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Differential Equation - 1 1
10 MAT 21 Dr. V. Lokesha 2012
Engineering Mathematics – II
(10 MAT21)
LECTURE NOTES (FOR II SEMESTER B E OF VTU)
VTU-EDUSAT Programme-16
Dr. V. Lokesha
Professor of Mathematics
ACHARYA INSTITUTE OF TECNOLOGY
Soldevanahalli, Bangalore – 90
Differential Equation - 1 2
10 MAT 21 Dr. V. Lokesha 2012
ENGNEERING MATHEMATICS – II
Content
CHAPTERS
UNIT I DIFFERENTIAL EQUATIONS – I
Unit-1
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10 MAT 21 Dr. V. Lokesha 2012
Differential Equation- I
Equations of first order and higher degree
Overview:
In this unit, we shall study differential equations of the first order and higher degree,
We study the differential equations solvable for and the problems involving in it, differential
equation solvable for and the problems involving in it, Differential equation solvable for and
some problems involving in it. We discuss the problems on special type called Clairaut’s
Equation and reducible to clairaut’s form involving both general solution and singular solution
and we discuss the application of the first order and first degree differential equation with
illustrative examples.
Objective:
At the end of this unit he will be able to understand
• To obtain the solution of non-linear differential equation.
• To obtain the solution of the differential equation of the form .
• The method of the solution is simple involving well known methods.
• Singular solution exists for higher degree equations of first order.
• Clairaut’s equation has numerous engineering applications like geodesics.
• Non linear equation of first order differential equation is reduced to linear differential
equations of first order.
• Mathematical models for some of the applications like Kirchoff’s law, Newton’s law of
cooling etc.
Equations solvable for .
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Introduction:
We are already familiar with differential equations of the first order and first degree , Now
we shall study differential equations of first order and degree higher than the first. For
convenience, we denote by . Such equations are of the form .
A differential equation of the first order but of the nth degree is of the form
Where are functions of and .
In several cases (2) can be solved by reducing (2) to first order and first degree (n)
equations by solving for p(b), solving for y(c), solving for x.
In this unit we discuss the following cases.
• Equations solvable for .
• Equations solvable for y.
• Equations solvable for x.
• Clairaut’s equation
Now we discuss the first case Equations solvable for p.
Splitting up the left hand side of (2) into n linear factors, we have
.
Equating each of the factors to Zero,
.
Solving each of these equations of the first order of first degree, we get the solutions
. These n solutions constitute the general
solution of (1).
Otherwise, the general solution of (1) may be written as
.
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Problems:
1. Solve :
Sol: Given equation is
From (1)
On integration, we get
From (2)
On integration, we get
Thus, , constitute the required solution.
On combining these into one, the required solution can be written as
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10 MAT 21 Dr. V. Lokesha 2012
.
2. Solve:
Sol: We have,
Adding on both sides,
=
From (1):
From (2):
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10 MAT 21 Dr. V. Lokesha 2012
Thus combining (3) and (4), the required general solution is
3. Solve:
Sol: we have,
On Integration, we get
On integration, we get
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Thus combining (3) and (4), the required general solution is
2
2 2
2
2
2
Given that 6 0 =
( 3 )( 2 ) 0
4. + 6 =0
3 0 or 2 0
3 0
Sol :
dy d
dyx p xyp y where p
d
yx xy y
dx d
xxp y xp yxp y xp y
dyx
x
x
yd
+ − =
⇒ +
−
− =
⇒ + = − =
⇒ + =
( ) ( )3 2
2
1 2
1 2
3
1 2
or 2 0
3 log or 2 log
log log or log log
The general solution yis y 0
dyx y
dxdy dx dy dx
C Cy x y x
yyx C C
x C Cxx
− =
⇒ + = −
− −
=
⇒ =
∴ =
=
2
2
Given that 5 6 0 =
( 6)( 1) 0 6 0 or 1 0
5. So
Sol
lv
6 0 or 1 0
e
:
5 6 0 dy
p p where pdx
p pp pdy dy
dx dx
p p
− − =
⇒ − + =
⇒ − = + =
=
⇒ = + =
− −
−
( )( )1 2
1 2
6 =0 or 0 6 = or
The general solution 0 6 is
dy dx dy dxy x C y x C
y x C y x C
⇒ − + =
⇒ − + =
∴ − − + − =
( )( )
2 ( 4)( 3) 0 4 =0 or 3 0
6. Solve
The ge
7 1
ner 4 3al solu
So
tion
l
is
:
0
2 0 p p
dy dx dy dx
y x C y x C
p p− −
− − −
=
⇒ − − =
∴
+
−
−
=
=
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Exercise:
Solve the following differential equations:
1.
2.
( ) ( )
2
2
Given that ( ) 0 =
Sol
(
7.
0
)( 1
( ) =0
) 0
The general solution i
:
s
dyxp y x p y where p
dxx
dy dyx y x y
dx
xy C
d
y x C
p y
x
p
− − − =
⇒−
− − −
+
∴ −
−
−
=
=
2 2 2
2
2 2
Given that ( ) 0 =
( ) ( ) 0 ( )( ) 0
S
0 or
ol
8. ( )
:
=0
dy dy
xy x y xy
dyxyp x y p xy where p
dxxp yp x y yp
dx dx
xxp y yp x
dyx y
dx
− + + =
⇒
− +
− − − =
⇒ − − =
⇒ −
=
+
( )( )
2 2
2
1 2
2
y 0
0 or 0
log log or
The general solutio 0is yn x
dyx
dxdy dx
ydy xdxy x
yC y x C
xC y x C
− =
⇒
− −
− = − =
⇒ −
∴ −
=
=
=
( )
( )2
2
2 2
2
Given that sinh sinh 1 0
sinh cosh
sinh cosh sinh cosh
sinh cosh or sinh
9.
cosh
Solve 2 sinh
Sol :
1 0
cosh
p x x
p x x
p x x or p x xdy dy
x x x xdx d
p x
xy
p
− − − =
⇒ − =
⇒ − = − = −
⇒ = + = −
⇒ =
+ − =
( ) ( )1 2
1 2
1 2
1 2
sinh or y cosh sinh
or y2
The required solut
2 2 2 or
ion is 0
x x x x x x x x
x
x
x
x
x x C x x C
e e e e e e e
y e C
ey C C
y e C y e
e
C
y C
− − − −
−
+ + = − +
+ − + −
∴
⇒ = + + = − +
⇒ = + = − +
− − + − =
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Equations solvable for
Introduction:
It is possible to express explicitly as a function of and
Differentiating equation (1) w.r.t. and Let ,
We get an equation of the form
Let the solution of this equation be
Eliminating between (1) and (2) we get a relation in and which is the required solution.
Note: It is not possible always to eliminate between (1) and (2). In that case (1) and (2)
together constitute the solution giving and in terms of the parameter .
Problems:
1. Solve:
Sol:
Differentiating (1) w.r.t.
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Consider
Integrating,
Substituting (3) in (1), which is the required solution.
2. Solve: .
Sol: Given equation is
Differentiating both sides w.r.t. ,
Or
Discarding the factor , we have
Integrating
Or .
Putting this value of in (1), we have which is the required solution.
3. Solve:
Sol: Given equation is
Differentiating both sides w.r.t. ,
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Or
Which is a linear equation.
or
Putting these values of in (1), we have
Or
Equations (3) and (4) together constitute the general solution of (1).
1
1
4. Solve 2 On differentiating w.r.to x, we get
2 2
2 2
This is a linear differential equation in
p
Integrati
Sol :
n
n
n
n n p
y px p
dy dp dpp p x np
dx dx dxdx dx x
p x np npdp dp p
−
− −
= +
= = + +
⇒ + = − ⇒ + = −
( )
2
2
2
g factor(IF) = =
the solution is ( ) ( )
dppdp p
n
e e p
x IF np IF dp c−
∫∫ =
∴ = − +∫( )
( )
2
12
1
( )
1
(1)1
Substituting this value of x in given equation, we get
12 y (2)
1Equations (1) and (
n
n
n
n
x p np dp c
npxp c
nnp
x cn
n pc
p n
+
−
⇒ = − +
−⇒ = +
+−
⇒ = + − − − −+
−= + − − − −
+
∫
2) together constitute the general solution.
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Exercise:
Solve the following differential equations:
1.
2.
3.
3
2
2
2
2
On differentiating w.r.to , we get
1 3
1 3 = =
3 1
1 1
6. Solve
So
3 = 1
l : x
dy dpp p
dx dx
dp p pdp dx
dx p p
pd x
x
p
p
p d
y
= = +
−⇒ ⇒
−
− +⇒
−
=
+
2
1 3 1 =
1
On integrating we get,
the solution is =3 log( 1) (1)2
This solution expresses x in terms of p and the given differential equation expresses y in terms of
p dp dxp
px p p c
⇒ + +
−
∴ + + − + − −
p. hence (1) and the given differential equation constitute itsgeneral solution.
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Equations solvable for .
Introduction:
If the given equation on solving for , take the form then
differentiating w.r.t. gives an equationof the form = .
Now it may be solve the new differential equation in . Let its solution be
. The elimination of from (1) and (2) gives the required solution.
Problems:
1. Solve: .
Sol: Solving for , we have
Differentiating both sides w.r.t. ,
Discarding the factor we have
Integrating
Putting this value of p in the given equation, we have
which is the required solution.
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2. Solve:
Sol: Solving for , we have
Differentiating both sides w.r.t. ,
Integrating,
Equations (1) and (2) together constitute the general solution.
3. Solve:
Sol: Writing the above equation in the form
Differentiating both sides w.r.t. ,
⇒
Integrating,
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2 2
2 2
4
2
2
2 3
3
1 Solving for , we get
2
Diffrentiating w.r.to
1 1 1
4.
Sol :
2 2 2
2 2 2
2
yx x y p
p
y
dx y dp dpyp y p
dy p p p dy dy
dp dpp p y yp y p
y px
y dy
p
d
y
= −
⇒ = = − − −
⇒
=
−
+
= − −
( ) ( )3 3 1+2 1+2 0dp
p yp y ypdy
⇒ + =
( )
2
3
33
1+2 0
Consider 0 0
log log log .
Putting this value of p in the given equation, we get
2 y+ 2
dpyp p y
dy
dp dy dpp y
dy y pc
y p c yp c py
cx c
ycy
ycx
⇒ + =
+ = ⇒ + =
⇒ + = ⇒ = ⇒ =
= += ⇒
( )
2
2
2
2
2
1
S o lving fo r , w e get 2 2
D ifferen tiating w .r.to ' '1 1
1 1 1
1
5 . 2 = 0
S ol :
1
x x y yp
ydx y dp dp
y pdy p p p dy dy
y px y
dpp y
p p d
p
y
pp
= −
⇒ = = − + +
⇒ − = −
⇒ −
−
+
2
11 0
dpy
p dy
+ − =
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10 MAT 21 Dr. V. Lokesha 2012
2 2
2
2
1 1 0
Consider 0 0
log log log .
Substituting this value of p in the given equation, we get
2 + 0 y 2=
dpp y
p dy
dp dy dpp y
dy y pc
y p c yp c py
cx cy cx c
y y
⇒ − + =
+ = ⇒ + =
⇒ + = ⇒ = ⇒ =
− +− ⇒ = 0
( )
( )
1
2
2
222
2
Solving for , we get
tan (1)1
Differentia
6. tan
ting w.r.to ' '
1 21 1
Sol :
1
1 1
xp
x ppy
dp dpp p p
dy dydx dp
dy p p dy p
pp x
p
−= + − −
= −
− −+
+ −
+
⇒ = = ++ +
( )
( )
( )
2 2
22
22
2
2 1 21
1
2
1
(2)
(1) 1
(2) .
p p dp
p dyp
pdy dp
p
and constitute the general sol
c
u
y cp
tion
= −+
+ −⇒ =
+
⇒ =+
⇒ − − − −
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10 MAT 21 Dr. V. Lokesha 2012
:
Solve the following differential equations:
1.
2.
3.
2 2
2 2
2 2
1
Solving for , we get log
Differentiating w.r.to ' '1 1 1 1
log (1 log )
1 1 1 lo
7. log
g log
Sol :
0
y y xypp
x x y yp y
ydx dp p dp
y y ydy p p dy p y y dy
dp p dpy y y
p dy
p
p y y dy
= +
−⇒ = = + + − +
−⇒ + − +
=
=
+
2
2
1 log 1 1 0
1 log 1 0
1 0 0
log log log
Substituting this value in
y dp p y dpy
p p dy y p dy
p y dpy
p y p dy
y dp dp dy
p dy p y
p y cp
c p cy
o yr
⇒ − − − =
⇒ − − =
⇒ − = ⇒ − =
⇒ − = ⇒ = =
(1) gives the general solution.
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CLAIRAUT’S EQUATION
Introduction:
An equation of the form is known as Clairaut’s equation.
Differentiating w.r.t , we get
Discarding the factor , we have
Integrating,
Putting , in (1), the required solution is
Thus, the solution of (1) is obtained by writing for p.
Note: 1.If we eliminate from and (1), we get an equation involving no
constant. This is the singular solution of (1) which gives the envelope of the family of straight
lines (3).
2. Equations which are not in the Clairaut’s form can be reduced to Clairaut’s form by
suitable substitutions (transformation).
To obtain the singular solution, we proceed as follows:
i) Find the general solution by replacing by i.e., (3).
ii) Differentiate this w.r.t. giving .
iii) Eliminate from (3) and (4) which will be the singular solution.
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Problems:
1. Solve:
Sol: The given equation can be written as
This is of Clairaut’s form. Hence putting c for p, the solution is .
2. Solve:
Sol: {In problems involving and ,
put }
Put and
So that
Or which is of Clairaut’s form.
Its solution is and hence .
3. Solve: .
Sol: Put
So that
Or which is of Clairaut’s form.
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Its solution is and hence .
4. Solve: . Also find its singular solutions.
Sol. Given equation can be written as which is the Clairaut’s
equation.
(1)
To find the singular solution, differentiate w.r.t. giving
0 = x + → (2)
To eliminate c from (1) and (2), we rewrite (2) as
C = N ( -1)/x
Now substituting this value of c in (1), we get
y = N ( -1) + {N( -1)/x}
which is the desired singular solution.
5. Solve:
Sol: Clairaut’s equation is . Its general solution is obtained by replacing by . Thus
is the required complete integral.
To obtain the singular solution, differentiate the general solution w.r.t. . Then
From
Eliminating from the general solution, we get
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Thus the singular solution (which is a parabola) is the envelope of the family of one
parameter family of straight lines (representing the general solution).
6) Solve the D.E
Sol:
Given
=> …..(1)
this is in the form
replacing ‘p’ by ‘c’ we get
is the general solution.
To find singular sol: consider
=>
Using this in (1) we get
is the singular solution.
7 : Solve
Sol :
….(1) is clairauts equation.
The general soln is
Now consider
Using this in (1) we get
tan( )p px y= −
tan( )p px y= −
1tan p px y−= − 1tany px p−
= −
( )y px f p= +
'[ ( )] 0x f p+ =
2
10
1x
p− =
+
1 xp
x
−=
( )1 1(1 ) tan
xy x x
x− −
= − −
1tany cx c−= −
log( ) 0p y px+ − =
log( )
p
p
y px p
y px e
y px e
−
−
− = −
⇒ − =
⇒ = +
cy cx e
−= +
( )
' ( ) 0
0
1log
p
p
x f p
x e
x e
px
−
−
+ =
⇒ − =
⇒ =
∴ =
( ) ( )
( )
( )
1log1log
1log
1log
xy x ex
y x xx
y x xx
−
= +
⇒ = +
⇒ − =
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10 MAT 21 Dr. V. Lokesha 2012
is the singular solution.
8) Solve use substitution
Sol : Let …..(1)
We have , where
(1) becomes
Divide by X we get ……(2)
is the clairauts Equ with
The solution of (2) becomes
The general Soln of (1) is
Consider
Using this in (2)
is singular solution of (1)
Exercise:
Solve the following differential equations:
1.
2.
3.
2 3 2 0yp x p x y+ − =2 2;X x Y y= =
12
2
dy dy dY dX dY xp x P
dx dY dX dx y dX y
= = = =
( )2
2 3 22
2 2 4 2 2
2 2
0
0
0
x xy p x p x yyy
x p x p x y
XP X P XY
+ − =
⇒ + − =
⇒ + − =
2 3 2 0yp x p x y+ − =
2Y XP P= +
2( ) .f P P=
2Y Xc c= +2 2 2y x c c= +
dYP
dX=
'( ) 0
2 0
2
X f P
X P
XP
+ =
⇒ + =
−⇒ =
( )2
2
2 4 4
XX XY X−−= + =
42
4
xy
−=
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Miscellaneous Problems
Problems:
1. Solve:
Sol: Given equation is
On factorizing, we get
From (1):
On integration, we get
From (2):
On integration, we get
Thus combining (3) and (4), the required general solution is
2. Solve:
Sol:
Differentiating both sides w.r.t. ,
Integrating,
3. Solve:
Sol: is of Clairaut’s type whose solution is got by replzcing p by c.
.
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4. Solve:
Sol: Given equation is
Differentiating both sides w.r.t. ,
Discarding the factor , we have
Or
Or
Putting this value in (1), we get , which is the required solution.
Applications of Differential equations of first order
Overview:
In this section, we study the engineering applications in ordinary differential equation
of first order by illustrative examples.
Objective:
At the end of this section, you will be able to understand
• The physical problems in real situations can be mathematically modeled as a differential
equation.
• The application of differential equation is in various fields like research, Engineering,
Sciences, Management, Life science etc.
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Application to Electric Circuits:
If be the electrical charge on a condenser of capacity C and be the current, then
(a) .
(b) The potential drop across the inductance is .
(c) The potential drop across the resistance is .
(d) The potential drop across the capacitance is .
Also, by Kirchhoff’s Law, the total potential drop (voltage drop) in the circuit is equal to
the applied voltage (E.M.F.).
Example 1. A constant electromotive force E volts is applied to a circuit
containing a constant resistance R ohms in series and a constant inductance
L henries. If the initial current is zero, show that the current builds up to half
its theoretical maximum in seconds.
Sol. Let be the current in the circuit at any time t
By Kirchhoff’s Law, we have
Which is Leibnitz’s linear equation,
Thus (2) becomes,
This equation gives the current in the circuit at ant time .
Clearly, increases with and attains the maximum value .
Let the current in the circuit be half its theoretical maximum after a time T seconds.
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Then
Example 2. The equations of electromotive force in terms of current for an
electrical circuit having resistance R and a condenser of capacity C, in series,
is Find the current at any time , when .
Sol. The given equation can be written as .
Differentiating both sides w.r.t. t, we have
Or which is Leibnitz’s linear equation.
Which gives the current at any time .
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Newton’s Law of cooling
Example: According to Newton,s Law of cooling, the rate at which a substance cools in
moving air is proportional to the difference between the temperature of the substance
and that of the air. If the temperature of the air is and the substance cools from
in 15 minutes, find when the temperature will be 4 .
Sol. Let the unit of time be a minute and T the temperature of the substance of any instant time
t. Then by Newton,s Law of cooling, we have
Integrating,
Initially, when
Substituting the value of c in (1), we have
Or
Also, when
Dividing (2) by (3), we have
Now, when T=40, we have from (4),
Hence the temperature will be of 4 after 52.2 minutes.
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Conduction of Heat
The fundamental principle involved in the problems of heat conduction is that the
quantity of heat Q flowing per second across a slab of area A and thickness whose faces are
at temperature , is given by where k is the coefficient of thermal
conductivity and depends upon the material of the body.
Example: A long hollow pipe has an inner diameter of 10 cm and outer diameter of 20cm. The
inner surface is kept at 200 and the outer surface at 50 . The thermal conductivity is 0.12.
How much heat is lost per minute from a portion of the pipe 20 meters long? Find the
temperature at a distance x = 7.5cm from the centre of the pipe.
Sol. Here the isothermal surfaces are cylinders, the axis of each one of them is the axis of the
pipe. Consider one such cylinder of radius x cm and length 1 cm. The surface area of this
cylinder is . Let be quantity of heat flowing across this surface, then
Integrating, we have
Since
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Also
Subtracting (3) from (2), we have
Or
Hence the heat lost per minute through 20 metre length of the
pipe=60*2000Q=120000*163=1956000 cal.
Now, let T=t, when x = 7.5
From (1),
Subtracting (2) from (5),
Dividing (6) by (4), we have
.
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Multiple choice questions:
1. The general solution of the equations yp2+(X-Y)P-X=0 is
A) (x-y-c)(y2-x
2-c)=0 B) (yx-c)(x
2-y
2-c)=0
C) (y-x-c) )(y2+x
2-c)=0 D) (y-x
2-c)(x
2-y
2-c)=0
2.The given differential equation is solvable for x if it is possible to express x in term of
A) x & y B) x & p C) y & p D) None of these
3. The singular solution of the equation y=px+a/p is
A) y2=4ax B) x
2=4ay C)x
2=y D)y
2=x
4. The general solution of the clairaut’s equation is
A) y=cx + f(c) B) x=cy +f(c) c)y=c x-f(c ) D)none of these
5. The differential equation of first order but second degree ( solvable for p) has the general
solution as
A) f1(x,y,c)+f2(x,y,c)=0 B) f1(x,y,c)*f2(x,y,c)=0
C) f1(x,y,c) - f2(x,y,c)=0 D) f1(x,y,c)/f2(x,y,c)=0
6. If the given differential equation is solving for x then it is of the form
A) x= f(p/y) B) y= f( x,p) C) x= f(y/p) D) x=f( y,p)
7. Clairaut’s equation of P= sin(y-xp) is
A) y= p/x+ sin-1
p B) y=px+sinp C) y=px+sin-1
p D) y=x+sin-1
p
8. The differential equation of R.L series circuit is
A) +Ri =E B) L +I =E C) +Ri = D) L + Ri =E
9. The general solution of the equation - = -
A) (x2+y
2+c)(y-c)=0 B) (x
2+y
2+c)(xy) =0
C) (x2-y
2-c)(xy-c)=0 D) none of these
10. The general solution of the equation y= 2px+ y2p
3 is
A) x2 =2cx + c B) y
2=2cx+c
3 C) y=2cx+c
3 D) none of these
Differential Equation - 1 32
10 MAT 21 Dr. V. Lokesha 2012
11. The generally clairaut’s equation in the form
A) xy= px+ F(p) B) y2= px+ f(P) C) y = p x +f(p) D) y= pc+f(c)
12. The general solution of the clairaut’s equation is obtained by replacing
A) x by c B) y by c C) x by y D) p by c
13. The given differential equation is solvable for y if it is possible to express y in term of
A) x & y B) x & p C) y & p D) None of these
14. The general solution of the equation p= sin(y-xp) is
A) x= cy + sin-1
c B) y = cx + sin-1
x C) y = cx + sin-1
c D) y = px + sin-1
p
15. The general solution of the equation y= 3x+logp
A) y= 3x+logc B) y= 3x+logy C) y= 3x+log(3+cey) D) y= 3x+log(3+e
x)
16. In general, singular solution of the equation in the form
A) f(x, p, c)=0 B) f( y,p,c)=0 C) f ( x,y,c) =0 D) f( x,y,p)=0
17. The singular solution of the equation Y=Px +
A) y-k=4ax B) (y+k)2=4ax C) (y- k)
2=4ax D) y+k=4ax
18. Clairaut’s equation of (px-y)(py+x)=2p by taking substitutions X=x2,Y=y
2 is
A) y2=cx
2- B) y=px- C) y=px+ D) none of these
19. The given differential equation is solvable for x then general solution of the equation in the
form
A) f(x, p, c)=0 B) f( y,p,c)=0 C) f ( x,y,c) =0 D) f( x,y)=0
20. The singular solution of the equation is y=cx+ is
A) x+y=4x B) (x+y)2=4x C) (x+y)
2=4 D) none of these
Answers:
1-C 2-C 3-A 4-A 5-B 6-D 7-C 8-D 9-C 10-C
11-C 12-D 13-B 14-C 15-A 16-D 17-B 18-A 19-C 20-B