form𝑥 𝑛 + P1 + − − − − − + pn = ø(x)
dx n dx n −1
82 | P a g e
Where p1 , p2 ,……… pn are real constants and ø(x) is function of x is called a homogeneous linear
equation or Euler- Cauchy’s linear equation of order n
n −1 n −1
The equation in the operator form is ( x D + p1 x D + − − − − − + Pn )=ø(x)
n n
𝑑
Where 𝑑𝑥= D Cauchy’s differential equation can be transformed into a linear equation with constant
coefficents by change of independent variable with the substitution
dz 1 xdy dy
X= e z and = and =
dx x dx dz
3
d 2 y d 2 y dy x3d d3y d2y dy
x2
2
= 2
− similarily 3
y = 3
− 3 2
+2
dx dz dz dx dx dz dx
d d
Let us denote = Dand =  can be written as 𝑥 2 𝐷2 =  (  -1) and xD= 
dx dz
x3 D3 =  ( − 1)( − 2) etc.
Example; 1 Solve ( x D − 4 xD + 6) y = x
2 2 2
Solution: Given equation ( x D − 4 xD + 6) y = x this is homogeneous differential equation

2 2 2
dz 1 xdy dy
Let
z
X= e logx = Z and = and =
dx x dx dz
d 2 y d 2 y dy
= 2 − substitution in equation we get  ( − 1) − 4 + 6 = e a differential equations
2z
x2 2
dx dz dz
with constant coefficients A .E is m − 5m + 6 = 0 The root are m=3 and m=2
2
e2 z
C.F is yc = c1e + c2 e and P.I is given by y p = = − ze
2x 3x 2z
( − 1)( − 2)
General solution is y= 𝑦𝑐 + y p = yc = c1e + c2 e − ze or y= yc = c1e + c2 e - (log x) x

2x 3x 2z 2x 3x 2
Example-2 solve ( x D − XD + 1) y = log x

2 2
83 | P a g e
Solution; Given differential equations is ( x 2 D2 − XD + 1) y = log x
dz 1 xdy dy
X=logz and X= e z and = and =
dx x dx dz
d 2 y d 2 y dy d d
x2 2
= 2 − Let us denote = Dand =  can be written as 𝑥 2 𝐷2 =  (  -1) and xD=  so
dx dz dz dx dz
that equation becomes ( − 2 + 1) y = z and A. E IS m 2 − 2m + 1 = 0 and m=1 and m=1repeated
2
z
root C.F= yc = (c1 + c2 x)e and P.I = y p = = (1 −  ) −2 z = z + 2 General solution is
z
( − 1) 2
Y= yc + y p =
Legendre`s Linear equation

n −1
dny n −1 d y
( a + bx ) + ( + ) + .......... + Pn y = Q ( x )
n
n
P 1 a bx n −1
dx dx
where P1 , P2 , P3 ,....... Pn are real constants and Q ( x ) is a function of x in called Legendre`s
linear equation.
d
This can be solved by the substitution ( a + bx ) = e z , z = log ( a + bx ) and  =
n
dy
Then ( a + bx ) Dy = b y, ( a + bx ) D 2 y = b 2 ( − 1) y , and so on
2
Problems:
d2y dy
1) Solve ( x + 1) − 3 ( x + 1) + 4 y = x 2 + x + 1
2
2
dx dx
2
2 d y dy
Solution : the given D E is ( x + 1) 2
− 3 ( x + 1) + 4 y = x 2 + x + 1
dx dx
The operator form is (( x +1) 2
)
D − 3 ( x + 1) D + 4 y = x 2 + x + 1
This is Legendre`s differential equation
du
( x + 1) Dy = u, so that x = u − 1, =1
dx
84 | P a g e
du
Now ( x + 1) Dy = u , so that x = u − 1, =1
dx
dy dy du dy
= , =
dx du dx du
d2y dy
− 3u + 4 y = ( u − 1) + ( u − 1) + 1
2
Then the equation becomes u 2 2
du du
d2y dy
u2 2
− 3u + 4 y = u2 − u +1 (2)
du du
Let u = e so that z = log u
z
(3)
d d2y dy
=  , Then u 2 2 =  ( − 1) y and u =  y (4)
dz du du
Substituting in (2) , we get (  ( − 1) − 3 + 4) y = e2 z − e z + 1
( 2
− 4 + 4 ) y = e 2 z − e z + 1 (5)
The A E is m − 4m + 4 = 0, m = 2, 2
2
C.F = yc = ( c1 + c2 z ) e2 z
z2 2z 1
P.I = e − ez +
2 4
z2 1
y = ( c1 + c2 z ) e2 z + e2 z − e z +
2 4
( log u )
2
1
y = ( c1 + c2 log u ) u 2
+ u2 − u +
2 4
( log ( x + 1) )
2
y = ( c1 + c2 log ( x + 1) ) ( x + 1)
1
+ ( x + 1) − ( x + 1) +
2 2
2 4
85 | P a g e
TUTORIAL QUESTIONS
1. Solve the D.E (D2 + 5D +6) y = ex
2. Solve (D2+9) y = cos3x
3. Solve y111 + 2y11 - y1 – 2y = 1-4x3
4. Solve the D.E (D3 - 7 D2 + 14D - 8) y = ex cos2x
5. Solve ( D − 2 D − 3D + 4 D + 4 ) y = 0
4 3 2
6. Solve ( 4 D − 4 D + 1) y = 100
2
7. Solve ( D − 6 D + 11D − 6 ) y = e + e
3 2 −2 x −3 x
8. Solve ( D − 4 ) y = 2 cos x
2 2
9. Solve ( D + 2 D + D ) y = e + x + x + sin 2 x
3 2 2x 2
10. Solve y + 2 y − y − 2 y = 1 − 4 x
lll ll l 3
( )
11. Solve D 2 − 4 D + 4 y = x 2 sin x + e 2 x + 3
d3y  1
2
2 d y
12. Solve x3 + 2 x + 2 y = 10  x + 
 x
3 2
dx dx
d2y dy
− 3x + 4 y = (1 + x )
2
13. Solve x 2 2
dx dx
d2y dy
14. Solve x 2 2
− 2x − 4 y = x4
dx dx
d2y dy
15. Solve ( x + 1) − 3 ( x + 1) + 4 y = x 2 + x + 1
2
2
dx dx
86 | P a g e
DESCRIPTIVE QUESTIONS
1. Explain Linear differential equations with constant coefficients.
2. Define Auxilary equation.
3. Explain method of variation of parameters.
4. Explain Legendre’s linear differential equation.
5. Define particular integral.

7. Solve +4)y = +3
8. . Solve -8)y = cos2x
9. Solve y111+2y11 - y1-2y= 1-4x3
10.Solve (D2+9)y = cos3x
87 | P a g e
OBJECTIVE QUESTIONS
1. The solution of (D2+9)y = cos3x is-------------------------
2. The solution of y111+2y11 - y1-2y= 1-4x3 is-------------------
3. If the roots are real and distinct then the complementary function is----------
----------
4. If the roots are real and equal then the complementary function is--------------
------
5. If the roots are complex conjugate then the complementary function is--------
------------
6. Legendre’s linear equations is of the form-------------------
7. Cauchy’s linear equation is of the form--------------------
8. The solution of ( D 2 − 4 ) y = 2 cos 2 x
is---------------------
d2y dy
9. The solution of 2
− 3 + 2 y = e5 x
dx dx is--------------------
10. The solution of y + 4 y + 4 y = 4cos x + 3sin x is ------------------

ll l
88 | P a g e
UNIT TEST PAPERS

TEST- II
SET NO-I
1. A)Explain Linear differential equations with constant coefficients
B)Solve +4)y = +3
(OR)

1. The solution of (D2+9)y = cos3x is-------------------------
----------
------
------------
is---------------------
d2y dy
− 3 + 2 y = e5 x
dx dx is--------------------

ll l
89 | P a g e

TEST- II
SET NO-II
1. A)Define Auxilary equation.
d3y  1
2
2 d y
x3 + 2 x + 2 y = 10  x + 
 x
3 2
B) Solve dx dx
(OR)
2
d y dy
2. Solve ( x + 1) − 3 ( x + 1) + 4 y = x 2 + x + 1
2
2
dx dx
2
1. The solution of (D +9)y = cos3x is-------------------------
3. If the roots are real and distinct then the complementary function is-------
-------------
4. If the roots are real and equal then the complementary function is----------
----------
5. If the roots are complex conjugate then the complementary function is----
----------------
is---------------------
d2y dy
− 3 + 2 y = e5 x
dx dx is--------------------

ll l
90 | P a g e

TEST- II
SET NO-III
1. A)Solve
(D 3
+ 2 D 2 + D ) y = e 2 x + x 2 + x + sin 2 x
B) Solve the D.E (D3 - 7 D2 + 14D - 8) y = ex cos2x

(OR)
2. Solve -8)y = cos2x
2
3. If the roots are real and distinct then the complementary function is-------
-------------
4. If the roots are real and equal then the complementary function is----------
----------
5. If the roots are complex conjugate then the complementary function is----
----------------
is---------------------
d2y dy
− 3 + 2 y = e5 x
dx dx is--------------------

ll l
91 | P a g e

TEST- II
SET NO-IV
2
d y dy
( x + 1) − 3 ( x + 1) + 4 y = x 2 + x + 1
2
2
1. A) Solve dx dx
d2y dy
− 3x + 4 y = (1 + x )
2
B) Solve x 2 2
dx dx
(OR)
2. Solve y111 + 2y11 - y1 – 2y = 1-4x3

2
----------
------
------------
is---------------------
d2y dy
− 3 + 2 y = e5 x
dx dx is--------------------

ll l
92 | P a g e
SEMINAR TOPICS
TOPIC 1:
Linear de with constant coefficients
TOPIC 2:
Linear de with variable coefficients
TOPIC 3:
Method of variation of parameters.
TOPIC 4:
Legendre’s linear equations
TOPIC 5:
Euler’s equations.
93 | P a g e
Assignment Questions
1. (
Solve D 2 + 5 D + 6 y = e x )
d2y dy
2. Solve 2
− 3 + 2 y = e5 x
dx dx
3. (
Solve D 2 − 4 D + 3 y = cos 2 x)
Solve y + 4 y + 4 y = 4cos x + 3sin x
ll l
4.
d 2 y dy
5. Solve 2
+ = x2 + 2 x + 4
dx dx
6. ( )
Solve D 3 + 2 D 2 + D y = e 2 x + x 2 + x + sin 2 x
7. Solve ( D 2
+ 5D + 4 ) y = x 2
8. Solve ( D 2
+ 1) y = x 2 e3 x
d2y dy
9. Solve x 2
2
− x + 2 y = x log x
dx dx
10. (
Solve x 2 D2 − 4 xD + 6 y = ( log x ) ) 2
94 | P a g e
APPLICATIONS
second-order linear differential equations are used to model many situations
in physics and engineering.
second-order linear differential equations works for systems of an object with
mass attached to a vertical spring and an electric circuit containing a resistor, an
inductor, and a capacitor connected in series. Models such as these can be used
to approximate other more complicated situations; for example, bonds between
atoms or molecules are often modeled as springs that vibrate, as described by
these same differential equations.
Simple Harmonic Motion
Consider a mass suspended from a spring attached to a rigid support. (This is
commonly called a spring-mass system.) Gravity is pulling the mass
downward and the restoring force of the spring is pulling the mass upward.
when these two forces are equal, the mass is said to be at the equilibrium
position. If the mass is displaced from equilibrium, it oscillates up and down. This
behavior can be modeled by a second-order constant-coefficient differential
equation.
95 | P a g e
A spring in its natural position (a), at equilibrium with a mass m attached (b),
and in oscillatory motion (c).
96 | P a g e
NPTEL VIDEOS
https://www.youtube.com/watch?v=OBhZvyhc8JQ
https://nptel.ac.in/courses/111/108/111108081/
97 | P a g e
BLOOMS TAXONOMY
TOPIC: 1.Complementary Function
ANALYSIS:
Analyze the Complementary Function.
It is the general solution of the homogeneous part of the l.d.e with constant
coefficients f(D)y=Q(x) i.e. it should have the no. of arbitrary constants same as its
order.
SYNTHESIS:
Explain the synthesis of complementary function.
the l.d.e with constant coefficients f(D)y=Q(x).
Where f(D) = Dn + P1 Dn-1 + P2 Dn-2 +-----------+Pn is a polynomial in D.

Now consider the auxiliary equation : f(m) = 0
i.e f(m) = mn + P1 mn-1 + P2 mn-2 +-----------+Pn = 0
where p1,p2,p3 ……………pn are real constants.
Let the roots of f(m) =0 be m1, m2, m3,…..mn.
Depending on the nature of the roots we write the complementary function
as follows:
Consider the following table
S.No Roots of A.E f(m) =0 Complementary function(C.F)
1. m1, m2, ..mn are real and distinct. yc = c1em1x+ c2e m2x +…+ cnemnx
2. m1, m2, ..mn are and two roots are
equal i.e., m1, m2 are equal and yc = (c1+c2x)em1x+ c3em3x +…+ cnemnx
real(i.e repeated twice) &the rest
are real and different.
98 | P a g e
3. m1, m2, ..mn are real and three yc = (c1+c2x+c3x2)em1x + c4em4x+…+ cnemnx
roots are equal i.e., m1, m2 , m3 are
equal and real(i.e repeated thrice)
&the rest are real and different.
4. Two roots of A.E are complex say yc = (c1 cos x + c2sin x)+ c3em3x +…+ cnemnx
+i -i and rest are real and
distinct.
5. If ±i are repeated twice & rest yc = [(c1+c2x)cos x + (c3+c4x) sin x)]+ c5em5x
are real and distinct +…+ cnemnx
6. If ±i are repeated thrice & rest yc = [(c1+c2x+ c3x2)cos x + (c4+c5x+ c6x2) sin
are real and distinct x)]+ c7em7x +……… + cnemnx
7. If roots of A.E. irrational say  
yc = ex c1 cosh  x + c2 sinh  x + c3em3x + ....... + cnemn x
   and rest are real and
distinct.
EVALUATION
Evaluate the complementary function of -3 + 2y = 0
: Given equation is of the form f(D).y = 0

Where f(D) = (D3 -3D +2) y = 0
Now consider the auxiliary equation f(m) = 0
f(m) = m3 -3m +2 = 0  (m-1)(m-1)(m+2) = 0
 m = 1 , 1 ,-2
Since m1 and m2 are equal and m3 is -2
We have yc = (c1+c2x)ex + c3e-2x
2) Topic: Particular integral

ANALYSIS:
Analyze the particular integral.
99 | P a g e
The p.I of l.d.e with constant coefficients f(D)y=Q(x) is the particular solution
of the R.H.S Q(x) and it is the part of the complete solution. P.I consists of no
arbitrary constants
SYNTHESIS:
Explain the synthesis of Particular integral of L.D.E with constant
coefficients.
P.I consists of no arbitrary constants and P.I of f (D) y = Q(x)
Is evaluated as P.I = . Q(x)
Depending on the type of function of Q(x).
EVALUATION:
2. Evaluate the particular integral of (D2+5D+6)y=ex
Sol : Given equation is (D2+5D+6)y=ex
Here Q( x) =e x
Particular Integral = yp= . Q(x)
= ex = ex
Put D = 1 in f(D)
P.I. = ex
Particular Integral = yp= . ex
100 | P a g e
y=c1e-2x+c2 e-3x +
101 | P a g e

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I B.

S.NO TOPIC PAGE NUMBER

3 LECTURE NOTES 9-51 67-85

6 OBJECTIVE BANK 54-55 88

7 UNIT TEST PAPERS 56-59 89-92

10 REAL TIME APPLICATIONS 62 95-96

12 BLOOM’S TAXONOMY 64-66 98-101

• Methods of solving the differential equations of first and higher order.

UNIT-I: First Order ODE

UNIT-III: Multivariable Calculus (Integration)

UNIT-IV: Vector Differentiation

UNIT-V: Vector Integration

Designation: Asst. Professor Branch:

S.N UNIT CLASS TOPIC T/R DATE DATE Remarks

1. LH1 Overview of differential T1/R1

5. LH5 non-exact diff. equations T1/R1

6. LH6 non-exact diff. equations T1/R1

19. LH19 Decay

with constant coefficients

31. LH31 Method of variation of T1/R2

34 LH34 Cauchy-Euler equations T1/R2

35 LH35 Legendre’s equations T1/R2

36 LH36 Eigen values and Eigen vectors

38 LH38 Test T1/R2

43 LH43 Change of variables T1/R2

44 LH44 Change of variables T1/R2

46 LH46 Change of order of integration T1/R1

47 LH47 Change of order of integration T1/R2

48 LH48 Evaluation of Triple integral T1/R1

49 LH49 Evaluation of Triple integral T1/R1

50 LH50 Change of variables in Triple T1/R1

55 LH55 Volumes in double integral T1/R2

56 LH56 Volumes in Triple integral T1/R2

57 LH57 Centre of mass T1/R2

59 LH59 Centre of gravity T1/R2

61 LH61 Active Learning(Flipped Class T1/R2

69 LH69 Problems on Irrotational vectors T1/R2

70 LH70 Problems T1/R2

71 LH71 Vector operators T1/R2

72 LH72 Vector operators T1/R2

73 LH73 Vector Identities T1/R2

75 LH75 PPT T1/R2

77 LH77 Test T1/R2

79 LH79 Line integral T1/R2

80 IV LH80 Line integral T1/R2

81 LH81 Surface integral T1/R2

85 LH85 Green’s theorem T1/R2

87 LH87 Gauss divergence theorem T1/R2

88 LH88 Gauss divergence theorem T1/R2

89 LH89 Stoke’s theorem T1/R2

90 LH90 Stoke’s theorem T1/R2

91 LH91 Stoke’s theorem T1/R2

92 LH92 PPT T1/R2

93 LH93 Active learning(Muddiest point) T1/R2

1. Higher Engineering Mathematics by B.S. Grewal, Khanna Publishers.

2. Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley& Sons,

REFERENCES: 1.Paras Ram, Engineering Mathematics, CBS publishes

FACULTY H.O.D PRINCIPAL/DIRECTOR

& FIRST DEGREE

is the highest derivative in that equation.