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DIFFERENTIAL EQUATION x y dx dx solution x dx y x dy y x y dx
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DIFFERENTIAL EQUATION x y dx dx solution x dx y x dy y x y dx
 
 
 
 
 
ORDINARY DIFFERENTIAL EQUATION [ODE]  
Step(i) 
 
Lecture Plan 
Part 1: 
 Basic definitions / terms: order, degree, formation, linearity, homogeneous etc. 
 First order, degree one: How to solve! 
                 Methods: 
                    (i) Variables seperable 
                    (ii) Homogeneous differential equations 
                    (iii) Exact differential equations  
                    (iv) Linear differential equations  
                    (v) By changing independent variable 
 First order, not of degree one: How to solve! 
                Methods:  
(i) 
Solvable for 
,
,
p x y   
(ii) Clairaut’s form reduction into it 
(iii) Singular solutions, Extraneous loci 
 Orthogonal trajectories: Cartesian form, Polar form  
 Decay Problems (Formation of differential equations and then solve them) 
 
Part 2: 
    Higher Order ODEs 
                      Introduction:  
                     Existence and uniqueness of solution of an IVP,  
Decoding thesyllabus


Differential Equation Introduction
ODE
PDE
First Order


Higher Order Order 2

Of degree one
Not of degree one
With constant coefficients
With variable coefficients
 
                     Linear combination of solutions of a Linear homogeneous differential equation,  
                     Linear dependence of solutions of a differential equation, 
                  Wronskian of solutions, Abel’s formula for calculating Wronskian,  
 
                  All possible solutions of a linear homogeneous differential equations (C.F.) 
    
    ODEs with constant coefficients 
 
1
2
0
1
2
1
1
2
...
n
n
n
n
n
n
n
n
d y
d
y
d
y
dy
y
f x
dx
dx
dx
dx












 


 
….(1) 
where 
0
1
,
,....,
n
 
  all are constants. 
Solving above differential equations through two steps  
Step1- Finding C.F. of 
1
0
1
1
1
1
...
0
n
n
n
n
n
n
d y
d
y
dy
y
dx
dx
dx









 


 
….(2) 
Step2- Finding particular integral (P.I.) of differential equation (1) through different methods. 
 
Methods to Find P.I. 
(i) When 
 
ax
f x
e

 
(ii) When 
 
1
0
1
1
...
m
m
n
n
f x
a x
a x
a x a




 

 
(iii) When 
 


sin
f x
ax b


 or 


cos ax b

 
(iv) When 
 
 
ax
f x
e
v x


 
(v) When 
 
f x is some ordinary function 
 
      ODEs with variable coefficients  
               Case (i): Restricted / special differential equations in some particular forms like – 
                         1-Cauchy-Euler’s differential equations  
                       
 
1
1
0
1
1
1
...
n
n
n
n
n
n
n
n
d y
d
y
dy
x
x
x
y
f x
dx
dx
dx









 


 
                        where 
0
1
,
,.... n
 

  all are constants. 
                 To solve 
                               we change independent variable x  into some other independent variable z                           
through a substitution 
z
x e

. Basically we try to reduce the given differential equation into a differential 
equation with constant coefficients. 
 
                           2- Legendre’s differential equations  






 
1
1
0
1
1
1
...
n
n
n
n
n
n
n
n
d y
d
y
dy
a bx
a bx
a bx
y
f x
dx
dx
dx











 



 
Putting a bx
t
 
 and then 
z
t
e

 for solving. 
                  Case (ii): ODEs of the form 
 
 
 
2
2
d y
dy
p x
Q x y R x
dx
dx



 
….(1) 
Notice here that 
 
p x  and 
 
Q x  may be any function here. We need to solve such differential equations. 
             Methods:  
                              (i) Solving (1) by using some known integral / solution of  
                              
 
 
2
2
0
d y
dy
p x
Q x y
dx
dx



 
….(2) 
                           by different kind of assumptions like that known integral is  
2
2
2
,
,1
,
,
...
x
x
x x
x e e
Actually 
we remember rules to get idea about known integral of (2). 
                             (ii) By reducing (1) into normal form (a differential equation in which first order 
derivative is removed by some way) 
2
2
d v
Iv S
dx
 
 form. 
                             (iii) Method of variation of parameters: This method is designed to find particular integral 
of (1) with the help of two linearly independent solution of (2). 
                              (iv) By changing independent variable: Here we have a set method to change x  into 
some new independent variable z . 
 
 
 
 
 
 
 
 
 
 
 
 
BASIC DEFINITIONS  
 
Differential equation: An equation which involves derivatives, dependent variable/s, independent 
variable/s. 
e.g. (1) 


2
3
2
,
1,1
x
dy
x y
x e x
dx


 
. Here x  is independent variable and y  is depending on x  so 
dependent variable. 
(2) 


3
2
3
2
2
cos ,
,
x
d y
dy
d y
e
x x
a b
dx
dx
dx




 
(3) 


,
2
cos ;
,
t
dz
dy
e
y
z
t t
a b
dt
dt
 



 is a system of differential equations. 
(4) 




2
2
3
2
;
,
,
,
z
z
z
x y
x
x
a b y
c d
x
x
y

 







 
 
here x  and y  are independent variables and z  is dependent. 
Note:   Variables for which domain is defined are called independent variables. 
 
Classification of differential equations  
ODE: If number of independent variable is one. 
PDE: If number of independent variable are more than one. 
e.g. (1), (2), (3) are ODEs, (4) is a PDE. 
 
Point: Now next definitions / explanations we will have about ODEs. 
 Order of differential equation: The order of the highest order derivatives present in that differential 
equation is called order of that differential equation. 
e.g. Differential equation (1) is of order 1. 
Differential equation (2) is of order 3. 
 
 Degree of a Differential equation: The power of the highest order derivative, present in that differential 
equation is called degree of that differential equation provided that derivatives are radial free. 
e.g. 
(1) 
3
4
1
dy
y
dx


 




, order = 1 
For degree, we try to make derivatives radial free. 
 given differential equation can be written as  
 


4
3
1
dy
y
dx
 
 
 degree = power of highest order derivative = 1 
e.g. (2) 
2
2
3
4
2
3
4
cos
2
d y
d y
d y
e
x
y
dx
dx
dx




 



 
here order of differential equation = 4 
degree of differential equation is NOT DEFINED (since derivatives are functions of exponential 
trigonometric, so not radical free)   
e.g. (3) 
2
5
2
5
cos
cos
d y
x y
y
dx








 
 given differential equation can be managed as 


2
5
1
2
5
cos
cos
d y
x y
y
dx










 
now derivatives are radial free. So degree of differential equation is 2. 
 Linear differential equation: A differential equation is said to be linear ‘in the variable y ’ if it satisfies 
following three conditions: 
(i) power of y  does not exceed by 1 
(ii) power of derivatives does not exceed by 1 
(iii) No term present as product of derivatives or derivatives and y  
Note: here powers are either 0 it 1. 
e.g. (1) 
 
 
dy
p x y Q x
dx


 ; is a liner differential equation of order one in the variable y . 
(2) 
2
4
4
2
x
d y dy
y e
dx
dx



 ; is a linear differential equation of order 4 in the variable y . 
Note: Linearity has nothing to do with order of derivatives. It only cares about degree of the variable and 
equation for which we are checking the linearity. 
 
 Homogeneous differential equation:  If some differential equation can be written in the form 
 
Ly
f x

 ; where L is the differentiation operator and 
 
 
0 . .
0
f x
i e L y
 

 
e.g. (1) 
3
2
3
3
2
2
3
5
d y
d y
dy
y
x
dx
dx
dx




 can be written as 
3
2
3
3
2
2
3
5
d
d
d
y x
dx
dx
dx










 
 
3
L y
x


 is non homogeneous. 
 
(2) 
3
2
3
2
2
3
5
0
d y
d y
dy
y
dx
dx
dx




 is homogeneous  
Note: The linearity or non linearity of L may again be classified as : linear homo, linear non homo non-
linear homogeneous, Non linear non homogeneous. 
 Solution of a differential equation: A function or curve which satisfies/represents the given differential 
equation, is known as solution of that differential equation. 
e.g. (1) 
2
2
1
2
x
y


 satisfies the differential equation 
0
xdx ydy


. So it is a solution of it. 
(2) 
 
1
2
cos
sin
y x
c
x c
x


 with 
1c  and 2c  as arbitrary constants, satisfies  
2
2
0
d y
y
dx
 
. So 
 
y x
 is 
the solution of this differential equation. 
 General solution of a differential equation: A solution which contains that many arbitrary constants 
as the order of that differential equation is known as the general solution. 
e.g. 
 
1
2
cos
sin
y x
c
x c
x


 with 
1c  and 
2c  as arbitrary constants is the general solution of the 
differential equation 
2
2
0
d y
y
dx
 
 
 Particular solution of a differential equation: by giving some particular values to arbitrary constants in 
the differential equation’s general solution, we get a particular solution of that differential equation. 
e.g. if we take 
1
2
1,
3
c
c

 
 in 
1
2
cos
sin
y c
x c
x


 we get 
  cos
3sin
y x
x
x


 as a particular 
solution of 
2
2
0
d y
y
dx
 
. 
 Singular solution:  A solution which does not contain any arbitrary constant and is not a particular 
solution. 
 
Note: As the general solution of a differential equation contains arbitrary constants. It means the general 
solution represents a family of curves. 
e.g. 
2
2
2
x
y
c


 with c  as arbitrary constants represents family of concentric circles and it is the general 
solution of differential equation 
0
xdx ydy


. 
Differential equation represented by a given family of curves  
As we know the general solution of a differential equation contains arbitrary constants. Now if we 
remove/eliminate these arbitrary constants with the help of differentiation, we get a differential equation 
represented by this given family of curves i.e. the general solution. 
 
Order of differential equation = Number of arbitrary constants   
 e.g. Let us consider a family of circles which touches the y -axis at the origin. 
 
 
 
Observe centres of circles : 

,0
h
 where h  is arbitrary constant with respective radius h . 
 equation of circles is represented by  




2
2
2
0
x h
y
h




 ; h  is arbitrary constant. 
Now to get differential equation represented by (1);  
Differentiating (1) w.r.t x , we get 


2
2
0
dy
x h
y
dx
 

 
….(2) 
Note: Here we may eliminate h  from (2), if we differentiate one more time. But we have to maintain that 
order of differential equation must be equal to arbitrary constant which is 1 here. 
So using value for h  from (2) in (1), we get required differentials equation. 
 from (2) 
dy
h x y
dx
 
 
 from (1), we have  
2
2
2
dy
dy
x
x y
y
x y
dx
dx























; which is a differential equation of order one but [not linear in 
the variable y ] 
Initial Value Problem (IVP): A differential equation with given conditions at initial point of domain of 
independent variable. 
e.g. (1) 




0
0
,
;
dy
f x y
y x
y
dx


 where 


0
,
x
a b

 with 


,
x
a b

 is an IVP 
x
y

 
(2) 
 
 
2
2
,
0
1,
' 0
2
x
d y
y e
y
y
dx
 


 where 


1,1
x 
 is an IVP. 
Solution of an IVP: A solution / curve which satisfies the differential equation as well as given initial 
conditions is known as solution of that IVP. 
Boundary Value Problems: A differential equations with conditions at boundary points. 
e.g. 


3
3
2
;
2,2
x
d y
dy
y e
x
dx
dx

 
 
 with 


 
2
0,
' 2
0
y
y
 

 
 
CHAPTER – 1 
DIFFERENTIAL EQUATIONS OF ORDER ONE AND DEGREE ONE 
  To solve above differential equations, we have different following methods: 
Method (I): Let’s consider a differential equation 


,
dy
f x y
dx

 
….(1) 
Now we try to reduce (1) into the form  
 
 
dy
x
y
dx




 
 
dy
dx
y
x




 
Now on integrating above differential equation, we get the general solution as 
 
 
1
1
dy
dx c
y
x






 
where c  is some integration constant. 
e.g. 
2
2
2
dy
dy
dy
dx
x
y
x y
dx
dx
y
x

 

 

  
….(1) 
on integrating 
1
loge y
C
x
  
 as the general solution of (1) 
Method (II): Differential equation of the form 




1
2
,
,
f
x y
dy
dx
f
x y

  
….(1) 
where 


1
,
f
x y
 and 


2
,
f
x y
 both are homogeneous functions of some degree. 
Note: (1) A function is said to be homogeneous of degree if each term in that function of some degree n . 
e.g. 


3
3
1
,
n
n
n
f x y
x y
y
x
y



 

 
Note (2): Clearly a homogenous function of degree n  can be written as  
e.g. 


3
1
,
n
n
n
n
y
y
y
f x y
x
x
x
x



































n
x
y x



 
 
So differential equation (1) can be written as 






n
n
x
y x
dy
f y x
dx
x
y x




 
….(2) 
To solve (2): put y
vx

: as from this substitution (2) will be reduced into variables separable form as  
 
dv
v x
f v
dx


 
dy
dv
y vx
v x
dx
dx
   
 
i.e. 
 
dv
dx
f v
v
x


 
 Required general solution of (1) is given by  
 


1
1
dv
dx c
x
f v
v





: c  as arbitrary constant. 
Note: To solve differential equations of the form 
1
1
1
2
2
2
a x b y c
dy
dx
a x b y c





 
….(1) 
Case (i): If 



1
1
2
2
a x b y
a x b y




 then by substituting 
1
1
a x b y v


 in (1), we get a differential 
equation in variable separable form which can be easily solved then.     
Case (ii): If 



1
1
2
2
a x b y
a x b y




, then we try to reduce (1) in the homogenous form i.e.  


dy
f y x
dx

 
….(2) 
by a procedure. 
Procedure: 
Step (i) Put x X h
 
, y Y k
 
dx dX
 
, dy dY

 
with h  and k  so chosen that (1) is reduced into the form (2) i.e. we have to pick h  and k  from 
1
1
1
2
2
2
0
0
a h b k c
a h b k c

  


  
 
….(3) 
i.e. We have 


1
1
2
2
a X bY
dY
g Y X
dx
a X b Y




 
Now substitute Y vX

, solve like above usual process to solve homogenous differential equations 
[Don’t forget to substitute back as 
,
X x h Y
y k
 
 
 where h  and k  are found by solving (3)] 
 
Method (III) Exact differential equations  
 
Definition: A differential equation 


,
dy
f x y
dx

 is said to be exact if there exists 


,
U x y C

 such 
that the total differentiation of 


,
U x y C

 gives the differential equation 


,
dy
f x y
dx

; without any 
substitution or manipulation. 
e.g. the differential equation 
dy
x
dx
y
 
 is exact 
 there exists 
2
2
2
x
y
C


 such that  


2
2
2
x
y
d
d C








0
xdx ydy



dy
x
dx
y
  
 
The necessary and sufficient condition for exactness 
The necessary part: If the differential equation 




,
,
0
M x y dx N x y dy


 is exact then the condition 
M
N
y
x





 mercenarily holds. 
Proof: Let if 




,
,
0
M x y dx N x y dy


 is exact  
….(1) 
 by definition: 


,
u x y C


 
….(2)  
s.t. Total differential of (2) gives (1)  
From (2) 
0
0
u
u
du
dx
dy
x
y


 




 
….(3) 
Now (2) and (3) must represent same differential equation 
,
u
u
M
N
x
y







 
2
2
,
M
u
N
u
y
y x
x
x y








 

 
M
N
y
x






 holds 
Hence proved. u  is continuous 
2
2
u
u
y x
x y




 
 
 
The sufficient part: The condition 
M
N
y
x





 
is sufficient for the differential equation 




,
,
0
M x y dx N x y dy


 to be exact differential equation. 
 
Proof: Let if 
M
N
y
x





 
….(1) 
Holds well. Then we have to show 


,
U x y C


 s.t. it’s total differentiation gives 
0
Mdx Ndy


 for 
this. 
Let 


,
M x y dx u


  
….(2) 
treating y  as constant in the integration  
2
u
y
M
u
N
M
x
y x
y
x
y
x





 


 






 






 
….(3) 
Using (1) 
 On integrating w.r.t x  
 
u
N
f y
y




 
….(4) 
where 
 
f y
 is some integration constant. 
Note: Since we have treated y  as constant initially and integrating w.r.t x , so the integration constant may 
be function of y . 
Now let’s consider 
0
Mdx Ndy


 
On using (3), (4) ;  
 
0
u
u
dx
f y dy
x
y














 
 
0
u
u
dx
dy
f y dy
x
y














 
 


 
du d
f y dy
d c
 


 
 
 
d u
f y dy
d c








 
u
Mdx
N
dy C
y














; where c  is some arbitrary constant. 
(5) is known as the general solution of the differentiation equation 
0
Mdx Ndy


 
Have 
you 
noticed 
that, we 
have 
shown 
the 
existence 
of 


,
U x y C

; where 


,
u
U x y
Mdx
N
dy
y













such that its total differentiation gives 
0
Mdx Ndy


 
 
0
Mdx Ndy



 is exact differential equation hence proved. 
Exam Point 
To solve 
0
Mdx Ndy


 
Let if 
M
N
y
x





 i.e. exact differential equation  
The general solution is given by  
u
Mdx
N
dy C
y













 
where u
Mdx
 
 treating y  as constant. 
Note: The term 
u
N
y



 comes out as terms independent of x  in N. But here we have to take care in 
finding such terms. 
Note: If the differential equation 
0
Mdx Ndy


 is not an exact differential equation then we can try to 
make it exact. 
Integrating factor of 
0
Mdx Ndy


 
A factor / some functions f which makes 
0
Mdx Ndy


 as exact differential equation, is called an 
integrating factor for this differential equation. 
i.e. 
0
f Mdx
f Ndy




 is exact 




f M
f N
y
x








 
Exam point- Below mentioned Rules can be proved by using above logic. UPSC asked such theoretical 
questions as well. For example if Mx + Ny is an integrating factor of  Mdx + Ndy = 0 then by above logic 
take f = Mx + Ny and  proceed. 
Rules to find I.F. for 
0
Mdx Ndy


 
Rule (I): If 
0
Mdx Ndy


 is homogenous 
then I.F.
1
:
0
Mx Ny
Mx Ny




 
Rule (II): If the differential equation is of the form 




1
2
0
f
x y ydx
f
x y xdy




; then  
I.F. 
1
Mx Ny


 ; where 




1
2
,
M f x y y N
f x y x




 
Provided 
0
Mx Ny


 
 
Rule (III): If 
 
1 M N
f x
N
y
x












 i.e. function of x  alone. 
then I.F. 
 
f x dx
e

 
Rule (IV): If 
 
1 N M
g y
M
x
y












 i.e. function of y  alone  
then I.F. 
 
g y dy
e

 
Rule (V): If M and N are polynomials then we introduce x
y



 and 
,   such that the resulting 
differential equation becomes exact. 
i.e. 




x
y M
x y N
y
x


 








 
compare and get 
,  . 
 
Method (IV): Linear differential equations  
 
 
dy
P x y Q x
dx


 
….(1) 
Let’s try to find an integrating factor of (1) i.e. to find 
 
f x
 such that  
 
 
 
 
 
dy
f x
P x y f x
Q x
f x
dx





 is directly integrable (exact’s other definition)  
So, if we multiply by 
 
P x dx
e

 in (1), 
 
 
 
 
 
P x dx
P x dx
P x dx
dy
e
e
P x y e
Q x
dx











 
 
 
 
P x dx
P x dx
d
d
y e
e
Q x dx
dx
dx














, which is integrable; on integrating, we get  
 
 
 
P x dx
P x dx
y e
e
Q x C









; where c  is some integration constant. 
So the factor 
 
P x dx
e

 is known as an I.F. of differential equation (1) and (2) is known as the general 
solution of (1). 
Note: If the differential equation is of the form 
 
 
dx
P y x Q y
dy


 i.e. linear in the variable x . 
Then I.F. 
 
P y dy
e

 
General solution is  
 
 
 
 
P y dy
P y dy
x e
Q y e
c







; where c  is arbitrary constant. 
 
Existence and Uniqueness of solution of 1
st
 order ODE’s 
Let’s consider a definition of equation (IVP) 




0
0
'
,
;
y
f x y
y x
y


 
Now we are interested to know  
(1) The given IVP has some solution or not?i.e. existence of solution of IVP 
(2) if solution exists then ; is it unique? 
e.g. (i) 
 
'
2
,
0
1
y
x y


 (ii) 
 
'
1 ;
0
1
xy
y
y
 

 (iii) 
 
2
2
'
1
0 ;
0
1
y
y
y
 


 
(i) 
2
2
dy
x
dy
xdx
dx
  
2
y
x
c
  
 at 
0,
1
x
y


1 0
1
c
   
2 1
y
x
  
 
i.e. the given IVP has unique solution. 
(ii) 
1
dy
y
dx
x
x

 
 I.F. 
1
log
1
dx
x
x
e
e
x






2
1
y
dx c
x
x

 


1
c
x
 
1
y cx
  
  infinitely 
many solutions. 
(iii) 


2
2
1
dy
y
dx


 






2
1
dy
y
dx
   
 No solution over R  
Existence and Uniqueness Theorem 
Let’s consider an IVP 


,
dy
f x y
dx

; 


0
0
y x
y

  if 


,
f x y
 is continuous in a closed domain 




:
,
,
R
x y x
h y
k


 i.e. it is bounded in R . 
Let 


,
f x y
M

 
Then   at least one solution of 


'
,
y
f x y

; 


0
0
y x
y

 defined for 
0
x x 
 
, where 
min
,
k
h
m








. If 


,
f x y and 
f
y


 are continuous in the closed domain R  i.e., 


,
f x y
 and 
f
y


 are 
bounded in R . 
Then   at most one solution of 




0
0
'
,
;
y
f x y
y x
y


 defined for 
0
x x 
 
 where 
min
,
k
h
M








 
 
*On combining above two statements ; If 


,
,
f
f x y
y


 are bounded in  R  
 
Then 


'
,
y
f x y

; 


0
0
y x
y

 has unique solution defined for 
0
x x 
 
; where 
min
,
k
h
M








 
Note: 

0
0
,
x y
 initial point i.e. 
0
0
,
x y R  i.e. R  is of the form 




0
0
,
:
,
R
x y
x x
h
y y
k





 
Q. Let’s consider an ODE 
 
'
sin
;
0
2
y
xy
y y
 

. It has  
(a) Unique solution  
(b) No solution 
(c) Infinitely Many Solution  
(d) None of the above 
sin
dy
xy
y
dx


 


,
sin
f x y
xy
y
 
 is continuous in a closed region containing 

0,2  
cos
f
x
y
y

 

 is continuous. 
By Existence and uniqueness theorem (EOT) given IVP has uniqueness solution. 
sin
dy
xy
y
dx


sin
dy
y
xy
dx
 

 
Q. Consider the ODE 
 
2
'
1
;
0
0
y
y
y
 

 in the Rectangle 




,
100,
1
S
x y x
y



, then  
(a) The given IVP has a unique solution in nbd 

0,0  
(b) Given IVP has unique solution 
 
y x
 defined for 
1
2
x 
 
(c) The solution of above IVP is defined for 
,
2 2
x
 


 




 
(d) None of the above  


2
,
1
f x y
y
 
 




,
:
100 ,
1
S
x y
x
y



 


,
2
f x y 
 
,
f
f
y


 are continuous in S 
 By EUT ; Given IVP has unique solution defined for 
0
x

 
 where 
1
1
min 100,
2
2









 i.e. 
defined for 
1
2
x 
 
 
2
f
y
y



 
2
'
1
y
y
 
. 
2
1
dy
y
dx
 
1
tan y
x c


 


tan
y
x c
 

; 
 
0
0
y

 
1
0
tan
tan
c
c
x

 
 
0
; 
tan
y
x

 
Also can say; Given IVP has unique solution in nbd 

0,0  
2
'
1
y
y
 
 
2
1
dy
y
dx
 
2
1
dy
dx
y



1
tan y
x c


 


tan
y
x c
 

 
 
0
0
y

 
1
0
tan
tan
0
0
c
c

 
 

tan
y
x
 
y

 is defined for 
,
2 2
x
 


 




 
Q. Let’s consider an IVP 
 
'
;
1
0
y
x y
y


. It has  
(a) No solution  
(b) Unique solution in nbd of 

1,0
 
(c) finitely many solution (more than one) 
(d) infinitely many solution  
f  is continuous  and bounded Unique solution in nbd of 

1,0
 
Q. Let’s consider an IVP 
 
2
5
'
60
;
0
0
y
y
y


. Then discuss about solution of the given IVP. 
Ans. f  is continuous in a closed region having 

0,0   at least one solution of given IVP 
But 
3
5
2 1
60
5
f
y
y




i.e. 
f
y


 is not bounded for 
0
y   EUT fails  
i.e. we cannot say anything about uniqueness of given IVP. 
*In this case we have to solve the given IVP and then only we can state something about uniqueness of 
solution. 
(i) 
 
2
5
'
6
;
0
0
y
y
y


0
y 
 satisfies the given IVP
'
0
y


, 
0
y
 
 is a solution  
(ii) 
2
5
60
dy
y
dx

60
dy
dx
dx
 


2
5
3
5
1
2
5
3
60
60
1
5
y
x c
y
x c
 



 



3
5
3
36
5
y
x
c
 

 
0 0 c
  
3
5
36
y
x


i.e., the given IVP has more than one solution. 
Note:  Let’s consider an IVP 
'
; 0
1
y
y


 
, with 
 
0
0
y

. This IVP has infinite no. of solution. 
 
e.g. 
 
'
2
;
0
0
y
y
y


 
Let 
 


2
0 ;
;
x
r
y x
x r
x
r


 



 
r

R
 
 


0 ;
'
2
;
x
r
y x
x r
x
r


 



 
 
y x
 satisfies the given IVP also r  is arbitrary  
i.e., the given IVP has infinitely many solutions. 
Q. The IVP 
'y
y

, 
 
0
,
0
y
 


 has  
(a) At least two solution if 
0
 
 
(b) No solution for 
0
 
 
(c) At least one solution if 
0
 
 
(d) A unique solution for 
0
 
 
 
Trajectories  
It is a mathematical curve which cuts some given family of curves according to some given condition or 
some law. 
Introduction 
 
2
2
2
1F
x
y
a



 ; a  is parameter; 
2;
F y mx

; m  is parameter  
y
x

P
Q
 
2
2
2
1 ;
F x
y
a


2
2
0
dy
dy
x
x
y
dx
dx
y




 
 
  






 
2 ;
dy
F y mx
m
dx
 

 
Intersection points for 
2
2
2 ;
x
y
a
y mx



 
2
2 2
2
2
1
a
x m x
a
x
m

   

; 
2
2
2
2
,
,
,
1
1
1
1
a
am
a
am
P
Q
m
m
m
m



















 
At the point P  


1
2
2
2
1
1
1
F
F
dy
dy
a
m
m
dx
dx
am
m



 


 

 


 


 
At the point Q 


1
2
2
2
1
1
1
F
F
dy
dy
a
m
m
dx
dx
am
m


 



 

 


 



 
From the above example it can be seen that the family of straight lines passing through origin cuts every 
member of a given family of curves 

2
2
2
x
y
a


 at an angle of 90°. 
Determination of Oblique Trajectories of Some given family of curve 
 
 
1
2
  
  
2
1
  

    
2
1
2
1
tan
tan
tan
1 tan



 



 ….(1) 
Let 
1F  ; some given family of curves i.e. 
1
1
tan
F
dy
dx








 is known  
Suppose we have to find another family of curve 
2F  which cuts the family 1F  at an angle   (at the point) 
P . 
T
'
T
1
2

P
1F
2F
 
Note: If in equation (1); we put values of   and 
1
1
tan
F
dy
dx








 we get a differential equation of first 
order and in the equation 
2F
dy
dx






 will be present. 
On solving this differential equation we get the required trajectories. 
Q. Let’s consider a family of semi-cubical parabolas 
2
3
ay
x

 ; a  is some parameter. If family of ellipse 
2
2
2
3
x
y
c


; is parameter cuts the given family at an angle '
'  then what is the value of  . 
(a) 
2

 (b) 
3

 (c) 
4

 (d) 2  
2
3
ay
x

2
2
3
dy
ay
x
dx


1
2
3
3
2
2
2
2
3
3
3
3
2
2
2
2
dy
x
x
x
x
dx
ay
a
ax
x
y












3
2
y
x

 
For 
2F ; 4
cos
0
dy
x
y
dx


2
4
2
6
3
F
dy
x
x
dx
y
y











 
1
2
3
2
1
2
3
F
F
dy
dy
y
x
dx
dx
x
y











 












 
Solution: 
2
1
2
1
tan
1
dy
dy
dx
dx
dy
dy
dx
dx
















 

 
 


 

 
If 
2

 
; 1/0
2
1
2
1
1
dy
dy
dx
dx
dy
dy
dx
dx















 

 
 


 

2
1
1
dy
dy
dx
dx

 


 

 


 

 
2
3
ay
x

 
….(1) 
3
2
x
y
a

3
2
dy
x
dx
a
 
3
2
3
2
x
dy
x
dx
y
 
3
2
3
2
dy
xy
dx
x
 
 
3
2
dy
y
dx
x
 
 
….(2) 
 Now replacing 
dy
dx
 by 
dx
dy

 
 
3
3
2
2
dx
y
ydy
xdx c
dy
x
  


2
2
3
2
2
2
y
x
c



2
2
3
2
y
x
c



 
 1st curve is orthogonal to 2nd curve 
Q. xy
c

 is some given family of curves and 


1
,x y
c


 is some other family of curve which cuts the 
earlier family at an angle ‘
4

’ then  
(a) 

2
2
,
2
x y
y
xy x




 
(b) 

2
2
,
2
x y
y
xy x




 
(c) 

2
2
,
2
x y
y
xy x

  

 
(d) NOT 
0
dy
x
y
dx
 
dy
y
x
dx



 




2
2
tan
4
1
dy
y
dx
x
dy
y
x
dx





 
 



 
2
2
1
y
dy
dy
y
x
x
dx
dx





  














2
1
1
y
dy
y
x
x
dx



  








 


2
1
1
y
x
dy
y x
x y
y
dx
y x
x y
x








 








1
1
dv
vx
x C x
dx
vx

  


2
1
1
dv
v v v
x
dx
v
  



 
2
1 2
1
dv
v v
x
dx
v
 



 




2
2
2
2
x
y
x y dx
x y dy
xy
c

 
  

2
2
2
x
xy y
c
 
 
2 2
1
1
v
v
dx
dv
v
x
  



 


2
1
log
2
1
log
2
v
v
x

  
 


2 2
1
1
v
v
dv
x
dx
v









2
2
1
2
0
2
1
v
dv
dx
x
v
v




 


2
2
log
2
1
0
v
v
x

 

2
2
2
2
1
0
y
y
x
x
x










2
2
2
0
y
yx x
 
 
2
2
2
0
y
yx x
  
 
 
Note: A family of curves is said to be self orthogonal if orthogonal trajectories for the family is the family 
itself. 
 
*If on replacing 
dy
dx
 by 
dx
dy







  in the differential equation of the given family, we get the same differential 
equation then we say that family is self orthogonal. 
Q. The system of confocal conic 
2
2
2
2
1
x
y
a
b






; where   is self-orthogonal. 
Solution 
2
2
2
2
2
2
0
0
dy
dy
y
y
x
x
dx
dx
a
b
a
b





 






 




2
2
2
2
0
dy
dy
dy
x b
y a
x y
b x a y
dx
dx
dx








 

 

 









 
2
2 dy
dy
b x a y
x y
dx
dx


 

  



 


 

 


2
2
2
2
2
2
dy
b x a y
dy
dx
a
a
a
b x
x y
dy
dx
x y
dx




  



 




 






2
2
2
2
2
2
dy
b
b x a y
dx
b
b
a y dy dx x y dy dx
dy
x y
dx



 




 
Putting in equation (1)  















2
2
2
2
2
2
2
2
1
x x y dy dx
y x y dy dx
dy
dx
x y
x y
a
b
dx
dy
a
b x
a
b y dy dx




  

 


= 0 
Replacing 
dy
dx
 by 


dy
dx

 in (2) 
2
2
dx
dy
x y
x y
a
b
dy
dx























 
2 2
dy
dy
x y
x y
a b
dx
dx



 











 



 
Which is same as (2) 
 The given curve is self orthogonal. 
Orthogonal Trajectories in Polar Form 
 
tan
rd
dr

 
 
To get orthogonal Trajectories Re place 
d
r
dr

 by 
1
d
r
dr


 or 
2
dr
r d dr
d


 
 
Q. The orthogonal trajectories of the system 
sin
n
n
r
n
k
 
, k  being a parameter  
(a) 
cos
n
r
n
a
 
 (b) 
cos
n
r
a
 
 (c) 
2 cos
r
n
a
 
 (d) 
tan
n
r
n
a
 
 
Solution 
sin
n
n
r
n
k
 
1 sin
cos
0
n
n
d
nr
n
r n
n
dr







1 sin
cos
0
n
d
nr
n
r
n
dr



 








 
tan
d
r
n
dr



 
 
1
tan n
d
r
dr



 
cot
d
r
n
dr




cot
d
dr
n
r




tan
dr
n d
r
 




  
 
 
Page 1 of 14 
 
DIFFERENTIAL EQUATION AND THEIR SOLUTIONS  
Assignment - 1 
Classify each of the following differential equations as ordinary or partial differential equations; state 
the order of each equation; and determine whether equation under consideration is linear or nonlinear. 
1. 
2
x
dy
x y
xe
dx


 2. 
3
2
3
2
4
5
3
sin
d y
d y
dy
y
x
dx
dx
dx




 3. 
2
2
2
2
0
u
u
x
y






 
4. 
2
2
0
x dy y dx


 5. 
5
4
2
4
2
3
5
0
d y
d y
y
dx
dx









 6. 
4
2
2
2
2
2
2
0
u
y
u
u
x y
x
y





 
 


 
7. 
2
2
sin
0
d y
y
x
dx


 8. 
2
2
sin
0
d y
x
y
dx


 9. 
6
4
3
6
4
3
d x
d x
d x
x
t
dt
dt
dt




 






 
10. 
3
2
2
1
dr
d r
ds
ds








 
 
Assihnment-2 
1. Show that each of the function defined in Column I is a solution of the corresponding differential 
equation in Column II on every interval a x b
 
 of the x axis. 
I 
II 
(a) 
 
3
x
f x
x
e
 
 
1
dy
y
x
dx
  
 
(b) 
 
3
4
2
5
x
x
f x
e
e


 
2
2
7
12
0
d y
dy
y
dx
dx



 
(c) 
 
2
2
6
7
x
f x
e
x
x
 
 
 
(d) 

2
2
2
1
4
2
0
d y
dy
x
x
y
dx
dx




 
 
2.(a) Show 
that 
3
2
3
1
x
xy


 
is an 
implicit 
solution of 
the differential equation 


2
2
2
0
xy dy dx
x
y
 

 on the interval 0
1
x
 
. 
(b) Show that 
2 2
3 2
5
2
1
x y
x y


 
is an implicit solution of 
the differential equation 


3 3
x dy dx
y x y
 
 on the interval 
5
0
2
x
 
. 
3. (a) Show that every function f defined by   
 


3
3x
f x
x
c e


, 
where c is an arbitrary constant, is a solution of the differential equation  
2
3
3
3
x
dy
y
x e
dx



 
(b) Show that every function f defined by  
 
2
2
2
x
f x
ce
 
 
where c is an arbitrary constant, is solution of the differential equation  
4
8
dy
xy
x
dx


 
4. (a) Show that every function f defined by 
 
4
2
1
2
x
x
f x
c e
c e


, where 
1c  and 
2c  are arbitrary 
constants, is a solution of the differential equation 
2
2
2
8
0
d y
dy
y
dx
dx



 
 
Page 2 of 14 
 
(b) Show that every function g defined by 
 
2
2
2
1
2
3
x
x
x
g x
c e
c xe
c e



, where 
1
2
,
c c  and 3c  are 
arbitrary constants, is a solution of the differential equation  
3
2
3
2
2
4
8
0
d y
d y
dy
y
dx
dx
dx




 
 
Assignment-3 
1. Show that  
2
3
4
2
x
x
y
e
e


 is a solution of the initial-value problem
2
2
6
0
d y dy
y
dx
dx



,
 
0
6
y

,
 
' 0
2
y

 
Is 
2
3
2
4
x
x
y
e
e


 also a solution of this problem? Explain why or why not. 
2. Given that every solution of 
2
x
dy
y
xe
dx

 
may be written in the form 


2
x
y
x
c e


, for some 
choice of the arbitrary constant c, solve the following initial-value problems: 
(a) 
2
x
dy
y
xe
dx

 
, 
 
0
2
y

(b) 
2
x
dy
y
xe
dx

 
, 

1
3
y
e
  
 
3. Given that every solution of 
2
2
12
0
d y dy
y
dx
dx


 may be written in the form
4
3
1
2
x
x
y c e
c e


 
for some choice of the arbitrary constants 1c  and 2c , solve the following initial-value problems: 
(a) 
2
2
12
0
d y dy
y
dx
dx



,
 
0
5
y

,
 
' 0
6
y

 
(b) 
2
2
12
0
d y dy
y
dx
dx



,
 
0
2
y
 
,
 
' 0
6
y

 
4. Every solution of the differential equation  
2
2
0
d y
y
dx
  may be written in the form 
1
2
sin
cos
y c
x c
x


, for some choice of the arbitrary 
constants 
1c  and 
2c . Using this information, show that boundary problems (a) and (b) posses 
solutions but that (c) does not.(a) 
2
2
0
d y
y
dx
 
,
 
0
0
y

,


2
1
y 

 
(b) 
2
2
0
d y
y
dx
 
,
 
0
1
y

,


'
2
1
y 
   
5. Given that every solution of  
3
2
3
2
3
2
3
6
6
0
d y
d y
dy
x
x
x
y
dx
dx
dx



 may be written in the form 
2
3
1
2
3
y c x c x
c x



 for some 
choice of the arbitrary constants 1
2
,
c c  and 3c , solve the initial-value problem consisting of the above 
differential equation plus the three conditions
 
 
 
2
0,
' 2
2,
" 2
6
y
y
y



 
6. Apply Existence and Uniqueness Theorem to show that each of the following initial-value 
problems has a unique solution defined on some sufficiently small interval 
1
x
h
 
 about 0 1
x 
: 
(a) 
2 sin
dy
x
y
dx

,
 
2
y t  
 (b) 
2
2
dy
y
dx
x


,
 1
0
y

 
7. Consider the initial-value problem
 
 
2
dy
P x y Q x y
dx


,
 
2
5
y

 
where 
 
P x
 and 
 
Q x  are both third-degree polynomials in x. Has this problem a unique solution 
on some interval 
2
x
h
 
 about 0
2
x  ? Explain why or why not. 
 
Page 3 of 14 
 
8.The initial-value problem
1 3
dy
y
dx

,
 
0
0
y

,has infinitely many solutions. Verify that this is 
indeed the case by showing that 


3 2
0,
,
2
,
,
3
x c
y
x c
x c



 








 
is a solution of the stated problem for every real number 
0
c 
. 
 
Assignment-4 
1. Show that 
2
3
4
2
x
x
y
e
e


 is a solution of initial-value problem
2
2
6
0
d y du
y
dx
dx



,
 
0
6
y

, 
,
 
' 0
2
y

. Is 
2
3
2
4
x
x
y
e
e


 also a solution of this problem? Explain why and why not. 
2. Every solution of the differential equation 
2
2
0
d y
y
dx
 
 may be written in the form 
1
2
sin
cos
y c
x c
x


, for some choice of the arbitrary constants 1c  and 2c . Using this information, 
show that boundary problem (a) and (b) posses solution but that (c) does not. 
(a) 
 
2
2
0,
0
0,
1
2
d y
y
y
y
dx



 






(b) 
 
2
2
0,
0
1,
'
1
2
d y
y
y
y
dx



 

 




 
(c) 
 
 
2
2
0,
0
0,
1
d y
y
y
y
dx

 


 
 
Assignment-5 
1. 
2
2
3
dy
y
y
dx


 2. 
 
3
sin
dy
x y
x
dx


 3. 


2
dy
x
x y
dx


 4. 
2
1
dy
x
dx


 5. 
4
8
dy
y
dx


 
6. 
4
dy
xy
x
dx


 7. 
2
4
dy
y
x
dx


 8. 
3
2
6
dy
xy
x
y
dx




 9. 


sin
dy
x y
dx


 10. 
2
3
x
y
dy
y
e
dx


 
 
Assignment - 6 
1. 
4
dy
xy
x
dx


 2. 
 
2
2
3
sin
dy
y
y
x
dx


 3. 
3
2
6
dy
xy
x
y
dx




 4. 
 
tan
dy
y
dx

 
5. 
dy
y
dx
x

 6. 
2
2
6
4
3
4
dy
x
dx
y
y



 7. 

2
2
1
1
dy
x
y
dx



 8. 

2
2
1
4
dy
y
xy
dx


 9. 
y
dy
e
dx


 
10. 
1
y
dy
e
dx



 11. 
3
3
dy
xy
dx

 12. 
2
2
dy
x
dx
y



 13. 
2 2
2
3
3
dy
x y
x
dx

 
 14. 
2 2
2
3
3
dy
x y
x
dx


 
15. 
2
200
2
dy
y
y
dx


 
 
Assignment - 7 
1. 

0
x y dx xdy



 2. 

2
2
0
x
xy dx y dy



 3. 



2
2
0
x
xy
y dx
y
x dy






 
4. 



2
2
3
2
0
x
xy y dx
y
x y dy





 
 
Assignment- 8 
1. Find the general solution of 
2
2
dy
xy y
dx
x


. 
 
Page 4 of 14 
 
2. Solve 
2
2
2
dy
xy
x
y
dx


 given that 
0
y 
 at 
1
x 
. 
3. Solve 
dy
x y
dx
x


 and find the particular solution when 
 1
1
y

. 
4. Solve 
dy
x
x y
dx
 
 and find the particular solution when 
 
1
2
2
y

. 
5. Solve 
2
dy
x
y
dx
x


 and find the particular solution when 
 1
1
y
 
. 
6. Given that 
dy
x y
dx
x y



, prove that arctan 


2
2
1
ln
2
y
x
y
A
x









 where A is an arbitrary 
constant. 
7. Find the general solution of 
2
2
2
2
dy
x
x
y
dx


. 
8. Find the general solution of 

2
2
dy
x y
y x
dx



. 
Assignment- 9 
Note: If 
p
q
r
a
b
c
 
, suppose 
p
q
m
a
b
 
, then the differential equation 




0
ax by c dx
px qy r dy
 

 

 can be written as 






0
ax by c dx m ax by
r dy






. To solve this, let u ax by


. 
1. Find the solution of 



1
2
2
1
0
x y
dx
x
y
dy
 




. 
 
Assignment- 10 
1. Solve 



3
1
6
2
3
0
x y
dx
x
y
dy
 




. 
2. Solve the initial problem of 



2
3
1
4
6
1
0
x
y
dx
x
y
dy






; 


2
2
y  
. 
Part 3 
1. 
2
7
3
8
dy
x
y
dx
y
x



 2. 

3
2
2
1
6
6
dy
x
x y
x
dx



 3. 
2
2
2
2
2
dy
x
y
dx
xy x



 4. 



3
5
0
x
y dx
x y dy

 

 
5. 

 
2
2
2
0;
1
0
x
y dx
xydy
y




 6. 



5
2
1
2
1
0
x
y
dx
x y
dy



 

 
 
Assignment -11 
1. 



3
2
2
0
x
y dx
x y dy




 2. 



2 3
2
4
0
y
dx
xy
dy




 
3. 



2
2
1
4
0
xy
dx
x
y dy




 4. 



2
3
3
2
0
x y
dx
x
y dy




 
5. 



2
2
6
2
5
3
4
6
0
xy
y
dx
x
xy
dy






 6. 

2 1 cos
2 sin
0
rdr
rd






 
7. 



2
sec
sec tan
tan
2
0
y
x
x
x dx
x
y dy




 8. 
2
2
3
0
x
x
x dx
y dy
y
y
















 
9. 
2
2
2
1
0
s
s s
dx
dt
t
t
















 10. 
3
1
1
2
2
2
1
2
2
1
3
1
0
y
dx
x y
dy
x




















 
 
Assignment- 12 
 
Page 5 of 14 
 
1. 



2
2
3
4
0
x
xy dx
Ax
y dy




 2. 
2
2
3
1
1
1
0
Ax
dx
dy
x
y
y
















 
 
Assignment - 13 
1. 
2
1
0
y
dy
dx
x
x


 2. 
2
2
2
0
dy
xy
y
x
dx



 3. 



2
1
2
2
0
x
x
y
e dx
e
y dy




 
4. 



2
3
2
6
4
0
xy
x dx
x
y dy




 
 
Assignment- 14 
1. 
2
3
6
dy
y
x
dx
x


 2. 
4
3
2
1
dy
x
x y
dx


 3. 
2
3
3
3
x
dy
y
x e
dx



 4. 
4
8
dy
xy
x
dx


 
 
Assignment-15 
1. Consider the differential equation 
 
0
dy
P x y
dx


 
(a) Show that if f and g are two solutions of this equation and 1c and 
2c  are arbitrary constants, then 
1
2
c f
c g

 is also a solution of this equation. 
(b) Extending the result of (a), show that if 
1
2
,
,....., n
f
f
f  are n solutions of this equation and 
1
2
,
,...., n
c c
c  are n arbitrary constants, then   
1
n
k
k
k
c f


 
is also a solution of this equation. 
 
Assignment- 16 
1. (a) Let 1
f  be a solution of 
 
 
1
dy
P x y Q x
dx


and 
2
f
 be a solution of 
 
 
2
dy
P x y Q x
dx


, 
Where P, 
1Q , and 
2Q  are all defined on the same real interval I. Prove that 1
2
f
f

 is a solution of  
 
 
 
1
2
dy
P x y Q x Q x
dx



On I. 
(b) Use the result of (a) to solve the equation 
2sin
5sin 2
dy
y
x
x
dx
 

 
 
Assignment- 17 
1. 



2
2
5
4
1
2
0
xy
y
dx
x
xy dy





 2. 



2
2
tan
tan
0
x
y dx
x x
y dy

 

 
3. 




2
1
2
1
0
y x
y dx
xy
dy


 





 4. 



2
3
2
2
0
xy
y dx
y
x dy




 
5. Find an integrating factor of the form 
p
q
x y
 and solve. 



2
2
4
6
5
8
0
xy
y dx
x y
x dy




. 
 
Assignment - 18 
1. 
3
y cx

 2. 
2
y
cx

 3. 
2
2 1
cx
y


 4. 
cx
y
e

 
5. Find the value of K such that parabolas 
2
1
y
c x
K


 are the orthogonal trajectories of the family 
of ellipses 
2
2
2
2
x
y
y c

 
. 
6. Find the value of n such that curves 
1
n
n
x
y
c


 are orthogonal trajectories of the family  
 
Page 6 of 14 
 
2
1
x
y
c x


. 
7. Find a family of oblique trajectories that intersect the family of circles 
2
2
2
x
y
c


 at angle 45°. 
8. Find a family of oblique trajectories that intersect the family of parabolas 
2
y
cx

 at angle 60°. 
 
Assignment - 19 
1. A man is pushing a loaded sled across a level field of ice at the constant speed of 10 ft/sec. When 
the man in halfway across the ice field, he stops pushing and lets the loaded sled continue on. The 
combined weight of the sheld and its load is 80 lb 

80
w
lb

; the air resistance (in pounds) is 
numerically equal to 
3
4
v , where v is the velocity of the sheld (in feet per second); and the coefficient 
fraction of the runners of the ice is 0.04 

0.04
 
. How far will the sheld continue to move after 
the man stops pushing? 
 
Assignment - 20 
1. Assume that the population of a certain city increases at a rate proportional to the number of its 
inhabitants at any time. If the population doubles in 40 years, in how many years will it triple? 
2. The population of the city of Bingville increases at a rate proportional to the numbers of its 
inhabitants present ant any time t. If the population of Bingville was 30,000 in 1970 and 35,000 in 
1980, what will be the population of Bingville in 1990? 
3. The human population x of a certain island satisfies the logistic law 
2
dx
kx
x
dt



 
....(i)         
with 




2
8
0.03 3 10
,
3 10
k






, and time t measured in years. 
(a) If the population in 1980 is 200,000, find a formula for the population in future years. 
(b) According to the formula part (a), what will be the population in the year 2000? 
(c) What is the limiting value of the population as t  ? 
 
Assignment - 21 
1. A large tank initially contains 200 gal of brine in which 15 lb of salt is dissolved. Starting at 
0
t 
, 
brine containing 4 lb of salt per gallon flows into the tank at the rate of 3.5 gal/min. The mixture is 
kept uniform by stirring and the well-stirred mixture leaves the tank at the rate of 4 gal/min. 
(a) How much salt is in the tank at the end of one hour? 
(b) How much salt is in the tank when the tank contains only 50 gal of brine? 
 
Assignment- 22 
1. Prove that the differential equation  
2
2
2
0
d y dy
y
dx
dx



 
has two solutions of the form 
ax
y
e

, with a is a constant.  
2. If a differential equation has general solution of the form  
2
3
1
2
x
x
y c e
c e


, 
then determine that differential equation. 
3. Given that 
 
2
1y x
x

 is one of the solutions of the differential equation  
2
2
2
1
4
0
d y
dy
y
dx
x dx
x



 
....(i) 
Determine a general solution of that differential equation. 
 
Page 7 of 14 
 
 
Assignment 
In Exercises 1-10 determine whether or not each of the given equations is exact; solve those that are 
exact. 
1. 



3
2
2
0
x
y dx
x y dy




 2. 



2 3
2
4
0
y
dx
xy
dy




 
3. 



2
2
1
4
0
xy
dx
x
y dy




 4. 



2
3
3
2
0
x y
dx
x
y dy




 
5. 



2
2
6
2
5
3
4
6
0
xy
y
dx
x
xy
dy






 6. 

2 1 cos
2 sin
0
r dr
r d






 
7. 



2
sec
sec tan
tan
2
0
y
x
x
x dx
x
y dy




 8. 
2
2
3
0
x
x
x dx
y dy
y
y
















 
9. 
2
2
2
1
0
s
s s
ds
dt
t
t
















 10. 


3 2
1 2 1 2
1 2
2
1
3
1
0
y
dx
x y
dy
x




 
Solve the initial-value problems in Exercises 11-16. 
11. 



 
2
2
3
4
0,
1
2
xy
dx
x
y dy
y





 
12. 





2 2
3
3
3
3
2
2
3
1
0,
2
1
x y
y
x dx
x y
x y
dy
y






 
 
13. 



 
2
2
2 sin cos
sin
sin
2 cos
0,
0
3
y
x
x y
x dx
x
y
x dy
y





 
14. 



 
2
2
2
0,
0
6
x
x
x
ye
e
y dx
e
xy dy
y






 
15. 


2
2
2
3
2
0,
1
2
y
y
x
dx
dy
y
x
xy








 








 
16. 
 
2 3
4 3 2 3
1 3
2 3 1 3
4 3
1 8
2
0,
1
8
xy
x y
x
dx
dy
y
x y
y





 
17. In each of the following equations determine the constant A such that the equation is exact, and 
solve the resulting exact equation: 
(a) 



2
2
3
4
0
x
xy dx
Ax
y dy




 (b) 
2
2
3
1
1
1
0
Ax
dx
dy
x
y
y
















 
18. In each of the following equation determi0ne the constant A such that the equation is edact, and 
solve the resulting exact equation: 
(a) 



2
2
3
2
4
0
Ax y
y dx
x
xy dy




 (b) 
3
2
2
1
1
0
Ay
y
dx
dy
x
x
x
x
















 
19. In each of the following equations determine the most general function 


,
N x y
 such that the 
equation is exact: 
(a) 



3
2
,
0
x
xy dx N x y dy



 (b) 



2
2
3
,
0
x y
xy
dx N x y dy






 
20. In each of the following equations determine the most general function 


,
M x y
 such that the 
equation is exact: 
(a) 




2 3
4
,
2
0
M x y dx
x y
x y dy



 (b) 




2 3
,
2
0
x
x
M x y dx
ye
y e
dy



 
21. Consider the differential equation 

2
4
3
2
0
x
y dx
xy dy



 
(a) Show that this equation is not exact. 
(b) Find an integrating factor of the form 
n
x
, where n is a positive integer.    
(c) Multiply the given equation through by the integrating factor found in (b) and solve the resulting 
exact equation. 
22. Consider the differential equation 

2
2
2
0
y
xy dx x dy



 
 
Page 8 of 14 
 
(a) Show that this equation is not exact. 
(b) Multiply the given equation through by 
n
y
, where n is an integer, and then determine n so that 
n
y is an integrating factor of the given equation. 
(c) Multiply the given equation through by the integrating factor found in (b) and solve the resulting 
exact equation. 
(d) Show that 
0
y 
 is a solution of the original non-exact equation but is not a solution of the 
essentially equivalent exact equation found in step (c). 
(e) Graph several integral curves of the original equation, including all those whose equations are (or 
can be written) in some "special" form. 
23. Consider a differential equation of the form 






2
2
2
2
0
y xf x
y
dx
yf x
y
x dy














 
(a) Show that an equation of this form is not exact. 
(b) Show that 


2
2
1 x
y

 is an integrating factor of an equation of this form. 
24. Use the result of Exercise 23(b) to solve the equation  




2
2
2
2
2
2
0
y x x
y
dx
y x
y
x dy


















 
 
Assignment 
Solve each of the differential equation in Exercise 1-14. 
1. 


2
4
1
0
xy dx
x
dy



 2. 



2
2
2
2
0
xy
x y
dx
x
x dy

 



 
3. 




2
4
2
1
1
0
r s
dr
r
ds




 4. cos
sec
0
ydx
xdy


 
5. tan
2
0
dr
rd




 6. 



1 cos
sin
1
0
v
v
e
udu e
u
dv




 
7. 




2
2
4
1
3
2
0
x
y
dx y x
x
dy






 8. 

0
x y dx xdy



 
9. 



2
2
2
3
2
0
xy
y dx
xy x dy




 10. 


3
3
2
0
v du
u
uv dv



 
11. 
tan
0
y
x
y dx xdy
x









 12. 



2
2
2
2
2
2
2
0
s
st
t
ds
s
st
t
dt






 
13. 

3
2
2
2
2
2
0
x
y
x
y dx xy x
y dy





 14. 



0
x y
x y dx
x y
y dy
 


 

 
Solve the initial-value problems in Exercise 15-20. 
15. 





2
0,
3
1
y
dx y x y dy
y




  
 16. 
2
2
8cos
cos
0,
12
4
ydx
xdy
y











 
17. 




2
2
3
8
4
4
5
6
0
x
y
dx
y x
x
dy






, 
 1
2
y

 
18. 

 
2
2
3
2
0,
2
6
x
y dx
xy dy
y




 19. 



 
2
5
4
0,
1
4
x
y dx
x y dy
y





 
20. 



 
2
2
2
3
9
5
6
4
0,
2
6
x
xy
y dx
x
xy dy
y





 
 
21. (a) Show that the homogeneous equation  




0
Ax By dx Cx Dy dy




is exact if and only if B C

. 
(b) Show that the homogeneous equation 



2
2
2
2
0
Ax
Bxy Cy dx Dx
Exy Fy dy






 
is exact if and only if 
2
B
D

 and 
2
E
C

. 
22. Solve each of the following by two methods (see Example 21(a)): 
(a) 



2
2
0
x
y dx
x y dy




 (b) 



3
0
x y dx
x y dy

 

 
23. Solve each of the following by two methods (see Example 21(b)): 
 
Page 9 of 14 
 
(a) 



2
2
2
2
4
0
x
y dx
xy y dy




(b) 



2
2
2
2
2
2
0
x
xy y dx
x
xy dy





 
24. (a) Prove that if 
0
M dx N dy


 is homogeneous equation, then the change of variables x uy

 
transforms this equation into a separable equation in the variables u and x. 
(b) Use the result of (a) to solve the equation of Example 2.12 of the text. 
(c) Use the result of (a) to solve the equation of Example 2.13 of the text. 
25. Suppose the equation 
0
M dx N dy


 is homogeneous. Show that the transformation 
cos ,
sin
x
r
y
r




 reduces this equation to a separable equation in the variables r  and  . 
26. (a) Use the method of Exercise 25 to solve Exercise 8. 
(b) Use the method of Exercise 25 to solve Exercise 9. 
27. Suppose the equation 
0
M dx N dy


 
....(A) is homogeneous.   
(a) Show that Equation (A) is invariant under the transformation 
 
Assignment 
Solve the given differential equations in Exercises 1-18. 
1.  
2
3
6
dy
y
x
dx
x


 2. 
4
3
2
1
dy
x
x y
dx


 3. 
2
3
3
3
x
dy
y
x e
dx



 4. 
4
8
dy
xy
x
dx


 5. 
2
2
1
dx
x
dt
t
t


 
6. 

2 1
4
3
dv
u
uv
u
du



 7. 
1
1
1
dy
x
x
y
x
dx
x
 

 

 8. 



2
2
3
1
1
dy
x
x
x
y
x
dx
 


 
 
9. 

1
0
xdy
xy y
dx

 

 10. 


2
0
ydx
xy
x y dy

 

 11. 
tan
cos
dr
r
d





 
12. 


4
cos
sin
cos
0
d
r
d


 



 13. 



2
cos
cos
1 sin
0
x y
x dx
x dy

 

 
14. 



2
sin 2
cos
1 sin
0
y
x
x dx
x dy

 

 15. 
2
dy
y
y
dx
x
x
  
 16. 
6 4
2
dy
x
y
x y
dx
  
 
17. 


3
4
8
0
dy
y
y
xdx




 18. 
1
1
2
dx
t
t
x
dt
t
xt




 
Solve the initial-value problems in Exercises 19-30. 
19. 
 
4
2
2
,
2
8
dy
x
y
x
y
dx



 20. 
 
2
2
3
,
0
2
dy
x y
x
y
dx



 
21. 




 
2
3
1
1
0,
0
4
x
x
x
e
y
e
dx
e
dy
y












 
22. 




 
2
2
1
1
0,
1
5
x y
dx
x
dy
y




 
 
23. 
2
tan
cos
,
1
4
dr
r
r
d













 24. 
 
sin 2 ,
0
0
dx
x
t
x
dt
 

 
25. 
 
3
,
1
2
2
dy
y
x
y
dx
x
y



 26. 


 
3 2
,
1
4
dy
x
y
xy
y
dx
 

 
27. 
 
dy
y
f x
dx
 
, where 
 
 
2, 0
1,
0
0
0,
1,
x
f x
y
x
 






 
28. 
 
dy
y
f x
dx
 
, where 
 
 
5, 0
10
0
6
1,
10,
x
f x
y
x
 






 
29. 
 
dy
y
f x
dx
 
, where 
 
 
2
,
0
2
0
1
,
2
x
e
x
f x
y
e
x



 





 
30. 

 
2
dy
x
y
f x
dx

 
, where 
 
 
2 , 0
2,
0
4
4,
2,
x
x
f x
y
x
 






 
 
Page 10 of 14 
 
 
Exercise: Miscellaneous Review 
Solve each of the differential equation in Exercises 1-14. Several can be solved by at least two 
different methods. 
1.


2
3
6
1
0
x ydx
x
dy



 2. 



2 2
3
3
2
0
x y
x dy
xy
y dx




 
3. 



1
1
0
y
dx x x
dy




 4. 

2 2
0
x
y dx xdy



 5. 



3
5
0
x
y dx
x y dy

 

 
6. 


2
2
2
2
0
x
x
e y dx
e y
y dy



 7. 



3
3
4
8
12
1
0
x y
x dx
x
dy




 
8. 

2
2
2
2
2
0
x
xy y dx
x dy




 9. 
3 2
2
3
4
4
3
2
dy
x y
x y
dx
x
x y



 10. 
1
x
dy
x
xy
e
dx




 
11. 
2
7
3
8
dy
x
y
dx
y
x



 12. 
2
3
dy
x
xy
xy
dx


 13. 

3
2
2
1
6
6
dy
x
x y
x
dx



14. 
2
2
2
2
2
dy
x
y
dx
xy x



 
Solve the initial value problems in Exercises 15-24. 
15. 

 
2
2
2
0,
1
2
x
y dx
xydy
y




 
16. 



 
2
2
2
4
1
0,
3
0
y
dx
x
ydy
y

 


 
17. 

 
2
2
2
2
0,
0
2
x
x
e y
x dx e ydy
y




 
18. 



 
2
2
2
2
3
2
2
6
0,
1
2
x
xy dx
x y
y dy
y





 
19. 
 
2
4
1,
2
1
dy
xy
y
y
dx



 20. 
 
7
,
1
2
2
2
dy
x
y
y
dx
x
y
 



 21. 


2
,
15
2
1
dy
xy
y
dx
x



 
22. 
 
dy
y
f x
dx
 
, where 
 
 
1, 0
2,
0
0
0,
2,
x
f x
y
x
 






 
23. 

 
2
dy
x
y
f x
dx

 
, where 
 
 
2 , 0
2
0
4
4,
2,
x
x
f x
y
x
 






 
24. 
 
3
2
,
1
1
dy
y
x
xy
y
dx
x



 
 
Exercises 
Solve each differential equation in Exercises 1-4 by first finding an integrating factor. 
1. 



2
2
5
4
1
2
0
xy
y
dx
x
xy dy





 2. 



2
2
tan
tan
0
x
y dx
x x
y dy

 

 
3. 




2
1
2
1
0
y x
y dx
xy
dy


 





 4. 



2
3
2
2
0
xy
y dx
y
x dy




 
In each of Exercises 5 and 6 find an integrating factor of the form 
p
q
x y
 and solve. 
5. 



2
2
4
6
5
8
0
xy
y dx
x y
x dy




 6. 



2 3
4
3 2
3
8
2
5
8
0
x y
y dx
x y
xy dy




 
Solve each differential equation in Exercises 7-10 by making a suitable transformation. 
7. 



5
2
1
2
1
0
x
y
dx
x y
dy



 

 8. 



3
1
6
2
3
0
x y
dx
x
y
dy
 




 
9. 



2
3
2
1
0
x
y
dx
x y
dy



 

 10. 



10
4
12
5
3
0
x
y
dx
x
y
dy


 


 
Solve the initial-value problems in Exercises 11-14. 
11. 



1
6
4
1
4
2
2
0,
3
2
x
y
dx
x
y
dy
y









 
 
 
12. 



 
3
6
2
0,
2
2
x y
dx
x y
dy
y
 
  

 
 
13. 





2
3
1
4
6
1
0,
2
2
x
y
dx
x
y
dy
y






 
 
 
Page 11 of 14 
 
14. 



 
4
3
1
1
0,
3
4
x
y
dx
x y
dy
y
 
  

 
 
17. Show that if 


,x y

 and 

,
v x y
 are integrating factors of  




,
,
0
M x y dx N x y dy


 
....(A) 
such that 




,
,
x y v x y

 is not constant, then 




,
,
x y
cv x y


 
is a solution of Equation (A) for every constant c. 
18. Show that if the equation  




,
,
0
M x y dx N x y dy


 
....(A) 
is homogeneous and 




,
,
0
M x y x N x y y


, then 




1
,
,
M x y x HN x y y




  is an 
integrating factor of (A). 
19. Show that if the equation 




,
,
0
M x y dx N x y dy


 is both homogenous and exact and if 




,
,
M x y x N x y y

 
is not 
a 
constant, 
then 
the 
solution of 
this 
equation 
is 




,
,
M x y x N x y y c


, where c is an arbitrary constant.    
Exercises 
In Exercises 1-9 find the orthogonal trajectories of each given family of curves. In each case sketch 
several members of the family and several of the orthogonal trajectories on the same set of axes. 
1. 
3
y cx

 2. 
2
y
cx

 3. 
2
2 1
cx
y


 4. 
cx
y
e

 5. 
1
x
y
x
ce
  
 6. 
2
x y cx
 
 
7. 
2
2
3
x
y
cx


 8. 
2
2
2
1
y
x
y
ce

 
 9. 
2
2
4
y
c
x
y


 
10. Find the orthogonal trajectories of the family of ellipses having center at the origin, a focus at the 
point 

,0
c
, and semimajor axis of length 2c . 
11. Find the orthogonal trajectories of the family of circles which are tangent to the y axis at the 
origin.    
12. Find the value of K such that the parabolas 
2
1
y
c x
K


 are the orthogonal trajectories of the 
family of ellipses 
2
2
2
2
x
y
y c

 
. 
13. Find the value of n such that  the curves 
1
n
n
x
y
c


 are the orthogonal trajectories of the family  
2
1
x
y
c x


 
14. A given family of curves is said to be self-orthogonal if its family of orthogonal trajectories is the 
same as the given family. Show that the family of parabolas 
2
2
2
y
cx c


 is self-orthogonal. 
15. Find a family of oblique trajectories that intersect the family of circles 
2
2
2
x
y
c


 at angle 45°. 
16. Find a family of oblique trajectories that intersect the family of parabolas 
2
y
cx

 at angle 60°. 
17. Find a family of oblique trajectories that intersect the family of curves 
2
x y cx
 
 at angle  
such that tan
2
 
. 
 
Exercises 
1. Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei 
that are present in a given sample. In a certain sample 10% of the original number of radioactive 
nuclei have disintegration in a period of 100 years.   
(a) What percentage of the original radioactive nuclei will remain after 1000 years? 
(b) In how many years will only one-fourth of the original number remain? 
 
Page 12 of 14 
 
2. A certain chemical converted into another chemical by a chemical reaction. The rate at which the 
first chemical is converted is proportional to the amount of this chemical present at any instant. Ten 
percent of the original amount of the first chemical has been converted in 5 min. 
(a) What percent of the first chemical will have been converted in 20 min? 
(b) In how many minutes will 60% of the first chemical have been converted? 
3. A chemical reaction converts a certain chemical into another chemical, and the rate at which the 
first chemical is converted is proportional to the amount of this chemical present at any time. At the 
end of the one hour, 50 gm of the first chemical remain; while at the end of three hours, only 25 gm 
remain. 
(a) How many grams of the first chemical were present initially? 
(b) How many grams of the first chemical will remain at the end of five hours? 
(c) In how many hours will only 2 gm of the first chemical remain? 
4. A chemical reaction converts a certain chemical into another chemical, and the rate at which the 
first chemical is converted is proportional to the amount of this chemical present at any time. At the 
end of one hour, two-thirds kg of the first chemical remains, while at the end of four hours, only one-
third kg remains? 
(a) What fraction of the first chemical remains at the end of seven hours? 
(b) When will only one-tenth of the first chemical remain? 
5. Assume that the population of a certain city increases at a rate proportional to the number of 
inhabitants at any time. If the population doubles in 40 years, in how many years will it triple? 
6. The population of the city of Bingville increases at a rate proportional to the number of its 
inhabitants present at any time t. If the population of Bingville was 30,000 in 1970 and 35,000 in 
1980, what will be the population of Bingville in 1990? 
7. In a certain bacteria culture the rate of increase in the number of bacteria is proportional to the 
number present. 
(a) If the number triples in 5 hr, how many will be present in 10 hr? 
(b) When will the number present to 10 times the number initially present? 
8. Assume that the rate of change of the human population of the earth is proportional to the number 
of people on earth at any time, and suppose that this population is increasing at the rate of 2% per 
year. The 1979 World Almanac gives the 1978 world population estimate as 4,219 million; assume 
this figure is in fact correct. 
(a) Using this data, express the human population of the earth as a function of time. 
(b) According to the formula of part (a), what was the population of the earth in 1950? The 1979 
World Almanac gives the 1950 world population estimate as 2,510 million. Assuming this estimate is 
very nearly correct, comment on the accuracy of the formula of part (a) in checking such past 
populations. 
(c) According to the formula of part (a), what will be the population of the earth in 2000? Does this 
seem reasonable? 
(d) According to the formula of part (a), what was the population of the earth in 1900? The 1979 
World Almanac gives the 1900 world population estimate as 1,600 million. Assuming this estimate is 
very nearly correct, comment on the accuracy of the formula of part (a) in checking such past 
populations. 
(e) According to the formula of part (a), what will be the population of the earth in 2100? Does this 
seem reasonable? 
9. The human population of a certain 
island satisfies 
the 
logistic 
law (3.50) with 


8
0.03,
3 10
k




, and time t measured in years. 
(a) If the population in 1980 is 200,000 find a formula for the population in future years. 
(b) According to the formula of part (a), what will be the population in the year 2000? 
(c) What is the limiting value of the population as t  ? 
10. This is a general problem about the logistic law of growth. A population satisfies the logistic law 
(3.50) and has 0x  members at time 0
t
. 
(a) Solve the differential equation (3.50) and thus express the population x as a function of t. 
(b) Show that as t  , the population x approaches the limiting value k  . 
 
Page 13 of 14 
 
(c) Show that dx dt  is increasing if 
2
x k 

 and decreasing if 
2
x k 

. 
(d) Graph x as a function of t for 
0
t
t

. 
(e) Interpret the results of parts (b), (c) and (d). 
11. The human population of a certain small island would satisfy the logistic law (3.50) with 


8
1
,
10
400
k




, and t measured in years, provided the annual emigration from the island is 
neglected. However, the fact is that every year 100 people become disenchanted with island life and 
move from the island to the mainland. Modify the logistic differential equation (3.50) with the given k 
and  so as to include the stated annual emigration. Assuming that the population in 1980 is 20,000, 
solve the resulting initial-value problem and thus find the population of the island as a function of 
time. 
12. Under natural circumstances the population of mice on a certain island would increase at a rate 
proportional to the number of mice present at any time, provided the island had no cats. There were 
no cats on the island from the beginning of 1970 to the beginning of 1980, and during this time the 
mouse population doubled, reaching an all-time high of 100,000 at the beginning of 1980. At this time 
the people of the island, alarmed by the increasing number of mice, imported a number of cats to kill 
the mice. If the indicated natural rate of increase of mice was thereafter offset by the work of the cats, 
who killed 1000 mice a month, how many mice remained at the beginning of 1981? 
13. An amount of invested money is said to draw interest compounded continuously if the amount of 
money increases at a rate proportional to the amount present. Suppose $1000 is invested and draws 
interest compounded continuously, where the annual interest rate is 6%, 
(a) How much money will be present 10 years after the original amount was invested? 
(b) How long will it take the original amount of money to double? 
14. Suppose a certain amount of money is invested and draws interest compounded continuously. 
(a) If the original amount doubles in two years, then what is the annual interest rate? 
(b) If the original amount increases 50% in six months, then how long will it take the original amount 
to double? 
15. A tank initially contains 100 gal of brine in which there is dissolved 20 lb of salt. Starting at time 
0
t 
, brine containing 3 lb of dissolved salt per gallon flows into the tank at the rate of 4 gal/min. 
The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the 
tank at the same rate. 
(a) How much salt is in the tank at the end of 10 min? 
(b) When is there 160 lb of salt in the tank? 
16. A large tank initially contains 100 gal of brine in which 10 lb of salt is dissolved. Starting at 
0
t 
, pure water flows into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring 
and the well-stirred mixture simultaneously flows out at the slower rate of 2 gal/min. 
(a) How much salt is in the tank at the end of 15 min and what is the concentration at that time? 
(b) If the capacity of the tank is 250 gal, what is the concentration at the instant the tank overflows? 
17. A tank initially contains 100 gal of pure water. Starting at 
0
t 
, a brine containing 4 lb of salt per 
gallon flows into the tank at the rate of 5 gal/min. the mixture is kept uniform by stirring and the well-
stirred mixture flows out at the slower rate of 3 gal/min. 
(a) How much salt is in the tank at the end of 20 min? 
(b) When is there 50 lb of salt in the tank? 
18. A large tank initially contains 200 gal of brine in which 15 lb of salt is dissolved. Starting at 
0
t 
, brine containing 4 lb of salt per gallon flows into the tank at the rate of 3.5 gal/min. The 
mixture is kept uniform by stirring and well-stirred mixture leaves the tank at the rate of 4 gal/min. 
(a)   How much salt is in the tank at the end of one hour? 
(b) How much salt is in the tank when the tank contains only 50 gal of brine? 
19. A 500 liter tank initially contains 300 liters of fluid in which there is dissolved 50 gm of a certain 
chemical. Fluid containing 30 gm per liter of the dissolved chemical flows into the tank at the rate of 
4 liters/min. The mixture is kept uniform by stirring and the stirred mixture simultaneously flows out 
at the rate of 2.5 liters/min. How much of the chemical is in the tank at the instant it overflows? 
 
Page 14 of 14 
 
20. A 200 liter tank is initially fully of fluid in which there is dissolved 40 gm of a certain chemical. 
Fluid containing 50 gm per liter of this chemical flows into the tank at the rate of 5 liters/min. The 
mixture is kept uniform by stirring and the stirred mixture simultaneously flows out at the rate of 7 
liters/min. How much of the chemical is in the tank when it is only half full? 
21. The air in a room whose volume is 10,000 cu ft tests 0.15% carbon dioxide. Starting at 
0
t 
, 
outside air testing 0.05% carbon dioxide is admitted at the rate of 5000 cu ft/min. 
(a) What is the percentage of carbon dioxide in the air in the room after 3 min? 
(b) When does the air in the room test 0.1% carbon dioxide? 
22. The air in a room 50 ft by 20 ft by 8 ft tests 0.2% carbon dioxide. Starting at 
0
t 
, outside air 
testing 0.05% carbon dioxide is admitted to the room. How many cubic feet of this outside air must be 
admitted per minute in order that the air in the room test 0.1% at the end of 30 min? 
23. Newton's law of cooling states that the rate at which a body cools is proportional to the difference 
between the temperature of the body and that of the medium in which it is situated. A body of 
temperature 80°F is placed at time 
0
t 
 in a medium the temperature of which is maintained at 50°F. 
At the end of 5 min, the body has cooled to a temperature of 70°F. 
(a) What is the temperature of the body at the end of 10 min? 
(b) When will the temperature of the body be 60°F? 
24. A body cools from 60°C to 50°C in 15 min in air which is maintained to 30°C. How long will it 
take this body to cool from 100°C to 80°C in air that is maintained at 50°C? Assume Newton's law of 
cooling (Exercise 23). 
25. The rate at which a certain substance dissolves in water is proportional at the product of the 
amount undissolved and the difference 1
2
c
c

, where 1c  is the concentration in the saturated solution 
and 2c  is the concentration in the actual solution. If saturated, 50 gm of water would dissolve 20 gm 
of the substance. If 10 gm of the substance is placed in 50 gm of water and half of the substance is 
then dissolved in 90 min, how much will be dissolved in 3 hr?