NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations
NCERT Exemplar Class 12 Maths Solutions Chapter 9 provides an understanding of equations that relate to one or more functions and their derivatives. In this continually changing world, describing how things change with respect to several factors is very important. The representation of this information is termed as a differential equation. A differential equation is a great way to express a set of information, but it can be hard to solve or formulate. One of the essential languages of Science is that of differential equations.
Latest Updates for CBSE Class 12th
 5 days ago:
CBSE Board exam 2022 Class 12 date sheet released for term 1 exams. Check now.
 03 Sep 2021:
CBSE Class 12 sample papers 2022 are available to download for term 1 exams.
 30 Jul 2021:
CBSE 12th result 2021 released: Direct Link to download CBSE class 12 result
Stay upto date with CBSE Class 12th News
NCERT exemplar class 12 Maths chapter 9 solutions provided on this page would help you gain academic success, ensure an efficient and easy way of clearing doubts, aid in preparation for 12th board exams, and shape your perspective about how the world works. Also, read NCERT Class 12 Maths solutions
Question:1
Answer:
Given
To find: Solution of the given differential equation
Rewrite the equation as,
Integrating on both sides,
Formula:
Here c is some arbitrary constant
d is also some arbitrary constant = c In2
Question:2
Find the differential equation of all nonvertical lines in a plane.
Answer:
To find: Differential equation of all non vertical lines
The general form of equation of line is given by y=mx+c where, m is the slope of the line
The slope of the line cannot be or for the given condition because if it is so, the line will become perpendicular wihout any necessity.
So,
Differentiate the general form of equation of line
Formula:
Differentiating it again it becomes,
Thus we get the diferential euqation of all non vertical lines.
Question:3
Given that and y = 0 when x = 5. Find the value of x when y = 3.
Answer:
Given:
(5,0) is a solution of this equation
To find: Solution of the given differential equation
Rewriting the equation.
Integrate on both the sides,
Formula:
Given (5,0) is a solution so to get c, satisfying these values
Hence the solution is
e^{2y}=2x + 9
when y=3,
e^{2(3)} =2x + 9
e^{6}=2x + 9
e^{6}+ 9=2x
Question:4
Solve the differential equation
Answer:
Given:
To find: Solution of the given differential equation
Rewriting the equations as,
It is a first order liner differential equation Compare it with,
Calculate Integrating Factor,
Hence, solution of the differential equation is given by,
Question:5
Solve the differential equation
Answer:
To find: Solution of the given differential equation
Rewriting the given equation as,
Integrate on both the sides,
Question:6
Answer:
Given:
It is a first order differential equation. Comparing it with,
P(x) =a
Q(x)=e^{xm}
Calculating Integrating Factor
Hence the solution of the given differential equation is ,
Question:7
Solve the differential equation
Answer:
To find: Solution of the given differential equation
Assume x+y=t
Differentiate on both sides with respect to x
Substitute
in the above equation
Rewriting the equation,
Integrate on both the sides,
Is the solution of the differential equation
Question:8
Answer:
Given:
To find: solution of the differential equation
Rewriting the given equation,
Question:9
Solve the differential equation when y = 0, x = 0.
Answer:
Given:
and (0,0) is solution of the equation
To find: solution of the differential equation
Rewriting the given equation as,
Integrating on both the sides
Substitute(0,0) to find c’s value
0+0=c
c=0
Hence, the solution is
Question:10
Answer:
Given:
To find: Solution of the differential equation
Rewriting the equation as
It is a first order linear differential equation
Comparing it with
Calculation the integrating factor,
Therefore, the solution of the differential equation is
Question:11
If y(x) is a solution of and y(0) = 1, then find the value of
Answer:
Given:
To find: Solution of the differential equation
Rewriting the given equation as,
Integrating on both sides,
Let sinx=t and cos xdx= dt
ln(1+y)=ln(2+t)+logc
ln(1+y)+ln(2++sinx)=logc
(1+y)(2+sinx)=c
When x=0 and y=1
c=4
Question:12
If y(t) is a solution of and y(0) = –1, then show that
Answer:
and (0,1) is a solution
To find: Solution for the differential equation
Rewriting the given equation as,
It is a first order linear differential equation
Comparing it with,
Calculation Integrating Factor
Hence the solution for the differential equation is,
Substitution (0,1) to find the value of c
The solution therefore y(1) is
Question:13
Answer:
Given:
To find: Solution of the differential equation
Differentiating on both the sides,
Differntiate again on both the sides
Hence the solution is
Question:14
Answer:
To find: Differential equations of all circles which pass though origin and centre lies on x axis
Assume a point (0,k) on yaxis
Radius of the circle is
General form of the equation of circle is,
Here a, c is the center and r is the radius.
Substituting the values in the above equation,
Differentiate the equation with respect to x
Substituting the value of k in (i)
Question:15
Find the equation of a curve passing through origin and satisfying the differential equation
Answer:
Given:
and (0,0) is a solution to the curve
To find: Equation of the curve satisfying the differential equation
Rewrite the given equation
Comparing with
Calculating Integrating Factor
Calculating
Assume
Hence the solution is
Satisfying (0,0) in the equation of the curve to find the value of c
0+0=c
c=0
therefore equation of the curve is
Question:16
Answer:
Given:
To find: solution for the differential equation
Rewriting the given equation as
Clearly it is a homogenous equation
Assume y=vx
Differentiate on both sides
Substituting dy/dx in the equation
Integrating on both the sides
is the solution for the differential equation
Question:17
Find the general solution of the differential equation
Answer:
Given
To find: Solution of the given differential equation
Rewrite the given equation as,
It is a first order differential equation
Comparing it with
Calculating Integrating Factor
Hence the solution of the given differential equation is
Differentiate on both the sides
Question:18
Answer:
Given:
To find: Solution for the given differential equation
Rewrite the given equation
It is a homogenous differential equation
Assume x=vy
Differentiating on both the sides
Substitute dy/dx in the given equation
Substitute v=x/y
Integrating on both the sides
Question:19
Solve: (x + y) (dx – dy) = dx + dy. [Hint: Substitute x + y = z after separating dx and dy].
Answer:
Given:
To find: Solution of the given differential equation
Rewriting the given equation
Assume x+y=z
Differentiate on both sides with respect to x
Substituting the values in the equation
Integrate on both the sides
Substitute v=xy
Question:20
Answer:
Given:
To Find: Solution of the differential equation
Integrating on both sides
Substitute (2,1) to find value of c
Question:21
Solve the differential equation given that y = 2 when
Answer:
Given:
is a solution of the given differential equation
Rewriting the given equation
It is a first order differential equation
Calculate integrating factor
Therefore, the solution of the differential equation is
Substituting to find the value of c
Hence the solution is
Question:22
Form the differential equation by eliminating A and B in Ax^{2} + By^{2} = 1
Answer:
Given :
Ax^{2}+By^{2}=1
To find: Solution of the differential equation
Differentiate with respect to x
Differentiate the curve (i) again to get,
Substituting this in eq(i)
Question:23
Solve the differential equation (1+ y^{2}) tan^{1}x dx + 2y (1 + x^{2}) dy = 0
Answer:
Given:
To find: Solution for differential equation that s given
Rewriting the given equation as.
Integrate on the both sides
For LHS
Assume tan^{1} =t
For RHS
Assume 1+y^{2}=z
2ydy=dz
Substituting and integrating on both the sides
Substitute for t and z
Solution for the differential equation is
Question:24
Find the differential equation of system of concentric circles with centre (1, 2).
Answer:
To find: Differential equation of concentric circles whose center is (1,2)
Equation of the curve is given by
(xa)^{2}+(yb)^{2}=k^{2}
Where (a,b) is the center and k, radius.
Subsitute the values now,
(x1)^{2}+(y2)^{2}=k^{2}
Differentiate with respect to x
Question:25
Answer:
Now dx/dy (xy) refers to the differentiation of xy with respect to x
Using product rule
When we put it back originally in the differential equation given,
Divide by x
Compare
We get
The above equation is a linear differential equation with P and Q as functions of x
The first to find the solution of a linear differential equation is to find the integrating factor.
The solution of the linear differential equation is
Substituting values for Q and IF
Find the integrals individually,
Using uv for integration
Now
Use product rule
Substitute (i) and (ii) in (a)
Divide by
Question:26
Answer:
Divide throughout by dy
Divide by (1+tany)
Compare
We get
This is the linear differential equation with P and Q as functions of x
Put
Adding and subtracting siny in the numerator
Consider the integral
Let
Differentiate with respect to y
We get
The solution of the linear differential equation will be
Substitute values for Q and IF
Put and differentiate with respect to y
We get
Which means
Hence
Substitute t again
Question:27
Solve: [Hint: Substitute x + y = z]
Answer:
Using the given hint substitute x+y=z
Differentiate z x with respect to x
Integrate
As we know
And
Differentiate with respect to z
We get
hence
Again substitute t
Similarly substitute z
Question:28
Answer:
We get, P= 3 and Q= sin2x
The equation is a linear differential equation where P and Q are functions of x
For the solution of the linear differential equation, we need to find Integrating factor,
The solution of the linear differential equation is
Substitute values for Q and IF
Let
If are two functions, then by integration by parts.
after applying the formula we get,
Again, applying the above stated rule in
Put this value in (1) to get
Question:29
Answer:
Slope of the tangent is given by
Slope of the tangent of the curve
Put y=VX
Using product rule differentiate vx
Integrate
Put
Resubstitute 1
Resubstitute v
The curve is passing through (2,1)
Hence (2,1) will satisfy the equation (a)
Put x=1 and y=2 in (a)
Use loga+logb=logab
Put c in equation (a)
Question:30
Given: Slope of the tangent is
Slope of tangent of a curve
Integrate
Use partial fraction for
Equate the numerator
Put x=0
A=1
Put x=1
B=1
Hence
Hence equation (a) becomes,
Now it is given that the curve is passing through (1,0)
Hence (1,0) will satisfy the equation (b)
Put x=1 and y=0 in b
When we put y=0 in equation b the result is which is undefined
hence, we must simplify equation (b) further
using logalogb=loga/b
Constant c must be taken as log c to eliminate undefined elements in the
equation.(log cand not any other terms because taking logc completely
eliminates the log terms and we don't have to worry about undefined terms
in the equation)
Eliminate log
Substitute x=1 and y=0
c=2
put back c=2 in (c)
Question:31
Answer:
Abscissa refers to the x coordinate and ordinate refers to the y coordinate.
Slope of the tangent is the square of the difference of the abscissa and the ordinate.
Difference of the abscissa and ordinate is (xy) and its square is
Hence the Slope of the tangent is
The curve passes through the (0,0)
Question:32
Answer:
Points on the y axis and x axis are namely A(0,a), B(b,0). The midpoint of AB is P(x,y).
The x coordinate of the points is given by the addition of the x coordinates of A and B divided by 2.
Therefore, the coordinates of A and B are (0,2y) and (2x,0) respectively.
AB is the tangent to curve where P is the point of contact.
Slope of the line given with two points
Here respectively.
Slope of the tangent AB is
Hence the slope of the tangent is y/x
Slope of the tangent curve is given by,
Integrate
using logat logb=logab.
as given curve is passing through(1,a)
Hence (1,1) will satisfy the equation of the curve(a)
Putting
put c back in (a)
Hence the equation of the curve is
Question:33
Answer:
Using logalogb =loga/b
Put y=v x
Differentiate yx with respect to x using product rule
Now Integrate
Substitute log v =t
Differentiate with respect to v.
logt= logx + logc
Resubstitute value of t
log(log v)=log x + logc.
Resubstitute v
Therefore the solution of the differential equation is
Question:34
The degree of the differential equation is:
A. 1
B. 2
C. 3
D. Not defined
Answer:
Degree of differential equation is defined as the highest integer power of the highest order derivative in the equation.
Here’s the differential equation
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Differential means
The given differential equation is not a polynomial because of the term sin dy/dx and therefore degree of such a differential equation is not defined.
Option D is correct.
Question:35
The degree of the differential equation is:
A. 4
B. 3/4
C. not defined
D. 2
Answer:
Generally, for a polynomial degree is the highest power.
Differential equation is Squaring both the sides,
Now for the degree to exit the differential equation must be a polynomial in
some differentials.
The given differential equation is polynomial in differential is
Degree of differential equation is the highest integer power of the highest order
derivative in the equation.
Highest derivative is
There is only one term of the highest order derivative in the equation which is
Whose power is 2 hence the degree is 2
Option D is correct.
Question:36
The order and degree of the differential equation respectively, are
A. 2 and 4
B. 2 and 2
C. 2 and 3
D. 3 and 3
Answer:
The differential equation is
Order is defined as the number which represents the highest derivative in a differential equation.
Is the highest derivative in the given equation is second order hence the degree of the equation is 2 .
Integer powers on the differentials,
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Here differentials means
The given differential equation is polynomial in differentials
Degree of differential equation is the highest integer power of the highest
order derivative in the equation.
Observe that
Of differential equation (a) the maximum power
Highest order is and highest power is 4
Degree of the given differential equation is 4 .
Hence order is 2 and degree is 4
Option A is correct.
Question:37
Answer:
If is a solution of differential equation, then differentiating it will give the same differential equation.
Differentiate the differential equation twice. Twice because all the options have order as 2 and also because there are two constants A and B
Differentiating using product rule
But
Differentiating again with respect to x,
But
Also,
Means,
Question:38
The differential equation for are arbitrary constants is
Answer:
Let us find the differential equation by differentiating y with respect to x twice
Twice because we have to eliminate two constants .
Differentiating,
Differentiating again
Option B is correct.
Question:39
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
Answer:
is constant because e is a constant and c is the integration constant let it be denoted as k hence
is the equation of straight line and (0,0) satisfies the equation.
Option C is correct.
Question:40
Integrating factor of the differential equation is:
A. cosx
B. tanx
C. sec x
D. sinx
Answer:
Differential equation is
Compare
With
The IF integrating factor is given by
Substitute hence
Resubstitute the value of t
hence IF is sec x
Option C is correct.
Question:41
Solution of the differential equation is:
A. tanx + tany = k
B. tanx – tan y = k
C.
D. tanx . tany = k
Answer:
The given differential equation is
Divide it by tanx tany
Integrate
Put tanx=t hence,
Put tany =z hence
That is
Resubstitue t and z
is constant because e is a constant and c is the integration constant let it be denoted as
Option D is correct.
Question:42
Family of curves is represented by the differential equation of degree:
A. 1
B. 2
C. 3
D. 4
Answer:
let us find the differential equation representing it so we have to eliminate
the constant A
Differentiate with respect to x
Put back value of A in y
Now for the degree to exit the differential equation must be a polynomial in some differentials.
Here the differentials mean
The given differential equation is polynomial in differentials
Degree of differential equation is the highest integer power of the highest order derivative in the equation.
Highest derivative is
And highest power to it is 3 . Hence degree is 3 .
Option C is correct.
Question:43
Integrating factor of is:
A. x
B. logx
C.
D. –x
Answer:
Given differential equation
Divide though by x
Compare
We get
The IF integrating factor is given by el $^{\mathrm{iPdx}}$
Option C is correct.
Question:44
B.
C.
D.
Answer:
Integrate
now it is given that y(0)=1 which means when x=0, y=1 hence substitute x=0 and y=0 in (a)
put back in (a)
using
Hence solution of differential equation is
Option D is correct.
Question:45
The number of solutions of when y(1) = 2 is:
A. none
B. one
C. two
D. infinite
Answer:
using
Now as given y(1)=2 which means when x=1, y=2 Substitute x=1 and y=2 in (a)
So only one solution exists.
Option B is correct.
Question:46
Which of the following is a second order differential equation?
Answer:
Order is defined as the number which defines the highest derivative in a differential equation
Second order means the order should be 2 which means the highest
derivative in the equation should be or y Let's examine each of the option given
A.
The highest order derivative is is in first order.
B.
The highest order derivative is is in second order
C.
The highest order derivative is is in third order
D.
The highest order derivative is is in first order
Option B is correct.
Question:47
Integrating factor of the differential equation is:
A. x
B.
C.
D.
Answer:
Divide through by
Compare
We get
The IF factor is given by
Substitute hence
Which means
Resubstitute
Hence the IF integrating factor is
Option C is correct.
Question:48
is the general solution of the differential equation:
A.
B.
C.
D.
Answer:
If is a solution of differential equation then differentiating it will give the same differential equation.
To find the differential equation differentiate with respect to x.
Option C is correct..
Question:49
The differential equation represents:
A. Family of hyperbolas
B. Family of parabolas
C. Family of ellipses
D. Family of circles
Answer:
integrate
k is the integration constant
This is the equation of circle because there is no ‘xy’ term and and have the same coefficient.
This equation represents the family of circles because for different values of c and k we will get different circles.
Option D is correct.
Question:50
The general solution of is:
A.
B.
C.
D.
Answer:
Integrate
substitute cosy =t hence
Which means sinydy=dt
Option A is correct.
Question:51
The degree of the differential equation is
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
The answer is the option (a) 1 as the degree of a differential equation is the highest exponent of the order derivative.
Question:52
The solution of is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (b)
Explanation: 
This is a linear differential equation.
On comparing it with , we get
So, the general solution is:
Given that when x=0 and y=0
Eq. (i) becomes
Question:53
Integrating factor of the differential equation
(a)
(b)
(c)
(d)
Answer:
The answer is the option (b) Sec x
Explanation: 
Question:54
The solution of the differential equation is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (b) y – x = k(1 + xy)
Explanation: 
Question:55
The integrating factor of the differential equation is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (b)
Explanation: 
This is a linear differential equation.
On comparing it with we get
Question:58
The solution of is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (a)
Explanation: 
This is a linear differential equation. Dn comparing it with $\frac{d y}{d x}+P y=Q$, we get
So, the general solution is:
Question:59
The differential equation of the family of curves where a is arbitrary constant, is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (a)
Given:
Question:60
Family y = Ax + A^{3} of curves will correspond to a differential equation of order ,
(a) 3 (b) 2 (c) 1 (d) not defined.
Answer:
The answer is the option (c) 1.
Explanation: 
Putting the value of A in Eq. (i), we gt
Question:61
The general solution of is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (c)
Explanation: 
Question:62
The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is
(a) an ellipse (b) parabola (c) circle (d) rectangular hyperbola
Answer:
The answer is the option (d) Rectangular Hyperbola
Explanation: 
According to the question,
On integrating both sides, we get
which is an equation of rectangular hyperbola.
Question:63
The general solution of differential equation is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (c)
Explanation: 
This is a linear differential equation. On comparing it with we get
So, the general solution is:
Question:64
The solution of equation is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (c)
Explanation: 
Question:65
The differential equation for which is a solution, is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (a)
Explanation: 
On differentiating both sides w.r.t. x, we get
Again, differentiating w.r.t. x, we get
Question:66
The solution of is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (d)
Explanation: 
Here,
Given, when x=0 and y=0
Eq. (i) reduces to
Question:67
The order and degree of the differential equation
(a) 1,4
(b) 3,4
(c) 2,4
(d) 3,2
Answer:
Ans:  The answer is the option (d) 3, 2
Question:68
The order and degree of the differential equation are
(a)
(b) 2,3
(c) 2,1
(d) 3,4
Answer:
Ans: 
The answer is the option (c) 2, 1.
Question:69
The differential equation of family of curves is
(a)
(b)
(c)
(d)
Answer:
Ans:  The answer is the option (d)
Explanation: 
On differentiating both sides w.r.t. x, we get
On putting the value of a in Eq. (i), we get
Question:70
Which of the following is the general solution of ?
(a)
(b)
(c)
(d)
Answer:
Ans: 
The answer is the option (a)
Explanation: 
Question:71
General solution of is
(a)
(b)
(c)
(d)
Answer:
Ans:  The answer is the option (a) y sec x = tan x + C
Explanation: 
Here,
Question:72
Solution of the differential equation is
(a)
(b)
(c)
(d)
Answer:
Ans:  The answer is the option (a) x(y + cos x) = sin x + C
Explanation: 
Here,
Question:73
The general solution of differential equation is
(a)
(b)
(c)
(d)
Answer:
Ans:  The answer is the option (c)
Explanation: 
Question:74
The solution of the differential equation is
(a)
(b)
(c)
(d)
Answer:
The answer is the option (b)
Explaination:
Question:75
The solution of the differential equation
(a)
(b)
(c)
(d)
Answer:
Ans:  The answer is the option (a)
Explanation: 
Here,
Question:76
Answer:
(i) Given differential equation is
Degree of this equation is not defined as it cannot be expresses as polynomial of derivatives.
(ii) We have
So, degree of this equation is two.
(iii) Given that the general solution of a differential equation has three arbitrary constants. So we require three more equations to eliminate these three constants. We can get three more equations by differentiating given equation three times. So, the order of the differential equation is three.
(iv) We have
The equation is of the type
Hence it is linear differential equation.
(v) We have
For solving such equation we multiply both sides by
So we get
This is the required solution of the given differential equation.
(vi) We have,
This equation of the form
The general solution is
(vii) We have
This equation is of the form
So, the general solution is:
(viii) We have, $
(ix) We have,
Which is of the form
So, the general solution is:
(x) Given differential equation is
(xi) Given differential equation is
Which is linear differential equation.
Question:77
Answer:
i) Integrating factor of the differential of the form is given by . Hence given statement is true.
(ii) Solution of the differential equation of the type is given by .
Hence given statement is true.
iii) Correct substitution for the solution of the differential equation of the type is a homogeneous function of zero degree is y=v x.
Hence given statement is true.
(iv) Correct substitution for the solution of the differential equation of the type where g(x, y) is a homogeneous function of the degree zero is x=v y.
Hence given statement is true.
(V) There is no arbitrary constants in the particular solution of a differential equation. Hence given statement is Flase.
(vi) In thegiven equation the number of arbitrary constant is one. So the order order will be one.
Hence given statement is False.
(vii)
Hence the given statement is true.
(viii)
.
Hence the given statement is true.
ix) Given:
Compare with
Here ,
General solution
Hence the given statement is true.
x) Given:
Let y =vx
Hence the given statement is true.
xi) Assume equation of a nonhorizontal line in the plane
y = mx +c
Hence the given statement is true.
Question:56
satisfies which of the following differential equation.
Answer:
given
upon differentiation, we get
after differentiation again we get
Option c is correct
Main Subtopics of NCERT Exemplar Class 12 Maths Solutions Chapter 9
Below is the list of topics which are covered in class 12 Maths NCERT exemplar solutions chapter 9
9.1 Introductory Concepts
9.1.1 Ordinary differential equations
9.2 Order of a differential equation
9.3 Degree of a differential equation
9.4 General and particular solutions of a differential equation
9.5 Formation of a differential equation
9.5.1 Formation of a differential equation whose general solution is given
9.5.2 Formation of a differential equation that will represent a given family of curves
9.6 Methods of solving First order, First degree differential equations
9.6.1 Differential equations with separate variables
9.6.2 Homogeneous Differential equations
9.6.3 Linear differential equations.
What will the students learn in NCERT exemplar class 12 Maths solutions chapter 9?
 These equations have a variety of uses in academic subjects like Physics, Chemistry, Biology, Geology, etc., which makes it important to acquire detailed learning of these equations.
 A thorough understanding of differential equations is needed to solve Newton's laws of motion and cooling, the rate of spread of a pandemic or an epidemic, and also help measure market competition.
 If understood well, NCERT exemplar solutions for class 12 Maths chapter 9 would help you answer realworld questions like  How do you model an antibioticresistant bacteria's growth? How do you study everchanging online purchasing trends? At what rate, a radioactive material decays? What is the trajectory of a biological cell motion? How the suspension system of a car works to give you a smooth ride? These equations are a way to describe many things in the universe and model nearly anything around us. Scientists and geniuses understand the world through differential equations; you can too.
NCERT Exemplar Class 12 Maths Solutions
Important topics to cover for exams from NCERT exemplar class 12 Maths solutions chapter 9
 NCERT exemplar class 12 Maths chapter 9 solutions revolve around using the mathematical tool of Differentiation, analyses properties like the intervals of increment and decrement, study the local maximum and minimum of functions that are quadratic.
 In class 12 Maths NCERT exemplar solutions chapter 9, you will be introduced to the concepts related to differential equations, types of these equations, general and specific solutions to solve them, the formation and production of these equations, different forms of the equations, and a wide range of applications of modelling reallife situations by applying these equations
In NCERT exemplar class 12 Maths solutions chapter 9 pdf download, we would also look at the graphical aspects of differential equations, including a family of straight lines and curves, and have a look at the devised solutions and mathematical tools to solve the most complex equations over time.
NCERT Exemplar Class 12 Solutions
Also, check NCERT Solutions for questions given in book:
Chapter 1  
Chapter 2  
Chapter 3  
Chapter 4  
Chapter 5  
Chapter 6  
Chapter 7  
Chapter 8  
Chapter 9  
Chapter 10  
Chapter 11  
Chapter 12  
Chapter 13 
Frequently Asked Question (FAQs)  NCERT Exemplar Class 12 Maths Solutions Chapter 9 Differential Equations
Question: Are these solutions helpful in competitive exams?
Answer:
Yes, these NCERT exemplar class 12 Maths solutions chapter 9 can be highly useful in understanding the way the questions should be solved in entrance exams.
Question: How to make use of these NCERT exemplar class 12 Maths chapter 9 solutions?
Answer:
These solutions can be used for both getting used to the chapter and its topics and to also get an idea about how to solve questions in exams.
Question: What are the basic take away from the class 12 Maths NCERT exemplar solutions chapter 9?
Answer:
One can understand how to stepwise solve these questions through NCERT exemplar class 12 maths solutions chapter 9 and how the CBSE expects a student to solve in their final paper.
Question: Who prepare these solutions to the maths chapter?
Answer:
We have the best maths teachers onboard to solve the questions as per the students understanding and also CBSE standards. These teachers prepare the NCERT exemplar solutions for class 12 Maths chapter 9.
Latest Articles
AP Inter Supply Results 2021 Declared  Check Andhra Pradesh B...
AP Inter Supply Results 2021  Supplementary result for AP int...
TS Inter Supply Result 2021  Check Result and Statistics Here!
TS Inter Supply Result 2021  Telangana Board will release 2nd...
ISC Time Table 2022 (Released)  Check ISC 12th Board Exam New...
ISC Time Table 2022  CISCE has released the ISC new exam date...
Rashtriya Military School Result 2021 Date  Check RMS Result ...
Rashtriya Military School Result 2021  Rashtriya Military Sch...
HBSE 12th Result 2022  Check Haryana Board Class 12 Result @b...
HBSE 12th Result 2022  Class 12 HBSE result for Science, Arts...
GSEB HSC Result 2022  Check Gujarat Board Class 12 General & ...
GSEB HSC Result 2022  Gujarat Secondary and Higher Secondary ...
TN 12th Result 2022  Tamilnadu Class 12 Result for Science, C...
TN 12th Result 2022  Directorate of Government Examinations, ...
NSEP Syllabus 202122  Check IAPT NSEP Syllabus Here
NSEP Syllabus 202122  IAPT provides the syllabus of NSEP 202...
CBSE Passing Marks for Class 10th & 12th 2022
CBSE Passing Marks for Class 10th & 12th 2022  Check the pass...
Assam HS Result 2022  Assam 12th result 2022 @resultsassam.ni...
Assam HS result  Check Assam Class 12 result 2022 date, websi...
Explore Popular Degree, Branches and Courses
Browse by Degrees
Browse by Branches/Majors
Questions related to CBSE Class 12th
if any candidate has compartment in cbse any one subject but then candidate passed compartment exam then he / she is not eligible for upcet exam
Dear Annu,
If a candidate has cleared that compartment exam, then he/she is eligible for the exam. As the final result will not include your compartment result, it will depict your overall result after passing the compartment or reappear marks.
Hope this helps.
Good Luck :)
when cbse class 12 date sheet will be released for 202021
Dear,
Central Board of Secondary Education (CBSE) will release the CBSE term 1 exam dates 202122 on October 18 on the official website cbse.gov.in. The exams are expected to begin from November 15 and go on till December 2021. Once released, the CBSE Class 12 date sheet 2022 will be made available on our website for students and parents. Students preparing for the CBSE Class 12th board exams 2022 should download and save the date sheet and refer to it for preparation of different subjects.
Hope this helps!
CBSE class 12th exam date 2022
CBSE have not yet released the date sheet of class 12th board exam, it is expected that the board will release CBSE class 12th term 1 datesheet on 18th October 2021, the datesheet will be available on the official website for term 1 on 18th October 2021 on cbse.gov.in
The board will release the datesheet for all the steam ie. Arts, science and commerce, the exam for skill subjects will start from 15th November and the exam for mains subjects will start from 24th November, the board will conduct class 12th exam in two terms, where 50% syllabus will be covered in term 1 and another 50% will be covered in term2.
For more relevant information, you may visit here:
https://school.careers360.com/articles/cbseclass12datesheet
Hope this helps, all the best.
In Neet 2021 phase 2 details, the selection for code for class 12th is quite confusing. It is for CBSE class 12th 2021 passed out. so what code should be entered. can you clarify this
Hi
You need to choose the code 2 .
This is because,
Now your results are out and you have passed it and code 2 says it is for all those who have passed their 10+2/higher /senior secondary/ Indian School Certificate Examination with physics, chemistry, biology/biotechnology and English
Here's a short brief of Neet codes :
) Code 1 : code 1 is for class 12th board 2021 examination appearing students with physics, chemistry, biology and English or have have already appeared but results are awaited i. e have not been declared.
) Code 2 : code 2 is for all those who have passed their 10+2/higher /senior secondary/ Indian School Certificate Examination with physics, chemistry, biology/biotechnology and English
) Code 3 : code 3 says it is for those passed their intermediate / pre degree examination with physics, chemistry, biology/biotechnology and English ,your physics, chemistry and biology/biotechnology should also contain practical exam.
) Code 4 : pre  medical degree with Physics, chemistry and biology/biotechnology after passing hsc i. e 10+2 or puc or equivalent.
) Code 5 : code 5 is for all those who have passed 1 st year graduation with all the three of Physics, chemistry and biology/biotechnology
) Code 6 : code 6 is for all those students who have passed BSc with at least any two of following subjects :
>>>>Physics,
>>>>chemistry ,
>>>>biology/biotechnology
) Code 7 : code 7 is for those who have passed their examination from foreign board which is equivalent of Indian 10+2 board.
For more about codes visit
https://medicine.careers360.com/articles/neetqualificationcodes
Your neet 2021 application/registration is divided in 2 parts:

1st part/phase of registration to be filled before neet 2021 examination

2nd part/phase registration to be filled after neet 2021 examination
NTA has opened the window for neet phase 2 cum phase 1 correction on 1st october 2021 and the last date for it is 10th October upto 11:50 pm
To fill neet phase2 registration, you can follow the below steps:
) First of visit the official website i. e neet.nta.nic.in
) Then click on NEET phase 2 registration.
To make it even easier for you the direct link is given below : https://testservices.nic.in/NEET2021/Root/home.aspx?appFormId=101042111
) Now, log in using your log in credentials
) Fill the details asked , recheck everything.
) Finally, click on the submit button.
For details checkout
https://medicine.careers360.com/articles/neetphase2registration
) Documents to be uploaded in part/phase 2 of application form filling
>>>>>>Your Class X pass certificate

pdf format

size: 50 kb to 300kb
>>>>>>category certificate if you belong to ews/ obc /sc/st

pdf format

size: 50kb to 300kb
>>>>>>PwBD certificate ( if applicable)

pdf format

size: 50kb to 300kb
>>>>>> Citizenship certificate or Embassy certificate or any kind of Documentary proof for Citizenship certificate ( for foreign applicants)

PDF format

file size: 50 kb to 300 kb
For complete details about documents visit : https://medicine.careers360.com/articles/neetphotosizeformatdocuments
To help you further,
last 3 years neet qualifying cut off data is given below for reference
:
. Year wise cut off marks
Category 2020 2019 2018
General 720147 701134 691119
URPH 146129 133120 118107
Obc/sc/st 146113 133107 118  96
You can check the same at
https://medicine.careers360.com/articles/neetcutoff
To get the complete list of colleges in which you have chances, you can go through our college predictor at
https://medicine.careers360.com/neetcollegepredictor?utm_source=qna
It gives you a personalized report with top college in which you have chances for admission , with the help of predicted colleges you can make better choice while filling your choices of colleges during your counselling
Thank you
jac class 12 th Compamental result 2021 declare date
Hello Aspirant,
Hope you are doing well.
JAAC 12th compartment results will be announced in September 2021 by the Jharkhand Academic Council. The compartment results for the 12th board of examination will be available online at jacresults.com, the official result website. It is possible for students to check the JAC inter compartment result 2021 in the same way that they can check the annual JAC 12th result 2021.
To view the JAC 12th compartmental result 2021, they must first enter their roll number and roll code. July 2021 will be the date for the 12th compartment examination of the Joint Assessment Commission. The results of the JAC intermediate compartment were announced online on December 29, 2020, which was the same date as the previous season.
Hope you got the answer to your question. Incase of any queries, feel free to question. Have a great day!