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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations are provided here. In class 11th, you have already learned how to differentiate a given function (f) with respect to an independent variable. In this article, you will get NCERT solutions for class 12 maths chapter 9 for all major topics of NCERT Class 12 maths syllabus. The equation of function and its one or more derivatives is called a differential equation.
In this differential equations class 12 questions and answers, some basic concepts related to the differential equations solutions, particular solutions, and general solutions of differential equations class 12 will be comprehensively discussed. In NCERT solutions for chapter 9 class 12 maths, questions from all these topics are covered in this article. If you are interested in other subjects then you can refer to NCERT solutions for class 12
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You will also learn some methods to find the differential equations solutions, the formation of differential equations class 12, and applications of differential equations in different areas in this NCERT class 12 maths chapter 9 question answer are also explained in details. Questions related to these topics are also covered in the NCERT solutions for class 12 maths ch 9 differential equations article. You can refer to NCERT solutions from classes 6 to 12 to learn CBSE maths and science.
Also read :
>> Ordinary Differential Equations (ODEs): Ordinary Differential Equations involve derivatives of a function concerning a single independent variable. They are commonly used to model dynamic systems and phenomena.
>> Partial Differential Equations (PDEs): Partial Differential Equations involve derivatives of a function concerning multiple independent variables. They are frequently used in physics to describe phenomena like heat diffusion, wave propagation, and fluid dynamics.
>> Types of Differential Equations: Differential equations can be categorised based on their order, linearity, and specific properties. Common types include:
FirstOrder Differential Equations
SecondOrder Differential Equations
Linear Differential Equations
Nonlinear Differential Equations
Homogeneous Differential Equations
NonHomogeneous Differential Equations
>> Methods for Solving Differential Equations: Various techniques can be employed to solve differential equations, including:
Separation of Variables
Integrating Factors
Exact Differential Equations
Linear Differential Equations with Constant Coefficients
Method of Undetermined Coefficients
Variation of Parameters
Laplace Transforms
>> Applications of Differential Equations: Differential equations have widespread applications in science and engineering. Some examples include modelling population growth, describing electrical circuits, predicting radioactive decay, and simulating fluid flow.
Free download NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations for CBSE Exam.
NCERT differential equations class 12 solutions  Exercise: 9.1
Question:1 Determine order and degree (if defined) of differential equation
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation is 4
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined
Question:2 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation is 1
Now, the given differential equation is a polynomial equation in its derivatives and its highest power raised to y ' is 1
Therefore, it's a degree is 1.
Question:3 Determine order and degree (if defined) of differential equation
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation is 2
Now, the given differential equation is a polynomial equation in its derivatives and power raised to s '' is 1
Therefore, it's a degree is 1
Question:4 Determine order and degree (if defined) of differential equation.
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation is 2
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined
Question:5 Determine order and degree (if defined) of differential equation.
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:6 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 3 Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 2
Therefore, it's degree is 2
Question:7 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 3
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:8 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 1
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:9 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:10 Determine order and degree (if defined) of differential equation
Answer:
Given function is
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Now, the given differential equation is a polynomial equation in it's dervatives and power raised to is 1
Therefore, it's degree is 1
Question:11 The degree of the differential equation is
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Now, the given differential equation is a not polynomial equation in it's dervatives
Therefore, it's degree is not defined
Therefore, answer is (D)
Question:12 The order of the differential equation is
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, order of given differential equation is 2
Therefore, answer is (A)
NCERT differential equations class 12 solutions  Exercise: 9.2
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Substituting the values of y’ and y'' in the given differential equations,
y''  y' = e ^{ x }  e ^{ x } = 0 = RHS.
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y’ in the given differential equations,
.
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y’ in the given differential equations,
.
Therefore, the given function is not the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y in RHS,
.
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
.
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
Substituting the values of y in RHS.
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
y' + siny.y' = 1
y'(1 + siny) = 1
Substituting the values of y and y' in LHS,
= (x + cosy) = y = RHS
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y' in LHS,
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Substituting the values of y and y' in LHS,
Therefore, the given function is a solution of the corresponding differential equation.
Answer:
(D) 4
The number of constants in the general solution of a differential equation of order n is equal to its order.
Answer:
(D) 0
In a particular solution of a differential equation, there is no arbitrary constant.
Differential Equations Class 12 NCERT Solutions  Exercise: 9.3
Answer:
Given equation is
Differentiate both the sides w.r.t x
Now, again differentiate it w.r.t x
Therefore, the required differential equation is or
Answer:
Given equation is
Differentiate both the sides w.r.t x
(i)
Now, again differentiate it w.r.t x
(ii)
Now, divide equation (i) and (ii)
Therefore, the required differential equation is
Answer:
Given equation is
(i)
Differentiate both the sides w.r.t x
(ii)
Now, again differentiate w.r.t. x
(iii)
Now, multiply equation (i) with 2 and add equation (ii)
(iv)
Now, multiply equation (i) with 3 and subtract from equation (ii)
(v)
Now, put values from (iv) and (v) in equation (iii)
Therefore, the required differential equation is
Answer:
Given equation is
(i)
Now, differentiate w.r.t x
(ii)
Now, again differentiate w.r.t x
(iii)
Now, multiply equation (ii) with 2 and subtract from equation (iii)
(iv)
Now,put the value in equation (iii)
Therefore, the required equation is
Answer:
Given equation is
(i)
Now, differentiate w.r.t x
(ii)
Now, again differentiate w.r.t x
(iii)
Now, multiply equation (i) with 2 and multiply equation (ii) with 2 and add and subtract from equation (iii) respectively
we will get
Therefore, the required equation is
Question:6 Form the differential equation of the family of circles touching the yaxis at origin.
Answer:
If the circle touches yaxis at the origin then the centre of the circle lies at the xaxis
Let r be the radius of the circle
Then, the equation of a circle with centre at (r,0) is
(i)
Now, differentiate w.r.t x
(ii)
Put equation (ii) in equation (i)
Therefore, the required equation is
Answer:
Equation of perabola having vertex at origin and axis along positive yaxis is
(i)
Now, differentiate w.r.t. c
(ii)
Put value from equation (ii) in (i)
Therefore, the required equation is
Question:8 Form the differential equation of the family of ellipses having foci on yaxis and centre at origin.
Answer:
Equation of ellipses having foci on yaxis and centre at origin is

Now, differentiate w..r.t. x
(i)
Now, again differentiate w.r.t. x
(ii)
Put value from equation (ii) in (i)
Our equation becomes
Therefore, the required equation is
Answer:
Equation of hyperbolas having foci on xaxis and centre at the origin
Now, differentiate w..r.t. x
(i)
Now, again differentiate w.r.t. x
(ii)
Put value from equation (ii) in (i)
Our equation becomes
Therefore, the required equation is
Question:10 Form the differential equation of the family of circles having centre on yaxis and radius 3 units.
Answer:
Equation of the family of circles having centre on yaxis and radius 3 units
Let suppose centre is at (0,b)
Now, equation of circle with center (0,b) an radius = 3 units
Now, differentiate w.r.t x
we get,
Put value fro equation (ii) in (i)
Therefore, the required differential equation is
Question:11 Which of the following differential equations has as the general solution?
Answer:
Given general solution is
Differentiate it w.r.t x
we will get
Again, Differentiate it w.r.t x
Therefore, (B) is the correct answer
Question:12 Which of the following differential equations has as one of its particular solution?
Answer:
Given equation is
Now, on differentiating it w.r.t x
we get,
and again on differentiating it w.r.t x
we get,
Now, on substituting the values of in all the options we will find that only option c which is satisfies
Therefore, the correct answer is (C)
NCERT class 12 maths chapter 9 question answer  Exercise: 9.4
Question:1 Find the general solution:
Answer:
Given,
Question:5 Find the general solution:
Answer:
Given, in the question
Let,
This is the general solution
Question:7 Find the general solution:
Answer:
Given,
let logy = t
=> 1/ydy = dt
This is the general solution
Question:8 Find the general solution:
Answer:
Given, in the question
This is the required general equation.
Question:11 Find a particular solution satisfying the given condition:
Answer:
Given, in the question
Now,
Now comparing the coefficients
A + B = 2; B + C = 1; A + C = 0
Solving these:
Putting the values of A,B,C:
Therefore,
Now, y= 1 when x = 0
c = 1
Putting the value of c, we get:
Question:12 Find a particular solution satisfying the given condition:
Answer:
Given, in the question
Let,
Now comparing the values of A,B,C
A + B + C = 0; BC = 0; A = 1
Solving these:
Now putting the values of A,B,C
Given, y =0 when x =2
Therefore,
Question:13 Find a particular solution satisfying the given condition:
Answer:
Given,
Now, y =1 when x =0
1 = 0 + c
Therefore, c = 1
Putting the value of c:
Question:14 Find a particular solution satisfying the given condition:
Answer:
Given,
Now, y=1 when x =0
1 = ksec0
k = 1
Putting the vlue of k:
y = sec x
Question:15 Find the equation of a curve passing through the point (0, 0) and whose differential equation is .
Answer:
We first find the general solution of the given differential equation
Given,
Now, Since the curve passes through (0,0)
y = 0 when x =0
Putting the value of c, we get:
Question:16 For the differential equation , find the solution curve passing through the point (1, –1).
Answer:
We first find the general solution of the given differential equation
Given,
Now, Since the curve passes through (1,1)
y = 1 when x = 1
Putting the value of C:
Answer:
According to the question,
Now, Since the curve passes through (0,2).
x =0 and y = 2
Putting the value of c, we get
Answer:
Slope m of line joining (x,y) and (4,3) is
According to the question,
Now, Since the curve passes through (2,1)
x = 2 , y =1
Putting the value of k, we get
Answer:
Volume of a sphere,
Given, Rate of change is constant.
Now, at t=0, r=3 and at t=3 , r =6
Putting these value:
Also,
Putting the value of c and k:
Answer:
Let p be the principal amount and t be the time.
According to question,
Now, at t =0 , p = 100
and at t =10, p = 200
Putting these values,
Also,
,
So value of r = 6.93%
Answer:
Let p be the principal amount and t be the time.
According to question,
Now, at t =0 , p = 1000
Putting these values,
Also, At t=10
,
After 10 years, the total amount would be Rs.1648
Answer:
Let n be the number of bacteria at any time t.
According to question,
Now, at t=0, n = 100000
Again, at t=2, n= 110000
Using these values, for n= 200000
NCERT class 12 maths chapter 9 question answer  Exercise: 9.5
Question:1 Show that the given differential equation is homogeneous and solve each of them.
Answer:
The given diffrential eq can be written as
Let
Now,
Hence, it is a homogeneous equation.
To solve it put y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both side, we get;
Again substitute the value ,we get;
This is the required solution of given diff. equation
Question:2 Show that the given differential equation is homogeneousand solve each of them.
Answer:
the above differential eq can be written as,
............................(i)
Now,
Thus the given differential eq is a homogeneous equaion
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both sides, we get; (and substitute the value of )
this is the required solution
Question:3 Show that the given differential equation is homogeneous and solve each of them.
Answer:
The given differential eq can be written as;
....................................(i)
Hence it is a homogeneous equation.
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both sides, we get;
again substitute the value of
This is the required solution.
Question:4 Show that the given differential equation is homogeneous and solve each of them.
Answer:
we can write it as;
...................................(i)
Hence it is a homogeneous equation
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
integrating on both sides, we get
.............[ ]
This is the required solution.
Question:5 Show that the given differential equation is homogeneous and solve it.
Answer:
............(i)
Hence it is a homogeneous eq
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides, we get;
after substituting the value of
This is the required solution
Question:6 Show that the given differential equation is homogeneous and solve it.
Answer:
.................................(i)
henxe it is a homogeneous equation
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides,
Substitute the value of v=y/x , we get
Required solution
Question:7 Solve.
Answer:
......................(i)
By looking at the equation we can directly say that it is a homogenous equation.
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
integrating on both sides, we get
substitute the value of v= y/x , we get
Required solution
Question:8 Solve.
Answer:
...............................(i)
it is a homogeneous equation
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides we get;
Required solution
Question:9 Solve.
Answer:
..................(i)
hence it is a homogeneous eq
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
integrating on both sides, we get; ( substituting v =y/x)
This is the required solution of the given differential eq
Question:10 Solve.
Answer:
.......................................(i)
Hence it is a homogeneous equation.
Now, to solve substitute x = yv
Diff erentiating on both sides wrt
Substitute this value in equation (i)
Integrating on both sides, we get;
This is the required solution of the diff equation.
Question:11 Solve for particular solution.
Answer:
..........................(i)
We can clearly say that it is a homogeneous equation.
Now, to solve substitute y = vx
Diff erentiating on both sides wrt
Substitute this value in equation (i)
On integrating both sides
......................(ii)
Now, y=1 and x= 1
After substituting the value of 2k in eq. (ii)
This is the required solution.
Question:12 Solve for particular solution.
Answer:
...............................(i)
Hence it is a homogeneous equation
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i), we get
Integrating on both sides, we get;
replace the value of v=y/x
.............................(ii)
Now y =1 and x = 1
therefore,
Required solution
Question:13 Solve for particular solution.
Answer:
..................(i)
Hence it is a homogeneous eq
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
on integrating both sides, we get;
On substituting v =y/x
............................(ii)
Now,
put this value of C in eq (ii)
Required solution.
Question:14 Solve for particular solution.
Answer:
....................................(i)
the above eq is homogeneous. So,
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
on integrating both sides, we get;
.................................(ii)
now y = 0 and x =1 , we get
put the value of C in eq 2
Question:15 Solve for particular solution.
Answer:
The above eq can be written as;
By looking, we can say that it is a homogeneous equation.
Now, to solve substitute y = vx
Differentiating on both sides wrt
Substitute this value in equation (i)
integrating on both sides, we get;
.............................(ii)
Now, y = 2 and x =1, we get
C =1
put this value in equation(ii)
Question:16 A homogeneous differential equation of the from can be solved by making the substitution.
Answer:
for solving this type of equation put x/y = v
x = vy
option C is correct
Question:17 Which of the following is a homogeneous differential equation?
Answer:
Option D is the right answer.
we can take out lambda as a common factor and it can be cancelled out
NCERT class 12 maths chapter 9 question answer  Exercise: 9.6
Question:1 Find the general solution:
Answer:
Given equation is
This is type where p = 2 and Q = sin x
Now,
Now, the solution of given differential equation is given by relation
Let
Put the value of I in our equation
Now, our equation become
Therefore, the general solution is
Question:2 Solve for general solution:
Answer:
Given equation is
This is type where p = 3 and
Now,
Now, the solution of given differential equation is given by the relation
Therefore, the general solution is
Question:3 Find the general solution
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:4 Solve for General Solution.
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:5 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is where and
Now,
Now, the solution of given differential equation is given by relation
take
Now put again
Put this value in our equation
Therefore, the general solution is
Question:6 Solve for General Solution.
Answer:
Given equation is
Wr can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Let
Put this value in our equation
Therefore, the general solution is
Question:7 Solve for general solutions.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
take
Put this value in our equation
Therefore, the general solution is
Question:8 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of the given differential equation is given by the relation
Therefore, the general solution is
Question:9 Solve for general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of the given differential equation is given by the relation
Lets take
Put this value in our equation
Therefore, the general solution is
Question:10 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Lets take
Put this value in our equation
Therefore, the general solution is
Question:11 Solve for general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:12 Find the general solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Therefore, the general solution is
Question:13 Solve for particular solution.
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when
at
Now,
Therefore, the particular solution is
Question:14 Solve for particular solution.
Answer:
Given equation is
we can rewrite it as
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 1
at x = 1
Now,
Therefore, the particular solution is
Question:15 Find the particular solution.
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 2 when
at
Now,
Therefore, the particular solution is
Answer:
Let f(x , y) is the curve passing through origin
Then, the slope of tangent to the curve at point (x , y) is given by
Now, it is given that
It is type of equation where
Now,
Now,
Now, Let
Put this value in our equation
Now, by using boundary conditions we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
Our final equation becomes
Therefore, the required equation of the curve is
Answer:
Let f(x , y) is the curve passing through point (0 , 2)
Then, the slope of tangent to the curve at point (x , y) is given by
Now, it is given that
It is type of equation where
Now,
Now,
Now, Let
Put this value in our equation
Now, by using boundary conditions we will find the value of C
It is given that curve passing through point (0 , 2)
Our final equation becomes
Therefore, the required equation of curve is
Question:18 The Integrating Factor of the differential equation is
Answer:
Given equation is
we can rewrite it as
Now,
It is type of equation where
Now,
Therefore, the correct answer is (C)
Question:19 The Integrating Factor of the differential equation is
Answer:
Given equation is
we can rewrite it as
It is type of equation where
Now,
Therefore, the correct answer is (D)
Class 12 Maths Chapter 9 NCERT solutions  Miscellaneous Exercise
Question:1 Indicate Order and Degree.
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is
Therefore, the order of the given differential equation is 2
Now, the given differential equation is a polynomial equation in its derivative y '' and y 'and power raised to y '' is 1
Therefore, it's degree is 1
Question:1 Indicate Order and Degree.
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is y'
Therefore, order of given differential equation is 1
Now, the given differential equation is a polynomial equation in it's dervatives y 'and power raised to y ' is 3
Therefore, it's degree is 3
Question:1 Indicate Order and Degree.
Answer:
Given function is
We can rewrite it as
Now, it is clear from the above that, the highest order derivative present in differential equation is y''''
Therefore, order of given differential equation is 4
Now, the given differential equation is not a polynomial equation in it's dervatives
Therefore, it's degree is not defined
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Again, differentiating both sides w.r.t. x,
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given,
Now, differentiating both sides w.r.t. x,
Putting values in LHS
Therefore, the given function is the solution of the corresponding differential equation.
Answer:
Given equation is
we can rewrite it as
(i)
Differentiate both the sides w.r.t x
(ii)
Put value from equation (ii) in (i)
Therefore, the required differential equation is
Question:4 Prove that is the general solution of differential equation , where c is a parameter.
Answer:
Given,
Now, let y = vx
Substituting the values of y and y' in the equation,
Integrating both sides we get,
Now,
Let
Now,
Let v ^{ 2 } = p
Now, substituting the values of I _{ 1 } and I _{ 2 } in the above equation, we get,
Thus,
Answer:
Now, equation of the circle with center at (x,y) and radius r is
Since, it touch the coordinate axes in first quadrant
Therefore, x = y = r
(i)
Differentiate it w.r.t x
we will get
(ii)
Put value from equation (ii) in equation (i)
Therefore, the differential equation of the family of circles in the first quadrant which touches the coordinate axes is
Question:6 Find the general solution of the differential equation
Answer:
Given equation is
we can rewrite it as
Now, integrate on both the sides
Therefore, the general solution of the differential equation is
Question:7 Show that the general solution of the differential equation is given by , where A is parameter.
Answer:
Given,
Integrating both sides,
Let
Let A = ,
Hence proved.
Question:8 Find the equation of the curve passing through the point whose differential equation is
Answer:
Given equation is
we can rewrite it as
Integrate both the sides
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point
So,
Now,
Therefore, the equation of the curve passing through the point whose differential equation is is
Question:9 Find the particular solution of the differential equation , given that when .
Answer:
Given equation is
we can rewrite it as
Now, integrate both the sides
Put
Put again
Put this in our equation
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
Now, put the value of C
Therefore, the particular solution of the differential equation is
Question:10 Solve the differential equation
Answer:
Given,
Let
Differentiating it w.r.t. y, we get,
Thus from these two equations,we get,
Question:11 Find a particular solution of the differential equation , given that , when . (Hint: put )
Answer:
Given equation is
Now, integrate both the sides
Put
Now, given equation become
Now, integrate both the sides
Put again
Now, by using boundary conditions we will find the value of C
It is given that
y = 1 when x = 0
Now, put the value of C
Therefore, the particular solution of the differential equation is
Question:12 Solve the differential equation .
Answer:
Given,
This is equation is in the form of
p = and Q =
Now, I.F. =
We know that the solution of the given differential equation is:
Question:13 Find a particular solution of the differential equation , given that .
Answer:
Given equation is
This is type where and
Now,
Now, the solution of given differential equation is given by relation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when
at
Now, put the value of C
Therefore, the particular solution is
Question:14 Find a particular solution of the differential equation , given that when
Answer:
Given equation is
we can rewrite it as
Integrate both the sides
Put
put again
Put this in our equation
Now, by using boundary conditions we will find the value of C
It is given that y = 0 when x = 0
at x = 0
Now, put the value of C
Therefore, the particular solution is
Answer:
Let n be the population of the village at any time t.
According to question,
Now, at t=0, n = 20000 (Year 1999)
Again, at t=5, n= 25000 (Year 2004)
Using these values, at t =10 (Year 2009)
Therefore, the population of the village in 2009 will be 31250.
Question:16 The general solution of the differential equation is
Answer:
Given equation is
we can rewrite it as
Integrate both the sides
we will get
Therefore, answer is (C)
Question:17 The general solution of a differential equation of the type is
Answer:
Given equation is
and we know that the general equation of such type of differential equation is
Therefore, the correct answer is (C)
Question:18 The general solution of the differential equation is
Answer:
Given equation is
we can rewrite it as
It is type of equation where
Now,
Now, the general solution is
Therefore, (C) is the correct answer
If you want to get command on concepts then differential equations solutions of NCERT exercise are listed below
This class 12 differential equations NCERT solutions has 5 marks weightage in 12th board final examination. Generally, one question is asked from this Chapter 9 Class 12 Maths that can be studied in detail from the NCERT Class 12 maths book in the 12th board final exam. You can score these 5 marks very easily with the help of these Ncert Solutions For Class 12 Maths Chapter 9 Differential Equations.
Class 12 Maths ch 9 is very important for the students aspiring for the 12th board exam. This NCERT Class 12 Maths Chapter 9 solutions holds good weightage in competitive exams like JEE Main, VITEEE, BITSAT. In this chapter, there are 6 exercises with 95 questions. All these questions are prepared and explained in this class 12 differential equations NCERT solutions article.
9.1 Introduction
9.2 Basic Concepts
9.2.1. Order of a differential equation
9.2.2 Degree of a differential equation
9.3. General and Particular Solutions of a Differential Equation
9.4 Formation of a Differential Equation whose General Solution is given
9.4.1 Procedure to form a differential equation that will represent a given family of curves
9.5. Methods of Solving First Order, First Degree Differential Equations
9.5.1 Differential equations with variables separable
9.5.2 Homogeneous differential equations
9.5.3 Linear differential equations
So, what is basically a differential equation? A differential equation is an equation in which derivatives of the dependent variable with respect to independent variables involved. Let's understand it with an example from NCERT chapter 9 differential equations
From the above equations, we notice that equations (1), (2) and (3) involve dependent variable(variables) and/or independent only but equation (4) involves variables as well as derivative of the dependent variable (y) with respect to the independent variable (x). That type of equation is known as the differential equation.
Important terms used in class 12 chapter 9 differential equations
Differential equations class 12 ncert solutions are designed to help students understand the various concepts and techniques involved in solving differential equations. Some of the key features of these solutions are:
Comprehensive coverage: The class 12 maths ch 9 question answer cover all the topics included in the Class 12 Maths syllabus, ensuring that students are wellprepared for their exams.
Simple language: The class 12 maths ch 9 question answer are written in simple language, making it easy for students to understand the concepts and techniques involved in solving differential equations.
Stepbystep approach: The class 12 differential equations solutions follow a stepbystep approach, which helps students to understand the solution process in a structured way.
Wellillustrated solutions: The maths chapter 9 class 12 solutions are accompanied by diagrams and illustrations, which help students to visualize the solution process and understand the concepts better.
Conceptual clarity: The maths chapter 9 class 12 solutions aim to develop the conceptual clarity of students, rather than just providing them with the final answers. This helps students to build a strong foundation in the subject.
NCERT solutions for class 12 maths chapter 9 differential equations are very helpful for the preparation of this chapter. Here are some tips to get command on it.
Happy learning !!!
As CBSE board exam paper is designed entirely based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the NCERT very well to perform well in the exam. Only knowing the answer it not enough to perform well in the exam. In the NCERT solutions students will get know how best to write answer in the board exam in order to get good marks.
Generally, one question of 5 marks is asked from this chapter in the 12^{th }board final exam. you should refer NCER syllabus for it. NCERT textbook and NCERT Notes are recommended if you want to obtain meritious marks in the Board exam.
Basic concepts of differential equation, order and degree of the differential equation, general and particular solutions of a differential equation, formation of a differential equation, methods of solving first order,first degree differential equations, homogeneous differential equations and linear differential equations are the important topics of this chapter.
The NCERT class 12 maths differential equations are available in PDF format and have been created by subject experts in line with the textbook questions. These solutions for ch 9 maths class 12 adhere to the latest CBSE Syllabus for 2023 and encompass all the significant concepts for the board exam. The problems in the textbook are solved step by step in accordance with the marks weightage in the CBSE Board exams. Careers360 website offers both chapterwise and exercisewise PDF links that can be used by students to instantly clarify their doubts.
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