Class 12 Maths Chapter 9 Differential Equations

Exercise 9.1

Determine order and degree (if defined) of the differential equations given in Questions 1 to 10.

Question 1.
$\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } +({ sin }y^{ III })=0$
Solution:
Order of the equation is 4
It is not a polynomial in derivatives so that it
has not degree.

Question 2.
${ y }^{ I }+5y=0$
Solution:
${ y }^{ I }+5y=0$
It is a D.E. of order one and degree one.

Question 3.
${ \left( \frac { ds }{ dt } \right) }^{ 4 }+3s{ \left( \frac { { d }^{ 2 }s }{ { dt }^{ 2 } } \right) }=0$
Solution:
Order of the equation is 2.
Degree of the equation is

Question 4.
${ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }+cos\left( \frac { dy }{ dx } \right) =0$
Solution:
${ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }+cos\left( \frac { dy }{ dx } \right) =0$
It is a D.E. of order 2 and degree undefined

Question 5.
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =cos3x+sin3x$
Solution:
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =cos3x+sin3x$
It is a D.E. of order 2 and degree 1.

Question 6.
${ { (y }^{ III }) }^{ 2 }+{ { ( }y^{ II }) }^{ 3 }+{ { (y }^{ I }) }^{ 4 }+{ y }^{ 5 }=0$
Solution:
Order of the equation is 3
Degree of the equation is 2

Question 7.
${ { y }^{ III } }+{ 2y^{ II } }+{ { y }^{ I } }=0$
Solution:
${ { y }^{ III } }+{ 2y^{ II } }+{ { y }^{ I } }=0$
The highest order derivative is y.
Thus the order of the D.E. is 3.
The degree of D.E is 1

Question 8.
${ y }^{ I }+y={ e }^{ x }$
Solution:
${ y }^{ I }+y={ e }^{ x }$
The order of the D. E. = 1 (highest order derivative)
The degree of the D.E. = 1.

Question 9.
${ y }^{ III }+{ { (y }^{ I }) }^{ 2 }+2y=0$
Solution:
${ y }^{ III }+{ { (y }^{ I }) }^{ 2 }+2y=0$
The highest derivative is 2.
Order of the D.E. = 2.
Degree of the D. E = 1

Question 10.
${ y }^{ II }+{ { 2y }^{ I } }+siny=0$
Solution:
Order of the equation is 2
Degree of the equation is 1

Question 11.
The degree of the differential equation
${ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 3 }{ +\left( \frac { dy }{ dx } \right) }^{ 2 }+sin{ \left( \frac { dy }{ dx } \right) }+1=0$
(a) 3
(b) 2
(c) 1
(d) not defined
Solution:
${ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 3 }{ +\left( \frac { dy }{ dx } \right) }^{ 2 }+sin{ \left( \frac { dy }{ dx } \right) }+1=0$
The degree not defined.
Because the differential equation can not be written as a polynomial in all the differential coefficients.
Hence option (d) is correct.

Question 12.
The order of the differential equation
${ 2x }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } +y=0$
(a) 2
(b) 1
(c) 0
(d) not defined
Solution:
${ 2x }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } +y=0$
Thus order of the D.E. = 2
Hence option (a) is correct.

Exercise 9.2

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation

Question 1.
$y={ e }^{ x }+1:{ y }^{ II }-{ y }^{ I }=0$
Solution:
$y={ e }^{ x }+1:{ y }^{ II }-{ y }^{ I }=0$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Question 2.
$y=x^{ 2 }+2x+c:{ y }^{ I }-2x-2=0$
Solution:
$y=x^{ 2 }+2x+c:{ y }^{ I }-2x-2=0$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2

Question 3.
$y=cosx+c:{ y }^{ I }+sinx=0$
Solution:
$y=cosx+c:{ y }^{ I }+sinx=0$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Question 4.
$y=\sqrt { 1+{ x }^{ 2 } } :{ y }^{ I }=\frac { xy }{ 1+{ x }^{ 2 } }$
Solution:
$y=\sqrt { 1+{ x }^{ 2 } } :{ y }^{ I }=\frac { xy }{ 1+{ x }^{ 2 } }$
vedantu class 12 maths Chapter 9 Differential Equations 4

Question 5.
$y=Ax:x{ y }^{ I }=y(x\neq 0)$
Solution:
$y=Ax:x{ y }^{ I }=y(x\neq 0)$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 5

Question 6.
$y=x\quad sinx;{ xy }^{ I }=y+x\sqrt { { x }^{ 2 }-{ y }^{ 2 } } (x\neq 0\quad and\quad x>y\quad or\quad x<-y)$
Solution:
$y=x\quad sinx;{ xy }^{ I }=y+x\sqrt { { x }^{ 2 }-{ y }^{ 2 } } (x\neq 0\quad and\quad x>y\quad or\quad x<-y)$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Question 7.
xy = logy + C,
UP Board Solutions for Class 12 Maths Chapter 9 Differential Equations 7
Solution:
xy = logy + C,

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Question 8.
$y-cosy=x:(ysiny+cosy+x){ y }^{ I }=y$
Solution:
$y-cosy=x:(ysiny+cosy+x){ y }^{ I }=y$
vedantu class 12 maths Chapter 9 Differential Equations 8

Question 9.
$x+y={ ta }n^{ -1 }y;{ y }^{ 2 }{ y }^{ I }+{ y }^{ 2 }+1=0$
Solution:
$x+y={ ta }n^{ -1 }y;{ y }^{ 2 }{ y }^{ I }+{ y }^{ 2 }+1=0$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9

Question 10.
$y=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\in (-a,a);x+y\frac { dy }{ dx } =0,(y\neq 0)$
Solution:
$y=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\in (-a,a);x+y\frac { dy }{ dx } =0,(y\neq 0)$
vedantu class 12 maths Chapter 9 Differential Equations 10

Question 11.
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(a) 0
(b) 2
(c) 3
(d) 4
Solution:
(b) The general solution of a differential equation of fourth order has 4 arbitrary constants.
Because it contains the same number of arbitrary constants as the order of differential equation.

Question 12.
The number of arbitrary constants in the particular solution of a differential equation of third order are:
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) Number of arbitrary constants = 0
Because particular solution is free from arbitrary constants.

Exercise 9.3

In each of the following, Q. 1 to 5 form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

Question 1.
$\frac { x }{ a } +\frac { y }{ b } =1$
Solution:
Given that $\frac { x }{ a } +\frac { y }{ b } =1$ …(i)
differentiating (i) w.r.t x, we get
$\frac { 1 }{ a } +\frac { 1 }{ b } { y }^{ I }=0$ …(ii)
again differentiating w.r.t x, we get
$\frac { 1 }{ b } { y }^{ II }=0\Rightarrow { y }^{ II }=0$
which is the required differential equation

Question 2.
y² = a(b² – x²)
Solution:
given that
y² = a(b² – x²)…(i)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2

Question 3.
y = ae3x+be-2x
Solution:
Given that
y = ae3x+be-2x …(i)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Question 4.
y = e2x (a+bx)
Solution:
y = e2x (a+bx)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 4

Question 5.
y = ex(a cosx+b sinx)
Solution:
The curve y = ex(a cosx+b sinx) …(i)
differentiating w.r.t x
byjus class 12 maths Chapter 9 Differential Equations 5

Question 6.
Form the differential equation of the family of circles touching the y axis at origin
Solution:
The equation of the circle with centre (a, 0) and radius a, which touches y- axis at origin
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Question 7.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution:
The equation of parabola having vertex at the origin and axis along positive y-axis is
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Question 8.
Form the differential equation of family of ellipses having foci on y-axis and centre at origin.
Solution:
The equation of family ellipses having foci at y- axis is
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 8

Question 9.
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Solution:
Equation of the hyperbola is $\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Differentiating both sides w.r.t x
byjus class 12 maths Chapter 9 Differential Equations 9
which is the req. differential eq. of the hyperbola.

Question 10.
Form the differential equation of the family of circles having centre on y-axis and radius 3 units
Solution:
Let centre be (0, a) and r = 3
Equation of circle is
x² + (y – a)² = 9 …(i)
Differentiating both sides, we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10
which is required equation

Question 11.
Which of the following differential equation has $y={ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }$ as the general solution ?
(a) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0$
(b) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0$
(c) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +1=0$
(d) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -1=0$
Solution:
(b) $y={ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\Rightarrow \frac { dy }{ dx } ={ c }_{ 1 }{ e }^{ x }-{ c }_{ 2 }{ e }^{ -x }$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0$

Question 12.
Which of the following differential equations has y = x as one of its particular solution ?
(a) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=x$
(b) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=x$
(c) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0$
(d) $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=0$
Solution:
(c) y = x
$\frac { dy }{ dx } =1,\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0$

Exercise 9.4

For each of the following D.E in Q. 1 to 10 find the general solution:

Question 1.
$\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx }$
Solution:
$\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx }$
$\frac { dy }{ dx } =\frac { 1-cosx }{ 1+cosx } =\frac { { 2sin }^{ 2 }\left( \frac { x }{ 2 } \right) }{ { 2cos }^{ 2 }\left( \frac { x }{ 2 } \right) } ={ tan }^{ 2 }\left( \frac { x }{ 2 } \right)$
integrating both sides, we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Question 2.
$\frac { dy }{ dx } =\sqrt { 4-{ y }^{ 2 } } (-2<y<2)$
Solution:
$\frac { dy }{ dx } =\sqrt { 4-{ y }^{ 2 } } \Rightarrow \int { \frac { dy }{ \sqrt { { 4-y }^{ 2 } } } } =\int { dx }$
$\Rightarrow { sin }^{ -1 }\frac { y }{ 2 } =x+C$
$\Rightarrow y=2sin(x+C)$

Question 3.
$\frac { dy }{ dx } +y=1(y\neq 1)$
Solution:
$\frac { dy }{ dx } +y=1\Rightarrow \int { \frac { dy }{ y-1 } } =-\int { dx }$
$\Rightarrow log(y-1)=-x+c\Rightarrow y=1+{ e }^{ -x }.{ e }^{ c }$
$Hence\quad y=1+{ Ae }^{ -x }$
which is required solution

Question 4.
sec² x tany dx+sec² y tanx dy = 0
Solution:
we have
sec² x tany dx+sec² y tanx dy = 0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 4

Question 5.
$\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right) dx=0$
Solution:
we have
$\left( { e }^{ x }+{ e }^{ -x } \right) dy-\left( { e }^{ x }-{ e }^{ -x } \right) dx=0$
Integrating on both sides
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 5

Question 6.
$\frac { dy }{ dx } =\left( { 1+x }^{ 2 } \right) \left( { 1+y }^{ 2 } \right)$
Solution:
$\frac { dy }{ { 1+y }^{ 2 } } =\left( { 1+x }^{ 2 } \right) dx$
integrating on both side we get
${ tan }^{ -1 }y={ x+\frac { 1 }{ 3 } }x^{ 3 }+c$
which is required solution

Question 7.
y logy dx – x dy = 0
Solution:
$\because \quad y\quad logy\quad dx=x\quad dy\Rightarrow \frac { dy }{ y\quad logy } =\frac { dx }{ x }$
integrating we get
tiwari academy class 12 maths Chapter 9 Differential Equations 7

Question 8.
${ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }$
Solution:
${ x }^{ 5 }\frac { dy }{ dx } =-{ y }^{ 5 }\Rightarrow \int { { y }^{ -5 }dy } =-\int { { x }^{ -5 }dx }$
$\Rightarrow -\frac { 1 }{ { y }^{ 4 } } =\frac { 1 }{ { x }^{ 4 } } +4c\Rightarrow { x }^{ -4 }+{ y }^{ -4 }=k$

Question 9.
solve the following
$\frac { dy }{ dx } ={ sin }^{ -1 }x$
Solution:
$\frac { dy }{ dx } ={ sin }^{ -1 }x\Rightarrow \int { dy } =\int { { sin }^{ -1 }xdx }$
integrating both sides we get
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9

Question 10.
${ e }^{ x }tany\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0$
Solution:
${ e }^{ x }tany\quad dx+{ (1-e }^{ x }){ sec }^{ 2 }dy=0$
we can write in another form
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10

Find a particular solution satisfying the given condition for the following differential equation in Q.11 to 14.

Question 11.
$\left( { x }^{ 3 }+{ x }^{ 2 }+x+1 \right) \frac { dy }{ dx } ={ 2x }^{ 2 }+x;y=1,when\quad x=0$
Solution:
here
$dy=\frac { { 2x }^{ 2 }+x }{ \left( { x }^{ 3 }+{ x }^{ 2 }+x+1 \right) } dx$
integrating we get
tiwari academy class 12 maths Chapter 9 Differential Equations 11

Question 12.
$x\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } =1,y=0\quad when\quad x=2$
Solution:
$x\left( { x }^{ 2 }-1 \right) \frac { dy }{ dx } =1,y=0\quad when\quad x=2$
$\Rightarrow \int { dy } =\int { \frac { dy }{ x(x+1)(x-1) } }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12

Question 13.
$cos\left( \frac { dy }{ dx } \right) =a,(a\epsilon R),y=1\quad when\quad x=0$
Solution:
$cos\left( \frac { dy }{ dx } \right) =a\quad \therefore \frac { dy }{ dx } ={ cos }^{ -1 }a$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Question 14.
$\frac { dy }{ dx } =ytanx,y=1\quad when\quad x=0$
Solution:
$\frac { dy }{ dx } =ytanx\Rightarrow \int { \frac { dy }{ y } } =\int { tanx\quad dx }$
=> logy = logsecx + C
When x = 0, y = 1
=> log1 = log sec0 + C => 0 = log1 + C
=> C = 0
∴ logy = log sec x
=> y = sec x.

Question 15.
Find the equation of the curve passing through the point (0,0) and whose differential equation ${ y }^{ I }={ e }^{ x }sinx$
Solution:
${ y }^{ I }={ e }^{ x }sinx$
$\Rightarrow dy={ e }^{ x }sinx\quad dx$
tiwari academy class 12 maths Chapter 9 Differential Equations 15

Question 16.
For the differential equation $xy\frac { dy }{ dx } =(x+2)(y+2)$ find the solution curve passing through the point (1,-1)
Solution:
The differential equation is $xy\frac { dy }{ dx } =(x+2)(y+2)$
or xydy=(x + 2)(y+2)dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 16

Question 17.
Find the equation of a curve passing through the point (0, -2) given that at any point (pc, y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point
Solution:
According to the question $y\frac { dy }{ dx } =x$
$\Rightarrow \int { ydy } =\int { xdx } \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c$
0, – 2) lies on it.c = 2
∴ Equation of the curve is : x² – y² + 4 = 0.

Question 18.
At any point (x, y) of a curve the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3) find the equation of the curve given that it passes through (- 2,1).
Solution:
Slope of the tangent to the curve = $\frac { dy }{ dx }$
slope of the line joining (x, y) and (- 4, – 3)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 18
tiwari academy class 12 maths Chapter 9 Differential Equations 18.1

Question 19.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and offer 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Solution:
Let v be volume of the balloon.
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 19

Question 20.
In a bank principal increases at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years
Solution:
Let P be the principal at any time t.
According to the problem
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 20

Question 21.
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years
Solution:
Let p be the principal Rate of interest is 5%
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 21

Question 22.
In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present
Solution:
Let y denote the number of bacteria at any instant t • then according to the question
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 22

Question 23.
The general solution of a differential equation $\frac { dy }{ dx } ={ e }^{ x+y }$ is
(a) ${ e }^{ x }+{ e }^{ -y }=c$
(b) ${ e }^{ x }+{ e }^{ y }=c$
(c) ${ e }^{ -x }+{ e }^{ y }=c$
(d) ${ e }^{ -x }+{ e }^{ -y }=c$
Solution:
(a) $\frac { dy }{ dx } ={ e }^{ x }.{ e }^{ y }\Rightarrow \int { { e }^{ -y }dy } =\int { { e }^{ x }dx }$
$\Rightarrow { e }^{ -y }={ e }^{ x }+k\Rightarrow { e }^{ x }+{ e }^{ -y }=c$

Exercise 9.5

Show that the given differential equation is homogeneous and solve each of them in Questions 1 to 10

Question 1.
(x²+xy)dy = (x²+y²)dx
Solution:
(x²+xy)dy = (x²+y²)dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Question 2.
${ y }^{ I }=\frac { x+y }{ x }$
Solution:
${ y }^{ I }=\frac { x+y }{ x }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2

Question 3.
(x-y)dy-(x+y)dx=0
Solution:
$\frac { dy }{ dx } =\frac { x+y }{ x-y } =\frac { 1+\frac { y }{ x } }{ 1-\frac { y }{ x } }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Question 4.
(x²-y²)dx+2xy dy=0
Solution:
$\frac { dy }{ dx } =\frac { { y }^{ 2 }-{ x }^{ 2 } }{ 2xy }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 4

Question 5.
${ x }^{ 2 }\frac { dy }{ dx } ={ x }^{ 2 }-{ 2y }^{ 2 }+xy$
Solution:
$\frac { dy }{ dx } =1-2{ \left( \frac { y }{ x } \right) }^{ 2 }+\frac { y }{ x }$
vedantu class 12 maths Chapter 9 Differential Equations 5

Question 6.
$xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx$
Solution:
$xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Question 7.
$\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} ydx=\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} xdy$
Solution:
$\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} ydx=\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} xdy$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Question 8.
$x\frac { dy }{ dx } -y+xsin\left( \frac { y }{ x } \right) =0$
Solution:
$x\frac { dy }{ dx } -y+xsin\left( \frac { y }{ x } \right) =0\Rightarrow \frac { dy }{ dx } =\frac { y }{ x } -sin\frac { y }{ x }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 8

Question 9.
$ydx+xlog\left( \frac { y }{ x } \right) dy-2xdy=0$
Solution:
$\frac { dy }{ dx } =\frac { y }{ 2x-xlog\frac { y }{ x } } =\frac { \frac { y }{ x } }{ 2-log\frac { y }{ x } }$
vedantu class 12 maths Chapter 9 Differential Equations 9

Question 10.
$\left( { 1+e }^{ \frac { x }{ y } } \right) dx+{ e }^{ \frac { x }{ y } }\left( 1-\frac { x }{ y } \right) dy=0$
Solution:
$\frac { dx }{ dy } =-\frac { { e }^{ \frac { x }{ y } }\left( 1-\frac { x }{ y } \right) }{ { 1+e }^{ \frac { x }{ y } } } =\frac { \left( \frac { x }{ y } -1 \right) { e }^{ \frac { x }{ y } } }{ { 1+e }^{ \frac { x }{ y } } } =f(x,y)$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 10

For each of the following differential equation in Q 11 to 15 find the particular solution satisfying the given condition:

Question 11.
(x + y) dy+(x – y)dx = 0,y = 1 when x = 1
Solution:
given
(x + y) dy+(x – y)dx = 0
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 11
vedantu class 12 maths Chapter 9 Differential Equations 11.1

Question 12.
x²dy+(xy+y²)dx=0, y=1 when x=1
Solution:
$\frac { dy }{ dx } =\frac { xy+{ y }^{ 2 } }{ { x }^{ 2 } } =f(x,y)$
f(x,y) is homogeneous
∴ put y = vx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12

Question 13.
$\left( x{ sin }^{ 2 }\frac { y }{ x } -y \right) dx+xdy=0,y=\frac { \pi }{ 4 } ,when\quad x=1$
Solution:
$\left( x{ sin }^{ 2 }\frac { y }{ x } -y \right) dx+xdy=0$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Question 14.
$\frac { dy }{ dx } -\frac { y }{ x } +cosec\left( \frac { y }{ x } \right) =0,y=0\quad when\quad x=1$
Solution:
$\frac { dy }{ dx } -\frac { y }{ x } +cosec\left( \frac { y }{ x } \right) =0$
which is a homogeneous differential equation
vedantu class 12 maths Chapter 9 Differential Equations 14

Question 15.
$2xy-{ y }^{ 2 }-{ 2x }^{ 2 }\frac { dy }{ dx } =0,y=2,when\quad x=1$
Solution:
$\frac { dy }{ dx } =\frac { y }{ x } +\frac { 1 }{ 2 } { \left( \frac { y }{ x } \right) }^{ 2 }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 15

Question 16.
A homogeneous equation of the form $\frac { dx }{ dy } =h\left( \frac { x }{ y } \right)$ can be solved by making the substitution,
(a) y=vx
(b) v=yx
(c) x=vy
(d) x=v
Solution:
(c) option x = vy

Question 17.
Which of the following is a homogeneous differential equation?
(a) (a) (4x + 6y + 5)dy-(3y + 2x + 4)dx = 0
(b) $(xy)dx-({ x }^{ 3 }+{ y }^{ 3 })dy$
(c) $({ x }^{ 3 }+{ 2y }^{ 2 })dx+2xydy=0$
(d) ${ y }^{ 2 }dx+{ (x }^{ 2 }-xy-{ y }^{ 2 })dy=0$
Solution:
(d)

Exercise 9.6

Find the general solution of the following differential equations in Q.1 to 12

Question 1.
$\frac { dy }{ dx } +2y=sinx$
Solution:
Given equation is a linear differential equation of the form $\frac { dy }{ dx } +Py=Q$ ;
Here, P = 2, Q = sin x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 1

Question 2.
$\frac { dy }{ dx } +3y={ e }^{ -2x }$
Solution:
$\frac { dy }{ dx } +3y={ e }^{ -2x }$
Here P = 3, $IF={ e }^{ \int { p.dx } }={ e }^{ 3x }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 2
which is required equation

Question 3.
$\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }$
Solution:
$\frac { dy }{ dx } +\frac { y }{ x } ={ x }^{ 2 }$
$IF={ e }^{ \int { \frac { 1 }{ x } dx } }={ e }^{ logx }=x$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 3

Question 4.
$\frac { dy }{ dx } +(secx)y=tanx\left( 0\le x<\frac { \pi }{ 2 } \right)$
Solution:
Here, P = secx, Q = tanx; $IF={ e }^{ \int { p.dx } }={ e }^{ \int { secx.dx } }$
$={ e }^{ log|secx+tanx| }$
= sec x + tan x
i.e., The solu. is y.× I.F. = ∫Q × I.F. dx + c
or y × (secx+tanx) = ∫tanx(secx+tanx)dx+c
Reqd. sol. is
∴ y(secx + tanx) = (secx + tanx)-x + c

Question 5.
${ cos }^{ 2 }x\frac { dy }{ dx } +y=tanx\left( 0\le x\le \frac { \pi }{ 2 } \right)$
Solution:
$\frac { dy }{ dx } +{ y\quad sec }^{ 2 }x={ sec }^{ 2 }x\quad tanx$
⇒ integrating factor = ${ e }^{ \int { { sec }^{ 2 }xdx } }={ e }^{ tanx }$
byjus class 12 maths Chapter 9 Differential Equations 5

Question 6.
$x\frac { dy }{ dx } +2y={ x }^{ 2 }logx$
Solution:
$\frac { dy }{ dx } +\frac { 2 }{ x } y\quad =\quad x\quad logx$
Here P = $\frac { 2 }{ x }$ and Q = x logx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 6

Question 7.
$xlogx\frac { dy }{ dx } +y=\frac { 2 }{ x } logx$
Solution:
$\frac { dy }{ dx } +\frac { 1 }{ xlogx } y=\frac { 2 }{ { x }^{ 2 } }$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 7

Question 8.
(1+x²)dy+2xy dx = cotx dx(x≠0)
Solution:
(1+x²)dy+2xy dx = cotx dx
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 8

Question 9.
$x\frac { dy }{ dx } +y-x+xy\quad cotx=0(x\neq 0)$
Solution:
$x\frac { dy }{ dx } +y-x+xy\quad cotx=0$
$x\frac { dy }{ dx } +(1+xcot x)y=x$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 9

Question 10.
$(x+y)\frac { dy }{ dx } =1$
Solution:
$(x+y)\frac { dy }{ dx } =1$
$\frac { 1 }{ (x+y) } \frac { dx }{ dy } =1\Rightarrow \frac { dx }{ dy } =x+y$
byjus class 12 maths Chapter 9 Differential Equations 10

Question 11.
$ydx+(x-{ y }^{ 2 })dy=0$
Solution:
$ydx+(x-{ y }^{ 2 })dy=0$
$\Rightarrow y\frac { dx }{ dy } +x-{ y }^{ 2 }=0$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 11

Question 12.
$\left( { x+3y }^{ 2 } \right) \frac { dy }{ dx } =y(y>0)$
Solution:
$y\frac { dx }{ dy } =x+{ 3y }^{ 2 }\quad or\quad \frac { dx }{ dy } -\frac { x }{ y } =3y$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 12

For each of the following Questions 13 to is find a particular solution, satisfying the given condition:

Question 13.
$\frac { dy }{ dx } +2ytanx=sinx,y=0\quad when\quad x=\frac { \pi }{ 3 }$
Solution:
$\frac { dy }{ dx } +(2tanx)y=sinx,P=2tanx$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 13

Question 14.
$\left( 1+{ x }^{ 2 } \right) \frac { dy }{ dx } +2xy=\frac { 1 }{ 1+{ x }^{ 2 } } ,y=0\quad when\quad x=1$
Solution:
$\frac { dy }{ dx } +\frac { 2x }{ 1+{ x }^{ 2 } } y=\frac { 1 }{ { \left( { 1+x }^{ 2 } \right) }^{ 2 } }$
byjus class 12 maths Chapter 9 Differential Equations 14

Question 15.
$\frac { dy }{ dx } -3ycotx=sin2x,y=2\quad when\quad x=\frac { \pi }{ 2 }$
Solution:
Here P = -3cot x
Q = sin 2x
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 15

Question 16.
Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point
Solution:
$\frac { dy }{ dx } =x+y\Rightarrow \frac { dy }{ dx } -y=x\Rightarrow P=-1$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 16

Question 17.
Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5
Solution:
By the given condition
$x+y-\left| \frac { dy }{ dx } \right|=5$
NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations 17

Question 18.
The integrating factor of the differential equation $x\frac { dy }{ dx } -y={ 2x }^{ 2 }$
(a) ${ e }^{ -x }$
(b) ${ e }^{ -y }$
(c) $\frac { 1 }{ x }$
(d) x
Solution:
(c) $P=\frac { -1 }{ x } \therefore IF={ e }^{ -\int { \frac { 1 }{ x } dx } }={ e }^{ -logx }=\frac { 1 }{ x }$

Question 19.
The integrating factor of the differential equation $\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay$ (-1<y<1) is
(a) $\frac { 1 }{ { y }^{ 2 }-1 }$
(b) $\frac { 1 }{ \sqrt { { y }^{ 2 }-1 } }$
(c) $\frac { 1 }{ 1-{ y }^{ 2 } }$
(d) $\frac { 1 }{ \sqrt { { 1-y }^{ 2 } } }$
Solution:
(d) $\left( { 1-y }^{ 2 } \right) \frac { dx }{ dy } +yx=ay$
byjus class 12 maths Chapter 9 Differential Equations 19