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Differen-
tial
equations
Definition, order and degree, general and particular solutions of a
differential equation. Solution of differential equations by method of
separation of variables, solutions of homogeneous differential equations of
first order and first degree. Solutions of linear differential equation of the
type:
dy dx
- py = q, (^) where p and q are the functions of x or constants
dx dy
- px = q, (^) where p and q are the functions of y or constants
In this chapter you will study The concept of differential equations using the idea of differential calculus. Identify order and degree of a differential equations when solution is not given. Various method to solve differential equations. How to solve linear differential equation and homogenous differential equation of first order and first degree.
Topic-
Basic Differential Equations
Concepts Covered Order of differential Equation
Degree of differential Equation
Revision Notes
Orders and Degrees of Differential Equation : We shall prefer to use the following notations for derivatives. dy dx
= y ', d y dx
2 2 =^ y '',^
d y dx
3 3 =^ y ''' For derivatives of higher order, it will be in convenient to use so many dashes as super suffix therefore, we use the notation y (^) n for n th order derivative d y dx
n n. Order and degree (if defined) of a differential equation are always positive integers.
Key Words
Differential Equation: In
Mathematics, a differential equation is an equation with one or more derivatives of a function. The derivative of the function is given by dy/dx. In other words, it is defined as the equation that contains derivatives of one or more dependent variables with respect to one or more independent variables.
syllabus
C H a P t e r
List of Topics Topic-1: Basic Differential Equations Page No. 208 Topic-2: Variable Separable Methods Page No. 212 Topic-3: Linear Differential Equations Page No. 217 Topic-4: Homogeneous Differential Equations Page No. 224
subjeCtive tyPe questions
Very Short Answer Type Questions (1 mark each)
Q. 1. Find the order and the degree of the differential
equation x^2 d y dx
dy dx
2 2
2 4 = 1 + æèç öø÷
ì í
ï îï
ü ý
ï þï
.
R [CBSE Delhi Set-I, 2019] Q. 2. Write the order and the degree of the following differential equation :
x d y dx
x dy dx
3 2 2
æ^2 èç
ö ø÷^
R [CBSE Delhi Set-III 2019] Q. 3. Find the order and degree (if defined) of the differential equation. d y dx
x dydx
2 2
2
2 x (^2) d y dx
log
æ èç
ö ø÷ R [CBSE OD Set-I, 2019] Sol. Order = 2, Degree not defined ½+ ½ [CBSE Marking Scheme, 2019] Detailed Solution : Given differential equation is d y dx
x dy dx
2 2
2
^ = 2 x
(^2) log^ d y dx
2 2
The highest order derivative present in the differential equation is d y dx
2 2 .. So, it is of order 2. Clearly, the differential equation is not expressible as polynomial in d y dx
2 2.^ So, its degree is not defined. Hence, order = 2 degree = not defined
Commonly Made Error
Mostly students write the degree as 1.
Answering Tip
Note the questions in which degree is not defined. Q. 4. Write the order of the differential equation:
log
d y dx
2 2
^ =^
dy dx
+ x 3 R&U [SPQ 2018]
Q. 5. Write the sum of the order and degree of the
differential equation 1 7
(^4 ) 2
3 ++ æèçæèç ööøø÷÷ ==
ææ èè
çç
öö øø
÷÷
dy dx
d y dx
R&U [Delhi Set I, II, III Comptt. 2015] Sol. Degree of the given differential equation = 3 Order of the given differential equation = 2 ½ Hence, the sum of order and degree = 2 + 3 = 5 ½ [CBSE Marking Scheme 2015]
Short Answer Type Questions-I (2 marks each)
Q. 1. Show below is a differential equation.
y e dy
dx
d y
= dx +
sin^33
2 4
Find the order and the degree of the given differential equation. Give reasons to support your answer. [CBSE Practice Questions 2021-22]
Sol. Order is 3 as highest order derivative present in the given differential equation is d y dx
3 3.^^1 Degree is not defined as the highest order derivative is the function of sine. 1 Q. 2. Write the sum of the order and degree of the following differential equation: d [SQP 2021-22] dx
dy dx
^
=^^5
Sol. Order = 2 1 Degree = 1 ½ Sum = 3 ½ [CBSE Marking Scheme, 2022] Detailed Solution: Given differential equation: d dx
dy dx
^
^ = 5
\ d y dx
2 2 = 5 Thus, degree = 2, order = 1 and sum = 2 + 1 = 3 Q. 3. Find the sum of the order and the degree of the following differential equations : d y dx
dy dx
x
2 2
R&U (^) [SQP Dec. 2016-17]
Topic-
Variable Separable Methods
Concepts Covered General Solution, Particular Solutions , Variable Separable
Method
Revision Notes
Solutions of differential equations : (a) General Solution : The solution which contains as many as arbitrary constants as the order of the differential equations, e.g. y = a cos x + b sin x is the
general solution of d y dx
y
2 2 +^ =^0.
(b) Particular Solution : Solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equation is called a particular solution e.g. y = 3 cos x + 2 sin x is a particular solution of the
differential equation d y dx
y
2 2 +^ =^0.
(c) Solution of Differential by Variable Separable Method : A variable separable form of the differential equation is the one which can be expressed in the form of f ( x )
dx = g ( y ) dy. The solution is given by (^) ∫ f x dx (^ )^^ =^ ∫ g y dy (^ )^ + k , where k is the constant of integration.
Key Words
Variable: A value that keeps on
changing is said to be variable. Variables are often represented by an alphabet like a, b, c , or x, y, z. Its value changes from time to time. e.g.: 3 x + 5 y = 7 where x and y are variables that are changed according to the expression.
Constant: As the name implies, the
constant is a value that remains constant ever. Constant has a fixed value and its value cannot be changed by any variable. Constants are represented by numbers. e.g.: 3 x + 5 y = 7, where 7 is the constant, we know its face value is 7 and it cannot be changed. But 3 x and 5 y are not constants because the variable x and y can change their value.
objeCtive tyPe questions
Multiple Choice Questions
Q. 1. The solution of differential equation xdy − ydx = 0 represents (A) a rectangular hyperbola (B) parabola whose vertex is at origin (C) straight line passing through origin (D) a circle whose centre is at origin Ans. Option (C) is correct. Explanation: Given that, xdy ydx xdy ydx dy y
dx x
Þ =
Þ =
On integrating both sides, we get log log log log log
y x y x y x
Þ =
Þ =
C
C
C
which is a straight line passing through the origin.
Q. 2. The general solution of dy dx
= 2 xe x^ − y
(^2) is
(A) (^) e x^^2 -^ y = C (B) e -^ y^ + e x^2 = C (C) e^ y^^ =^ e^ x + C
2 (D) e x^ y C
(^2) +
Ans. Option (C) is correct. Explanation: Given that, dy dx
xe xe e
e dy dx
xe
e dy xe dx
x y x y
y x
y x
2 −^2 −
2 2
2
2
On integrating both sides, we get
ò^ e dyy^ =^2 ò xe^ xdx
2
Put x^2 = t in RHS integral, we get 2
2
xdx dt e dy e dt e e C e e C
y t y (^) t y x
Þ = +
Þ = +
ò ò
Explanation:
dy dx
e
e e dy e
e dx
e dy e dx
x y
x y
y
x
y (^) x
= ×
Þ =
Þ =
Integrating both sides, we get:
e dy e dx e e k e e k e e C C k
y x y x x y x y
ò =ò Þ - = + Þ + = - Þ + = ( where, = -)
subjeCtive tyPe questions
Very Short Answer Type Questions (1 mark each) Q. 1. How many arbitrary constants are there in the particular solution of the differential equation dy dx
== -- 4 xy 2 ; y( ) 0 == (^1) R&U [CBSE SQP 2020-21]
Q. 2. Find the general solution of the differential equation dy dx
= e x^ +^ y.
U [CBSE OD Set-II, 2019] [CBSE SQP-2020]
Sol. Given differential equation can be written as: dy dx
= e x^. e y^ Þ e - ydy = exx^ dx
Integrating both sides, weget
- e - y^ = e x^ + c^1 [CBSE Marking Scheme, 2020]
Detailed Solution : Given differential equation is dy dx
= e x+y
Þ dy dx
= e x^ ey
Þ dy e y^
= ( e x ) dx Þ ( e –y ) dy = ( ex ) dx Integrating both sides, we get ò (^ e^ - y ) dy =^ ò(^ e^ x ) dx Þ – e –y^ = e x^ + c' Þ e –y^ = – e x^ + c [where c = – c' ] Q. 3. Find the solution of the differential equation dy dx
= x e^3 -^2^ y.
R&U [O.D. Set I, II, III Comptt. 2015]
Q. 4. Write the solution of the differential equation dy dx
== 2 - - y. R&U [Foreign 2015]
Sol. Given differential equation is dy dx
= 2 – y
On separating the variables, we get 2 y^ dy = dx On integrating both sides, we get ∫ 2 y^ dy =^ ∫ dx
or
y log
= x + C 1
or 2 y^ = x log 2 + C 1 log 2 \ 2 y^ = x log 2 + C , where C = C 1 log 2 1 [CBSE Marking Scheme 2015]
Short Answer Type Questions-I (2 marks each)
Q. 1. Solve the following differential equation: dy = x^3 cosec y, given that y(0) = 0. dx R&U [CBSE SQP 2020-21]
Sol. dy dx
= x^3 cosec y ; y (0) = 0
cosec
dy
∫ (^) y = (^) ∫^ x dx^3 ½
∫ sin^ y dy =^ ∫^ x dx^3
4
x + c 1
–1 = c (∵^ y = 0, when x = 0)
cos y =
4
x
[CBSE SQP Marking Scheme 2020]
Commonly Made Error
Students forget to find the particular solution after finding the general solution.
Answering Tip
Practice more problems based on finding particular solution. Q. 2. Find the general solution of the differential equation. xy dy dx
== (x ++ 2 )( y ++ 2 ). R&U [OD Comptt. 2017]
Sol. y y
dy
( x ) x
1 2 2
−
∫ (^) y dy (^) = æ 1 +^2 èç^
ö ò (^) x ø÷ dx^^1
y – 2 log| y + 2| = x + 2 log| x | + C 1 [CBSE Marking Scheme 2017]
Q. 3. Find the particular solution of the differential
equation dy dx
y x
2 2 ;^ given that^ y(0) =^ 3.
R&U [OD Comptt. 2017] Short Answer Type Questions-II (3 marks each)
Q. 1. Find the general solution of the differential equation given below. dy dx (^) x x
Show your steps. [CBSE Practice Questions 2021-22] Sol. Given differential equation is dy dx
x ( 1 + x^2 )
Þ (^) ∫ dy = dx ∫ x ( 1 + x (^2) )
y = dx ∫ (^) x ( 1 + x (^2) ) ...(i)
Now, 1 x ( 1 + x^2 )
A B C
x (^) x
x 1 2
...(ii)
[Using partial fraction] 1 = A(1 + x^2 ) + (B x + C) x Þ 1 = (A + B) x^2 + C x + A \ A + B = 0, C = 0 and A = 1 \ A = 1, B = – 1 and C = 0 From eqn. (ii), we get 1 x ( 1 + x^2 )
x 1 2
x x
Now, from eqn. (i), we get
y = 1 x 1 2
dx x x ∫ −^ ∫ dx
or, y = 1 1 2
x 1 2
dx x x ∫ −^ ∫ dx
or, y = (^) log x − 1 log( + x ) +log c 2
or, y = log x − log ( 1 + x^2 ) +log c
or, y = (^) log cx 1 + x^2
Q. 2. Find the general solution of the differential equation. xdy = (e y^ – 1)dx
R&U [CBSE OD Set III-2020] Sol. Given differential equation can be written as dy dx =^
x
( e y - )
Þ
dy ò (^) e y (^) - 1 = dx ò (^) x
Þ e e
dy
y y
ò (^1) - - =^
dx ò (^) x
Þ log 1^ -^^ e^ - y^ = log^ x^ +log C Þ 1 – e–y^ = Cx 2 [CBSE Marking Scheme 2020] (Modified) Detailed Solution: dy dx x
+ 1 =^
e x
y
dy dx
= e x
y (^) - 1
e 1 y dy
=^1
x
dx
Integrating both side 1 e 1 y dy ò (^) - = (^) ò x^1 dx
1 e 1 e y y dy ò (^) ( - - ) =^
x ò dx
e e
dy
y y
ò (^1) - - =^
ò x dx Let 1 – e –y^ = t 0 – e –y^ (–1) dy = dt e–y^ dy = dt 1 t ò dt =^ ò x^1 dx log t = log x + log C log(1 – e –y ) = log xC 1 – e –y^ = xC 1 – xC = e –y 1 – xC = 1 e y
ey^ = 1 1 - xC y = –log (1 – xC )
or 1 +^ y^^2 =^ -^ ò du^ -^ ò u - du
[put 1 + y^2 = t ]
or (^1) + y^2 = - - -
u u + u
1 C
log 1
∵ dx x a a
x a (^2 2) x a
é ëê^
ù ò (^) ûú log
\ 1 +^ y^^2 =^ -^ +^ -^
2 2 x (^) 2 x x
log C
which is the required solution. 1 [CBSE Marking Scheme 2015] (Modified)
Topic-
Linear Differential Equations
Concepts Covered Linear Differential Equations in x only and in y only
Revision Notes
Solutions of Differential Equations: Linear differential equation in y : It is of the form dy dx
- P x y ( ) = Q x ( ) , where
P ( x ) and Q ( x ) are functions of x only. Solving Linear Differential Equation in y : STEP 1 : Write the given differential equation in the form dy dx
STEP 2 : Find the Integration Factor ( I.F. ) = e^ ∫^ P x dx (^ ).
Key Word
Integrating Factor: An integrating
factor is a function by which an ordinary differential equation can be multiplied in order to make it integrable. STEP 3 : The solution is given by, y .( I.F. ) = (^) ò Q x ( ).(. .) I F dx + k ,where k is the constant of integration. Linear differential equation in x : It is of the form dx dy
- P y x ( ) = Q y ( ) (^) , where
P ( y ) and Q ( y ) are functions of y only. Solving Linear Differential Equation in x : STEP 1 : Write the given differential equation in the form dx dy
STEP 2 : Find the Integration Factor ( I.F. ) = e^ ∫^ P y dy (^ ).
STEP 3 : The solution is given by, x .( I.F. ) = (^) ò Q y ( ).(. .) I F dy + l , where l is the constant of integration.
Key Word
Constant of integration: A
constant that is added to the function obtained by evaluating the indefinite integral of a given function, indicating that all indefinite integrals of the given function differ by, at most, a constant.
Mnemonics
Linear Differential Equations
F O L D E - D Y B D X - P Y Q
F amily O f L l i adhar D ixit E njoying
D aily Y oga B y D r. X avier at
P urana Y og Q ila
First Order Linear Differential Equation
dy dx + Py = Q
SOLDE-YIF-EIQ-IFC
S on O L f iladhar D ixit E klavya (SOLDE)
& WIF e (Y I F) E xploring ndian I Q illa
for I nternational F itness C ertificate
y. IF = Q. I F + C
Integration
S — Solution O — Of L — Linear
D — Differential E — Equation
Interpretation :
Differential equation is of the form
dy dx
+py=Q ,
where P and Q are constants or the function
of 'x' is called a first order linear differential
equations. Its solution is given as
Y.IF= Q.IF+C∫
objeCtive tyPe questions
Multiple Choice Questions
Q. 1. The integrating factor of differential equation
cos x dy sin dx
(A) cos x (B) tan x (C) sec x (D) sin x Ans. Option (C) is correct. Explanation: Given that, cos sin
tan sec
x dy dx
y x dy dx
y x x
Þ + =
cos sin^1
tan sec
x dy dx
y x dy dx
y x x
Þ + =
Here, P = tan x and Q = sec x IF
IF
= ò = ò = \ =
e e e x
Pdx xdx x
tan ln sec sec
Q. 2. The integrating factor of differential equation
( 1 − x 2 )dy − = 1 dx
xy is
(A) – x (B)
x 1 + x^2
(C) (^) 1 - x^2 (D)^1 2
log ( 1 - x^2 )
Ans. Option (C) is correct. Explanation: Given that, ( 1 ) 1
2
2 2
Þ -
x dy dx
xy dy dx
x x
y x which is a linear differential equation.
IF =
ò
e - x t xdx dt xdx dt
x x
dx 1 2 1 2 2
Put
IF
IF
ò
= ò
=
e - x t xdx dt xdx dt
e
e
x x
dx
dt t
1 2 2
1 2 1 2
Put
Now, llog
log ( )
t
x e x
(^12 1) - 2
IF
IF
ò
= ò
=
e - x t xdx dt xdx dt
e
e
x x
dx
dt t
1 2 2
1 2 1 2
Put
Now, llog
log ( )
t
x e x
(^12 1) - 2
Q. 3. The solution of x dy dx
(A) y^
e x
k x
x = + (^) (B) y = xe x^ + Cx
(C) y = xe x^ + k (D) x e y
k y
y = + Ans. Option (A) is correct.
Explanation: Given that,
x dy dx
y e
dy dx
y x
e x
x
x
Þ + =
which is a linear differential equation.
\ = ò
=
IF e e x
x dx x
1
(log )
The general solution is
y x e x
x dx
y x e dx y x e k
y e x
k x
x
x x x
× = ×
æ èç
ö ø÷ Þ × = Þ × = +
Þ = +
ò
ò
subjeCtive tyPe questions
Very Short Answer Type Questions (1 mark each)
Q. 1. Write the integrating factor of the differential
equations (^) x dy dx
+ y = e -^2^ x.
R&U (^) [O. D. Set I, II, III Comptt. 2015]
Q. 2. Find the integrating factor of the differential
equation e x
y x
dx dy
R&U [NCERT][Delhi 2015]
Sol. We know, I.F. = e^ ∫ p dx
Here, P = 1 x
\ I.F. = e^ x
∫^1 dx
= e^^2 x ½ [CBSE Marking Scheme 2015]
Q. 3. Write the integrating factor of the following differential equation. (1 + y^2 ) + (2xy – cot y) dy dx
R&U [All India 2015] Sol. Given differential equation is ( 1 + y^2 ) + ( 2 xy - cot y ) dy dx
The above equation can be rewritten as (cot y xy ) dy dx
or
cot ( )
y xy y
1 2 =^
dx dy
or
dx dy =^
cot y y
xy 1 y
or
dx dy
y y
. (^) = cot^ y 1 + y^2
which is a linear differential equation of the form dx dy
y
cot y. 1 + y^2
Now, integrating factor = e P dy e
y y
dy ò (^) = ò^ +
2 1 2 Put 1 + y^2 = t or 2 y dy = dt
\ IF = (^) e e t y
dt ò (^) t (^) = log|| t (^) = = 1 + (^2) ½
[CBSE Marking Scheme 2015]
Short Answer Type Questions-I (2 marks each) Q. 1. Find the general solution of the differential equation dy dx
++ 2 y == e 3 x R&U [Delhi Comptt. 2017]
Sol. Integrating factor is e ∫^2 dx = e^2 x^ ½ \ Required solution is y. e^2 x^ = (^) ∫ e^^3 x^^. e^^2 xdx ½
y. e^2 x^ = e^ C
5 x 5
or y = e^ Ce
3 x (^) 2 x 5
+ −^ ½
[CBSE Marking Scheme 2017] Q. 2. Find the general solution of the differential equation dy dx x
+^2 y = x (^) R&U [Delhi Comptt. 2017]
Sol. Integrating factor is
e x ∫ 2 dx = x^2 ½ Solution is y. x^2 = (^) ò x x dx. 2 + C ½
y.x^2 = x
4 4
+ C
or y =
x C x
2 4 +^2^1 [CBSE Marking Scheme 2017]
Commonly Made Error
Many candidates do not express the answer in terms of constant 'C'.
Answering Tip
Give adequate practice on various type.
Q. 3. Find the integrating factor of the differential equation dy dx
++ y=^1 ++^ y x R&U [OD Comptt. 2017]
Short Answer Type Questions-II (3 marks each)
Q. 1. Find the particular solution of the following
differential equation, given that y = 0 when x = π 4 :^
dy dx
y x x
cot sin
[SQP 2021-2022]
Sol. The differential equation a linear differential equation IF = e^ ∫ cot xdx = e logsin x^ = sin x 1 The general solution is given by y sin x = 2 1
sin sin
x x
dx
Þ y sin x = 2 1 1 1
sin sin
x x
dx
Þ y sin x = 2 1
^
(^) sin x (^) dx (^) ½
Þ y sin x = 2 1 1 1 2
^
cos ^ x
dx
Þ y sin x = 2 1 1 2 4 2
2
^
cos ^ x^
dx
Þ y sin x = 2 1
^
^
(^) sec ^ x^ dx
Þ y sin x = 2^ x^ 4 2 x^ c ^
tan^
Given that y = 0, when x =
π 4
Hence, 0 2 4 8
^
(^) tan c
Hence, the particular solution is
y = cosec x^ x^ 2 x 4 2 2
^
^
tan tan (^) ½
[CBSE Marking Scheme 2022]
Q. 2. Find the general solution of the following
differential equation: x dy – (y + 2x^2 )dx = 0
R&U [CBSE SQP 2020-21]
Sol. The given differential equation can be written as dy dx
= y^ x x
Þ
dy dx x
Here P = -^1 x
Q = 2 x ½
IF = e^ ò Pdx
= e^ x
x^1 The solutions is : y x
x^1 x ò æèç^ ´ öø÷ dx 1 Þ y x
= 2 x + c
Þ y = 2 x^2 + cx ½ [CBSE SQP Marking Scheme 2020-21]
Commonly Made Error
Some students think that it is homogeneous and put y x
= v and goes wrong.
Answering Tip
In homogeneous equation, the degree of all terms will be the same.
Q. 3. Solve (^) 1+ x 2 dy+ 2 4 2 = 0 dx ( ) xy^ -^ x subject^ to^ the
initial condition y(0) = 0. R&U [CBSE Delhi Set I-2019] [Delhi Set I, II, III Comptt. 2016] Sol. Given differential equation can be written as : dy dx
x x
2 2
x
with P =
2 2
x x
Q x
I.F. (Integrating factor)
= e^ e
Pdx
x x
dx ∫ (^) = ∫^ +
2 1 2 = e log(1 +^ x^2 )^ = 1 + x^2 ½ \ General solution is :
y (1 + x^2 ) = 4 1
2 2
x 2 x
x dx C
∫ ·(^ +^ )^ +^1
or y· (1 + x^2 ) = 4 3
x^3
Putting x = 0 and y = 0, we get C = 0 \ Solution is :
y =
3 2
x ( + x )
[CBSE Marking Scheme 2019] (Modified)
Sol. x d dx
y (^) + y = x cos x + sin x
or dy dx x
sin x ½
\ I.F. = e x
∫ 1 dx
= e log^ x^ = x ½ \ Solution is xy = (^) ∫ (^ x^ cos^ x^ +sin^ x dx ) ½ or xy = x sin x + C 1 \ y = sin x + C^1 x x = π 2
, y = 1 or 1 = 1 + C^2 p
æ èç^
ö ø÷^ or^ C^ = 0 \ Solution is y = sin x. ½ [CBSE Marking Scheme 2017] (Modified)
Q. 8. Find the particular solution of the differential equation : dy dx
+ y cot x = 2x + x^2 cot x, x ≠ 0.
Given that y = 0, when x = 2
π ·
R&U [Delhi 2017] [NCERT] Sol. The given equation is a linear differential equation of the type dy dx
where P = cot x , Q = 2 x + x^2 cot x
\ I.F. = e ∫^ cot^ x dx^ = e log sin x
= sin x ½ Hence, the solution of the differential equation is given by : y. sin x = (^) ∫( 2 x + x^2 cot x ).sin xdx + C
= (^) ∫ 2 x .sin x dx + (^) ∫ x^2 cos x dx + C
= sin^ x.^^ cos
2 x (^) x x (^) dx 2
2 2 − (^) ∫
= x^2 sin x + C ...(i) 1
Substituting y = 0 and x = π 2
in the above equation (i), we get
0 = π
2 4
× 1 + C
or C = −
π^2 4
Now substituting the value of C in eq. (i), we get
y sin x = x^2 sin x − π
2 4
or y = x x
2 2 4
, (sin x ≠ 0)
This is the required particular solution of the given differential equation. [CBSE Marking Scheme 2017] (Modified) Q. 9. Solve the differential equation (tan–1^ **x – y)dx = (
- x**^2 ) dy R&U^ [O.D. Set I 2017]
Long Answer Type Questions (5 marks each) Q. 1. Solve the differential equation : dy dx
- 3y cot x = sin 2x given y = 2, when (^) x = 2
π.
R&U [NCERT]
[O.D. Set I, II, III Comptt. 2015] [Foreign, 2017]
Sol. dy dx
− 3cot x y · = sin 2 x^^1
I.F. = e^ ∫^ −3 cot x dx = e –^ 3 log (sin^ x )^ = (sin x )– = cosec 3 x 1 \ Solution is y ·cosec 3 x = (^) ∫sin^2^ x^ ·cosec^3 x dx
= (^) ∫ 2cosec^ x^ cot x dx 1 or y ·cosec 3 x = – 2 cosec x + C or y = – 2 sin 2 x + C sin 3 x 1 At x =
π 2
, y = 2
or C = 4 \ y = – 2 sin 2 x + 4 sin 3 x 1 [CBSE Marking Scheme 2017]
Q. 2. Solve the differential equation x dydx + y = x cos x
+ sin x, given that y = 1 when x = p 2 R&U [Delhi, 2017] Q. 3. Find the particular solution of the differential equation (1 + y^2 ) + (x – e tan–1y) dy dx
= 0, given that
y = 0 when x = 1. (^) R&U [Foreign 2017] Sol. Given differential equation can be written as dx dy
x y
1 +^2
e y
tan- y
1
1 2^1
I.F. = e
dy ò 1 + y (^2) = e tan –1 y (^) ½
Solution is given by
xe tan^ –1 y^ = e y
e dy
tan − y (^) tan− y
∫ ×
(^11)
1 2
e y
dy
2 1 y 1 2
tan- ò (^) +
or xe tan^ –1 y^ = e^ c
2 1 y 2
tan-
when x = 1, y = 0 or c = 1 2
\ Solution is given by xe tan^ –1 y^ = 1 2
e 2tan^
–1 y (^) + 1 2
or x = 1 2
( e tan^ –1 y^ + e –tan^ –1 y ) ½
[CBSE Marking Scheme, 2017] (Modified) Q. 4. Find the particular solution of the differential equation (tan – 1^ y – x) dy = (1 + y^2 ) dx, given that when x = 0, y = 0. R&U [OD 2015] Sol. Given (tan –1^ y – x ) dy = (1 + y^2 ) dx
or dx dy =^
tan−^ −
1 1 2
y x y
or
dx dy
x y
1 +^2
tan−
1 1 2
y y
...(i) ½
This is a linear differential equation with : P =^1 1 + y^2
and Q = tan
−
1 1 2
y y
or I.F. = e^ y^ dy^ e y
1 1 + 2 −^1
= tan ½ Multiplying both sides of eqn. (i) by
I.F = (^) e tan−^1 y , we get x. I.F. = (^) ∫ Q I F dy... 1
xe tan^ y
− (^1) = tan^ tan
− (^) −
∫ +
1 2
1 1
y y
e y^ dy c ½
or xe^
tan −^1 y = te dtt (^) + C ∫ ,^ where^ t =^ tan^
− (^1) y
or xe^
tan −^1 y = e t (^) ( t – 1) + C ½
or xe^ tan^ y
− 1 = e tan^ y
− (^1) (tan –1 (^) y – 1) + c ...(ii) It is given that y (0) = 0 i.e., y = 0 when x = 0 Putting x = 0, y = 0 in eqn. (ii), we get 0 = e^0 (0 – 1) + c or c = 1 ½ Putting c = 1 in eq.^ (ii), we get
xe tan^ y
− (^1) = e tan −^1 y (tan − (^1) y − 1 ) + 1 or ( x – tan–1 y + 1) e tan –1^ y^ = 1 ½ [CBSE Marking Scheme 2015] (Modified) Q. 5. Find the particular solutions of differential equation: dy dx
x y x x
cos 1 sin
given that y = 1 when x = 0.
Sol. Since, dy dx
x x
y x 1 sin 1 x
cos sin
or dy dx
y x x
cos 1 sin
x 1 sin x
, ...(i) ½
which is a linear differential equation with P = cos sin
x ; 1 + x
Q = −
x 1 sin x
\ Integrating factor
I.F. = e
x x dx
cos ∫ (^1) +sin = e log(1 + sin x ) = 1 + sin x 1 For general solution, we have y (1 + sin x ) = (^) ∫ − xdx +C ∵ y ( IF) = (^) ∫Q IF( ) dx +C
y (1 + sin x ) = −^^ x +
2 2
C 1
Now, we have y = 1, when x = 0 or 1(1 + sin0) = − 0 + 2
C
or C = +1 ½ Putting C = 1 in eqn. (ii), we get y (1 + sin x ) = −^ x +
2 2
or 2 y (1 + sin x ) + x^2 – 2 = 0 1
Topic-
Homogeneous Differential Equations
Concepts Covered Solution of Homogenous Differential Equation of first order and
first degree
Revision Notes
Homogeneous Differential Equations and their solution : Identifying a Homogeneous Differential equation : STEP 1 : Write down the given differential equation in the form dy dx =^ f x (^^ ,^ y )^.
STEP 2 : If f ( kx, ky ) = knf ( x, y ), then the given differential equation is homogeneous of degree ‘ n ’.
Key Words
Homogeneous: To be Homogeneous a
function must pass this test: f ( zx, zy ) = z nf ( x, y )
Q. 4. To solve the homogeneous differential equation of the form dy dx
f y x
^
^
, we put: U
(A) x = vy (B) y = vx (C) x = v (D) y = v Ans. Option (B) is correct. Q. 5. To solve the homogeneous differential equation the form dx dy
g x y
, we put
(A) y = vx (B) y = x (C) y = v (D) x = vy Ans. Option (D) is correct.
Q. 6. A function F(x, y) is said to be homogeneous function of degree n (a non-negative integer), if (^) U (A) F(l x , l y ) = l n^ F( x, y ) (B) F(l x , l y ) = 1 λ n^
F( x, y )
(C) F(l x , l y ) = lF( x, y ) (D) None of these Ans. Option (A) is correct. Q. 7. A function F(x, y) is a homogeneous function of degree n, if
(A) F( x, y ) = x nf
y x
^
^
(B) F( x, y ) = y g^
x y
n
(C) Both (A) and (B) (D) F( x, y ) = x^ f^
y x
− n ^
Ans. Option (C) is correct.
subjeCtive tyPe questions
Very Short Answer Type Questions (1 mark each) Q. 1. For what value of n is the following a homogeneous
differential equation : dy dx
x y x y xy
n == - - ++
3 2 2
U [CBSE SQP 2020-21] Sol. 3 1 [CBSE Marking Scheme 2020] Detailed Solution: Homogeneous differential equation must have same degree in both numerator and denominator, it means dy dx =^
x y x y xy
3 n 2 2
so, n = 3 Short Answer Type Questions-I (2 marks each) Q. 1. Find the general solution of the following differential equation:
x dy dx
y x y x
^
sin (^) [SQP 2021-2022]
Sol. We have the differential equation: dy dx
y x
y x
^
sin
The equation is a homogeneous differential equation, Putting y = vx Þ dy dx
v x dv dx
The differential equation becomes v x dv dx
Þ dv sin v
= − dx x
Þ cosec vdv = − dx x
Integrating both sides, we get log|cosec v – cot v | = –log| x | + log k , k > 0 (Here log| k | is constant an arbitrary) Þ log|(cosec v – cot v ) x | = log k 1 Þ |(cosec v – cot v ) x | = k Þ x (cosec v – cot v ) = ± k
Þ cosec^ yx − yx x ^
cot (^) = c ,
where is the required general solution ½ [CBSE Marking Scheme 2022] Q. 2. Solve the differential equation :
x y x
dy dx
x y y x
sin^ ææèèçç ööøø÷÷ ++ - - sin æèçæèç ööøø÷÷ == 0
Given that x = 1 when y = pp 2
R&U [CBSE Delhi Set I, II-2020] Q. 3. Find the general solution of the differential equation yex/y^ dx = (xe x/y^ + y^2 ) dy, y ¹ 0 Sol. Given differential equation can be written as dx dy =^
xe y y e
x y x y
/ /
Put
x y =^ v
Þ
dx dy =^
v y dv dy
Þ v^ y
dv dy
v v
Þ y
dv dy =^
y e v \ (^) ò e^ vdv = (^) ò dy Þ ev^ = y + C 1 Þ e x/y^ = y + C , ½ which is the required solution [CBSE Marking Scheme 2020] (Modified) Detailed Solution: Given ye x/ydx = ( xe x/y^ + y^2 ) dy , y ¹ 0
Þ dy dx
= ye xe y
x y x y
/ / (^) + 2
dx dy =^
xe y ye
x y x y
/ /
Put x = vy dx dy =^
v y dv dy
Þ v^ y
dv dy
v v
Þ y
dv dy =^
vye y ye
v
v v
Þ y
dv dy =^
vye y vye ye
v v v
Þ y
dv dy =^
y ye v
2
Þ
dv dy =^
e v Þ (^) ò e dvv = (^) ò dy + C Þ ev^ = y + C Þ ex/y^ = y + C is the required solution Q. 4. Solve the differential equation : x dy – y dx = x^2^ +^ y^2 dx, given that y = 0 when x = 1.
R&U [CBSE Delhi Set-I, 2019]
OR
Find the general solution of the differential equation: x dy – y dx = x 2 + y^2 dx R&U [NCERT][Comptt. Delhi-2016]
Sol. Writing
2 2 2 1 dy y^ x^ y y y dx x x x
Put y vx dy^ v xdv dx dx
Differential equation becomes v x dv v 1 v^2 dx
dv dx v x
∫ (^) + ∫
⇒ log v + 1 + v^2 = log| x | + log c ½
⇒ v + 1 + v^2^ = cx ⇒ y + x^2 + y^2^ = cx^2 1 when x = 1, y = 0 ⇒ c = 1, [ v = 0] ½
∴ y + x^2 + y^2 = x^2 ½ [CBSE Marking Scheme, 2019] (Modified) Detailed Solution : Given differential equation is xdy – ydx = x^^2 + y^^2 dx
Þ xdy = (^) ( x^^2 +^ y^^2 + y dx )
Þ
dy dx =^
x y y x
...(i)
The given differential equation is homogenous with zero degree So, put y = vx Þ dy dx
v x dv dx
From eq. (i), we get
v + x dv dx
x vx vx x
(^2) + ( )^2 +
Þ x dv dx
x v vx x
v
(^2) ( 1 + (^2) ) + −
Þ x dv dx
= (^1) + v^2 + v − v
Þ x dv dx
= (^1) + v^2
Þ
dv 1 + v^2
= dx x Integrating both sides, we get dv 1 + v^2
∫ =^ dx ∫ x
Þ log v + v^2 + 1 = log^ x^ + log^ c
Þ log v + v^2 + 1 = log ( cx )
Þ v + (^) v^2 + 1 = cx
Put v = y x
, we get
y x
y x
cx
^
^
2 1
Þ y + x^^2 +^ y^^2 =^ cx^2 ...(ii) Given, x = 1 when y = 0 \ 0 +^1 +^0 =^ c^ × 1 Þ c = 1
