class xii chapter 9 – differential equations maths question...

Class XII Chapter 9 – Differential Equations Maths

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Exercise 9.1

Question 1:

Determine order and degree(if defined) of differential equation

Answer

The highest order derivative present in the differential equation is . Therefore, its

order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its

degree is not defined.

Question 2:


Answer

The given differential equation is:

The highest order derivative present in the differential equation is . Therefore, its order

is one.

It is a polynomial equation in . The highest power raised to is 1. Hence, its degree is

one.

Question 3:


Answer


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The highest order derivative present in the given differential equation is . Therefore,

its order is two.

It is a polynomial equation in and . The power raised to is 1.

Hence, its degree is one.

Question 4:


Answer


its order is 2.

The given differential equation is not a polynomial equation in its derivatives. Hence, its

degree is not defined.

Question 5:


Answer


order is two.


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It is a polynomial equation in and the power raised to is 1.


Question 6:


Answer


order is three.

The given differential equation is a polynomial equation in .

The highest power raised to is 2. Hence, its degree is 2.

Question 7:


Answer


order is three.

It is a polynomial equation in . The highest power raised to is 1. Hence, its

degree is 1.

Question 8:


Answer


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The highest order derivative present in the differential equation is . Therefore, its order

is one.

The given differential equation is a polynomial equation in and the highest power

raised to is one. Hence, its degree is one.

Question 9:


Answer


order is two.

The given differential equation is a polynomial equation in and and the highest

power raised to is one.


Question 10:


Answer


order is two.

This is a polynomial equation in and and the highest power raised to is one.


Question 11:

The degree of the differential equation

is


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(A) 3 (B) 2 (C) 1 (D) not defined

Answer

The given differential equation is not a polynomial equation in its derivatives. Therefore,

its degree is not defined.

Hence, the correct answer is D.

Question 12:

The order of the differential equation

is

(A) 2 (B) 1 (C) 0 (D) not defined

Answer


its order is two.

Hence, the correct answer is A.


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Exercise 9.2

Question 1:

Answer

Differentiating both sides of this equation with respect to x, we get:

Now, differentiating equation (1) with respect to x, we get:

Substituting the values of in the given differential equation, we get the L.H.S.

as:

Thus, the given function is the solution of the corresponding differential equation.

Question 2:

Answer


Substituting the value of in the given differential equation, we get:

L.H.S. = = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.


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Question 3:

Answer



L.H.S. = = R.H.S.


Question 4:

Answer

Differentiating both sides of the equation with respect to x, we get:


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L.H.S. = R.H.S.


Question 5:

Answer

Differentiating both sides with respect to x, we get:



Question 6:

Answer


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Question 7:

Answer



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L.H.S. = R.H.S.


Question 8:

Answer

Differentiating both sides of the equation with respect to x, we get:

Substituting the value of in equation (1), we get:


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Question 9:

Answer




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Question 10:

Answer





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Question 11:

The numbers of arbitrary constants in the general solution of a differential equation of

fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Answer

We know that the number of constants in the general solution of a differential equation

of order n is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential

equation is four.


Question 12:

The numbers of arbitrary constants in the particular solution of a differential equation of

third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Answer

In a particular solution of a differential equation, there are no arbitrary constants.



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Exercise 9.3

Question 1:

Answer

Differentiating both sides of the given equation with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Hence, the required differential equation of the given curve is

Question 2:

Answer




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Dividing equation (2) by equation (1), we get:

This is the required differential equation of the given curve.

Question 3:

Answer



Multiplying equation (1) with (2) and then adding it to equation (2), we get:

Now, multiplying equation (1) with equation (3) and subtracting equation (2) from it, we

get:

Substituting the values of in equation (3), we get:


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Question 4:

Answer


Multiplying equation (1) with equation (2) and then subtracting it from equation (2), we

get:





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Question 5:

Answer


Again, differentiating with respect to x, we get:

Adding equations (1) and (3), we get:


Question 6:

Form the differential equation of the family of circles touching the y-axis at the origin.

Answer

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is


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Differentiating equation (1) with respect to x, we get:

Now, on substituting the value of a in equation (1), we get:

This is the required differential equation.

Question 7:

Form the differential equation of the family of parabolas having vertex at origin and axis

along positive y-axis.

Answer

The equation of the parabola having the vertex at origin and the axis along the positive

y-axis is:


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Question 8:

Form the differential equation of the family of ellipses having foci on y-axis and centre at

origin.

Answer

The equation of the family of ellipses having foci on the y-axis and the centre at origin is

as follows:


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Substituting this value in equation (2), we get:



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Question 9:

Form the differential equation of the family of hyperbolas having foci on x-axis and

centre at origin.

Answer

The equation of the family of hyperbolas with the centre at origin and foci along the x-

axis is:

Differentiating both sides of equation (1) with respect to x, we get:




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Question 10:

Form the differential equation of the family of circles having centre on y-axis and radius

3 units.

Answer

Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as

follows:


Substituting the value of (y – b) in equation (1), we get:


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Question 11:

Which of the following differential equations has as the general solution?

A.

B.

C.

D.

Answer

The given equation is:

Differentiating with respect to x, we get:



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This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

Question 12:

Which of the following differential equation has as one of its particular solution?

A.

B.

C.

D.

Answer

The given equation of curve is y = x.



Now, on substituting the values of y, from equation (1) and (2) in each of

the given alternatives, we find that only the differential equation given in alternative C is

correct.


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Hence, the correct answer is C.


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Exercise 9.4

Question 1:

Answer


Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

Question 2:

Answer



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Now, integrating both sides of this equation, we get:


Question 3:

Answer


Now, integrating both sides, we get:


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Question 4:

Answer


Integrating both sides of this equation, we get:


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Substituting these values in equation (1), we get:


Question 5:

Answer




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Let (ex + e–x) = t.




Question 6:

Answer




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Question 7:

Answer


Integrating both sides, we get:



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Question 8:

Answer





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Question 9:

Answer





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Question 10:

Answer




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Substituting the values of in equation (1), we get:


Question 11:

Answer




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Comparing the coefficients of x2 and x, we get:

A + B = 2

B + C = 1

A + C = 0

Solving these equations, we get:

Substituting the values of A, B, and C in equation (2), we get:

Therefore, equation (1) becomes:


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Substituting C = 1 in equation (3), we get:

Question 12:

Answer



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Comparing the coefficients of x2, x, and constant, we get:

Solving these equations, we get

Substituting the values of A, B, and C in equation (2), we get:



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Substituting the value of k2 in equation (3), we get:

Question 13:

Answer


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Question 14:

Answer



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y = sec x

Question 15:

Find the equation of a curve passing through the point (0, 0) and whose differential

equation is .

Answer

The differential equation of the curve is:



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Now, the curve passes through point (0, 0).

Substituting in equation (2), we get:

Hence, the required equation of the curve is

Question 16:

For the differential equation find the solution curve passing

through the point (1, –1).


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Answer

The differential equation of the given curve is:


Now, the curve passes through point (1, –1).

Substituting C = –2 in equation (1), we get:

This is the required solution of the given curve.

Question 17:

Find the equation of a curve passing through the point (0, –2) given that at any point

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.

Answer

Let x and y be the x-coordinate and y-coordinate of the curve respectively.


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We know that the slope of a tangent to the curve in the coordinate axis is given by the

relation,

According to the given information, we get:


Now, the curve passes through point (0, –2).

∴ (–2)2 – 02 = 2C

⇒ 2C = 4

Substituting 2C = 4 in equation (1), we get:

y2 – x2 = 4

This is the required equation of the curve.

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).

Answer


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It is given that (x, y) is the point of contact of the curve and its tangent.

The slope (m1) of the line segment joining (x, y) and (–4, –3) is

We know that the slope of the tangent to the curve is given by the relation,

According to the given information:


This is the general equation of the curve.

It is given that it passes through point (–2, 1).


y + 3 = (x + 4)2



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Question 19:

The volume of spherical balloon being inflated changes at a constant rate. If initially its

radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t

seconds.

Answer

Let the rate of change of the volume of the balloon be k (where k is a constant).


⇒ 4π × 33 = 3 (k × 0 + C)

⇒ 108π = 3C

⇒ C = 36π


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At t = 3, r = 6:

⇒ 4π × 63 = 3 (k × 3 + C)

⇒ 864π = 3 (3k + 36π)

⇒ 3k = –288π – 36π = 252π

⇒ k = 84π

Substituting the values of k and C in equation (1), we get:

Thus, the radius of the balloon after t seconds is .

Question 20:

In a bank, principal increases continuously at the rate of r% per year. Find the value of r

if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).


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Answer

Let p, t, and r represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of r% per year.


It is given that when t = 0, p = 100.

⇒ 100 = ek … (2)

Now, if t = 10, then p = 2 × 100 = 200.



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Hence, the value of r is 6.93%.

Question 21:

In a bank, principal increases continuously 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 .

Answer

Let p and t be the principal and time respectively.

It is given that the principal increases continuously at the rate of 5% per year.


Now, when t = 0, p = 1000.

⇒ 1000 = eC … (2)

At t = 10, equation (1) becomes:


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Hence, after 10 years the amount will worth Rs 1648.

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?

Answer

Let y be the number of bacteria at any instant t.

It is given that the rate of growth of the bacteria is proportional to the number present.


Let y0 be the number of bacteria at t = 0.

⇒ log y0 = C

Substituting the value of C in equation (1), we get:

Also, it is given that the number of bacteria increases by 10% in 2 hours.


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Now, let the time when the number of bacteria increases from 100000 to 200000 be t1.

⇒ y = 2y0 at t = t1

From equation (4), we get:

Hence, in hours the number of bacteria increases from 100000 to 200000.


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Question 23:

The general solution of the differential equation

A.

B.

C.

D.

Answer


Hence, the correct answer is A.


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Exercise 9.5

Question 1:

Answer

The given differential equation i.e., (x2 + xy) dy = (x2 + y2) dx can be written as:

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as:

y = vx


Substituting the values of v and in equation (1), we get:


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This is the required solution of the given differential equation.


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Question 2:

Answer


Thus, the given equation is a homogeneous equation.


y = vx


Substituting the values of y and in equation (1), we get:



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Question 3:

Answer


Thus, the given differential equation is a homogeneous equation.


y = vx



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Question 4:

Answer


Therefore, the given differential equation is a homogeneous equation.


y = vx


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Question 5:

Answer



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y = vx




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This is the required solution for the given differential equation.

Question 6:

Answer



y = vx


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Substituting the values of v and in equation (1), we get:



Question 7:

Answer



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y = vx



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Question 8:

Answer



y = vx



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Question 9:

Answer



y = vx


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Question 10:

Answer


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x = vy

Substituting the values of x and in equation (1), we get:



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Question 11:

Answer



y = vx



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Now, y = 1 at x = 1.

Substituting the value of 2k in equation (2), we get:



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Question 12:

Answer



y = vx



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Now, y = 1 at x = 1.



Question 13:


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Answer


To solve this differential equation, we make the substitution as:

y = vx




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Now, .

Substituting C = e in equation (2), we get:


Question 14:

Answer



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y = vx




Now, y = 0 at x = 1.

Substituting C = e in equation (2), we get:



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Question 15:

Answer



y = vx

Substituting the value of y and in equation (1), we get:



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Now, y = 2 at x = 1.



Question 16:

A homogeneous differential equation of the form can be solved by making the

substitution

A. y = vx

B. v = yx

C. x = vy

D. x = v

Answer


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For solving the homogeneous equation of the form , we need to make the

substitution as x = vy.


Question 17:

Which of the following is a homogeneous differential equation?

A.

B.

C.

D.

Answer

Function F(x, y) is said to be the homogenous function of degree n, if

F(λx, λy) = λn F(x, y) for any non-zero constant (λ).

Consider the equation given in alternativeD:

Hence, the differential equation given in alternative D is a homogenous equation.


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Exercise 9.6

Question 1:

Answer

The given differential equation is

This is in the form of

The solution of the given differential equation is given by the relation,


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Question 2:

Answer




Question 3:

Answer



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Question 4:

Answer


The general solution of the given differential equation is given by the relation,

Question 5:

Answer


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By second fundamental theorem of calculus, we obtain

Question 6:

Answer


This equation is in the form of a linear differential equation as:



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Question 7:

Answer


This equation is the form of a linear differential equation as:



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Question 8:

Answer

This equation is a linear differential equation of the form:


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Question 9:

Answer

This equation is a linear differential equation of the form:



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Question 10:

Answer

This is a linear differential equation of the form:



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Question 11:

Answer




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Question 12:

Answer




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Question 13:

Answer


This is a linear equation of the form:


Now,

Therefore,



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Hence, the required solution of the given differential equation is

Question 14:

Answer



Now, y = 0 at x = 1.

Therefore,


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Question 15:

Answer




Now,

Therefore, we get:


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This is the required particular solution of the given differential equation.

Question 16:

Find the equation of a 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.

Answer

Let F (x, y) be the curve passing through the origin.

At point (x, y), the slope of the curve will be





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The curve passes through the origin.


1 = C

⇒ C = 1


Hence, the required equation of curve passing through the origin is

Question 17:

Find the equation of a 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.

Answer

Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent

to the curve at (x, y) is




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The general equation of the curve is given by the relation,


The curve passes through point (0, 2).


0 + 2 – 4 = Ce0

⇒ – 2 = C

⇒ C = – 2



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Question 18:

The integrating factor of the differential equation is

A. e–x

B. e–y

C.

D. x

Answer



The integrating factor (I.F) is given by the relation,


Question 19:

The integrating factor of the differential equation.

is

A.


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B.

C.

D.

Answer



The integrating factor (I.F) is given by the relation,



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Miscellaneous Solutions

Question 1:

For each of the differential equations given below, indicate its order and degree (if

defined).

(i)

(ii)

(iii)

Answer

(i) The differential equation is given as:

The highest order derivative present in the differential equation is . Thus, its order is

two. The highest power raised to is one. Hence, its degree is one.

(ii) The differential equation is given as:


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one. The highest power raised to is three. Hence, its degree is three.

(iii) The differential equation is given as:


four.

However, the given differential equation is not a polynomial equation. Hence, its degree

is not defined.

Question 2:

For each of the exercises given below, verify that the given function (implicit or explicit)

is a solution of the corresponding differential equation.

(i)

(ii)

(iii)

(iv)

Answer

(i)



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Now, on substituting the values of and in the differential equation, we get:

⇒ L.H.S. ≠ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii)




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Now, on substituting the values of and in the L.H.S. of the given differential

equation, we get:

Hence, the given function is a solution of the corresponding differential equation.

(iii)




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Substituting the value of in the L.H.S. of the given differential equation, we get:


(iv)


Substituting the value of in the L.H.S. of the given differential equation, we get:


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Question 3:

Form the differential equation representing the family of curves given by

where a is an arbitrary constant.

Answer


From equation (1), we get:

On substituting this value in equation (3), we get:


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Hence, the differential equation of the family of curves is given as

Question 4:

Prove that is the general solution of differential

equation , where c is a parameter.

Answer

This is a homogeneous equation. To simplify it, we need to make the substitution as:



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Substituting the values of I1 and I2 in equation (3), we get:



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Hence, the given result is proved.

Question 5:

Form the differential equation of the family of circles in the first quadrant which touch

the coordinate axes.

Answer

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which

touches the coordinate axes is:


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Substituting the value of a in equation (1), we get:

Hence, the required differential equation of the family of circles is


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Question 6:

Find the general solution of the differential equation

Answer


Question 7:

Show that the general solution of the differential equation is given by

(x + y + 1) = A (1 – x – y – 2xy), where A is parameter

Answer



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Hence, the given result is proved.


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Question 8:

Find the equation of the curve passing through the point whose differential

equation is,

Answer

The differential equation of the given curve is:


The curve passes through point

On substituting in equation (1), we get:

Hence, the required equation of the curve is


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Question 9:

Find the particular solution of the differential equation

, given that y = 1 when x = 0

Answer


Substituting these values in equation (1), we get:

Now, y = 1 at x = 0.




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Question 10:

Solve the differential equation

Answer

Differentiating it with respect to y, we get:

From equation (1) and equation (2), we get:



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Question 11:

Find a particular solution of the differential equation , given that

y = – 1, when x = 0 (Hint: put x – y = t)

Answer

Substituting the values of x – y and in equation (1), we get:


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Now, y = –1 at x = 0.


log 1 = 0 – 1 + C

⇒ C = 1

Substituting C = 1 in equation (3) we get:


Question 12:

Solve the differential equation

Answer


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This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

Question 13:

Find a particular solution of the differential equation ,

given that y = 0 when

Answer


This equation is a linear differential equation of the form


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Now,




Question 14:

Find a particular solution of the differential equation , given that y = 0

when x = 0

Answer


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Now, at x = 0 and y = 0, equation (2) becomes:



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Question 15:

The population of a village increases continuously at the rate proportional to the number

of its inhabitants present at any time. If the population of the village was 20000 in 1999

and 25000 in the year 2004, what will be the population of the village in 2009?

Answer

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of

inhabitants at any instant.


log y = kt + C … (1)

In the year 1999, t = 0 and y = 20000.

Therefore, we get:

log 20000 = C … (2)

In the year 2004, t = 5 and y = 25000.

Therefore, we get:


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In the year 2009, t = 10 years.

Now, on substituting the values of t, k, and C in equation (1), we get:

Hence, the population of the village in 2009 will be 31250.

Question 16:

The general solution of the differential equation is

A. xy = C

B. x = Cy2

C. y = Cx

D. y = Cx2

Answer



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Question 17:

The general solution of a differential equation of the type is

A.

B.

C.

D.

Answer

The integrating factor of the given differential equation

The general solution of the differential equation is given by,



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Question 18:

The general solution of the differential equation is

A. xey + x2 = C

B. xey + y2 = C

C. yex + x2 = C

D. yey + x2 = C

Answer


This is a linear differential equation of the form



45

45

DIFFERENTIAL EQUATIONS

INTRODUCTION

1. Problems based on the order and degree of the differential equations

Working rule

(a) In order to find the order of a differential equations, see the highest derivative in the

given differential equation. Write down the order of this highest order derivatives.

(b) In order to find the degree of a differential equation write down the power of the

highest order derivative after making the derivatives occurring in the given differential

equation free from radicals and fractions.

2. Problems based on formation of differential equation.

Working rule

(a) Write the given equation.

(b) Differentiate the given equation w.r.t. independent variable of x as many times as the

number of arbitrary constants.

(c) Eliminate the arbitrary constants from given equation and the equations obtained by

differentiation.

3. Problems based on solution of differential equation in which variables are separable.

Working rule

This differential equation can be solved by the variable separable method which can be

put in the form

𝑑𝑦

𝑑𝑥 = 𝑓(𝑥). 𝑔(𝑦).

i.e., in which 𝑑𝑦

𝑑𝑥 can be expressed as the product of two functions, one of which is

a function of x only and the other a function of y only.

In order to solve the equation 𝑑𝑦

𝑑𝑥= 𝑓(𝑥). 𝑔(𝑦). Write down this equation in the

form 𝑑𝑦

𝑔(𝑦)= 𝑓(𝑥). 𝑑𝑥, then the solution will be ∫

𝑑𝑦

𝑔(𝑦)= ∫ 𝑓(𝑥)𝑑𝑥 + 𝑐, where C is an

arbitrary constant.

4. Problems based on solution of differential equations which are homogeneous.

Working rule

(a) Write down the given differential equation in the form 𝑑𝑦

𝑑𝑥 = 𝑓(𝑥, 𝑦)

(b) If 𝑓(𝑘𝑥, 𝑘𝑦) = 𝑓(𝑥, 𝑦). then differential equation is homogeneous.

(c) In order to solve, put 𝑦 = 𝑣𝑥, so that 𝑑𝑦

𝑑𝑥= 𝑣 + 𝑥

𝑑𝑣

𝑑𝑥 and then separate the

variables𝑥 and 𝑣.

(d) Now solve the obtained differential equation by the variable separable

method. At the end put 𝑦

𝑥 in place of 𝑣.

5. Working Rule for Linear Differential Equation of first degree:

(a) Type 1.𝑑𝑦

𝑑𝑥+ 𝑃𝑦 = 𝑄 , where P and Q are constants or function of x only

General solution is𝑦. 𝐼𝐹 = ∫(𝑄. 𝐼𝐹)𝑑𝑥 + 𝐶 , where 𝐼𝐹 = 𝑒∫𝑃𝑑𝑥.

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(b) Type 2.𝑑𝑥

𝑑𝑦+ 𝑃𝑥 = 𝑄 , where P and Q are constants or function of y only

General solution is𝑥. 𝐼𝐹 = ∫(𝑄. 𝐼𝐹)𝑑𝑦 + 𝐶 , where 𝐼𝐹 = 𝑒∫𝑃𝑑𝑦.

Note: Particular solution can be obtained after getting the value of parameter C by

substituting the given initial values of the variables.

IMPORTANT SOLVED PROBLEMS

1. Solve the differential equation (x + y ) dy + ( x – y ) dx = 0

Solution

(x + y ) dy + ( x – y ) dx = 0

𝑑𝑦

𝑑𝑥 =

𝑦−𝑥

𝑥+𝑦

Put y = vx

𝑑𝑦

𝑑𝑥 = 𝑣 + 𝑥

𝑑𝑣

𝑑𝑥

∴ 𝑣 + 𝑥 𝑑𝑣

𝑑𝑥 =

𝑣𝑥−𝑥

𝑥 + 𝑣𝑥

𝑣 + 𝑥 𝑑𝑣

𝑑𝑥 =

𝑣−1

𝑣+1

𝑥 𝑑𝑣

𝑑𝑥 =

𝑣−1

𝑣+1 − 𝑣

𝑥 𝑑𝑣

𝑑𝑥 =

−( 1+ 𝑣2)

1+𝑣

1+𝑣

1+ 𝑣2 𝑑𝑣 = −𝑑𝑥

𝑥

Integrating both sides we get

∫1+𝑣

1+ 𝑣2 𝑑𝑣 = - ∫𝑑𝑥

𝑥

∫𝑑𝑣

1+ 𝑣2 + 1

2∫

2𝑣

1+𝑣2 𝑑𝑣 = − log 𝑥 + 𝑐

tan−1 𝑣 +1

2log|1 + 𝑣2| = − log 𝑥 + 𝑐

tan−1 𝑦

𝑥 +

1

2 𝑙𝑜𝑔 (1 +

𝑦2

𝑥2) = −𝑙𝑜𝑔 𝑥 + 𝑐

tan−1 𝑦

𝑥 +

1

2 𝑙𝑜𝑔 (𝑥2 + 𝑦2) = 𝑐

2. Solve 𝑥√1 − 𝑦2 𝑑𝑥 + 𝑦 √1 − 𝑥2 𝑑𝑦 = 0

Solution

𝑥√1 − 𝑦2 𝑑𝑥 = - 𝑦 √1 − 𝑥2 𝑑𝑦

𝑥

√1− 𝑥2𝑑𝑥 = −

𝑦

√1− 𝑦2𝑑𝑦

Integrating both sides

∫𝑥

√1− 𝑥2𝑑𝑥 = - ∫

𝑦

√1− 𝑦2𝑑𝑦

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−1

2∫

𝑑𝑡

√𝑡 =

1

2∫

𝑑𝑢

√𝑢 ( put t = 1 – x2 and put u = 1 – y2 )

− √𝑡 = √𝑢 + C

√1 − 𝑥2 + √ 1 − 𝑦2 = 𝐶

3. Solve the differential equation 𝑑𝑦

𝑑𝑥−

1

𝑥 . 𝑦 = 2𝑥2

Solution

The diff.eqn is in the form 𝑑𝑦

𝑑𝑥 + 𝑃𝑦 = 𝑄

Where P = −1

𝑥 and Q = 2x2

I.F = 𝑒∫𝑃 𝑑𝑥 = 𝑒∫−1

𝑥𝑑𝑥 = 𝑒−𝑙𝑜𝑔 𝑥 =

1

𝑥

Multiplying both sides of diff.eqn by I.F we get

1

𝑥

𝑑𝑦

𝑑𝑥 −

1

𝑥2𝑦 = 2𝑥

𝑑

𝑑𝑥(𝑦.

1

𝑥) = 2𝑥

Integrating both sides w.r.t.x we get

𝑦.1

𝑥 = 𝑥2 + 𝑐

y = x3 + Cx

4. Solve the diff. eqn 𝑥 𝑑𝑦

𝑑𝑥 = 𝑦 − 𝑥 tan

𝑦

𝑥

Solution

𝑑𝑦

𝑑𝑥 =

𝑦 − 𝑥 tan𝑦

𝑥

𝑥

Put y = vx =>𝑑𝑦

𝑑𝑥 = 𝑣 + 𝑥

𝑑𝑣

𝑑𝑥

∴ 𝑣 + 𝑥 𝑑𝑣

𝑑𝑥 =

𝑣𝑥 − 𝑥 𝑡𝑎𝑛𝑣

𝑥

𝑣 + 𝑥 𝑑𝑣

𝑑𝑥 = v – tan v

𝑥 𝑑𝑣

𝑑𝑥 = −𝑡𝑎𝑛 𝑣

cot v dv = - 𝑑𝑥

𝑥

Integrating both sides ,we get

∫ 𝑐𝑜𝑡 𝑣 𝑑𝑣 = − ∫𝑑𝑥

𝑥

log sin v = - log x + log C

log [x. sin (𝑦

𝑥) ] = 𝑙𝑜𝑔 𝑐

x sin (𝑦

𝑥) = C

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5. 𝑥𝑦 𝑑𝑦

𝑑𝑥 = (𝑥 + 2 )(𝑦 + 2 ) , find the equation of the curve passing through the points (1, -

1)

Solution

𝑥𝑦 𝑑𝑦

𝑑𝑥 = (𝑥 + 2 )(𝑦 + 2 )

𝑦

𝑦+2 𝑑𝑦 =

𝑥+2

𝑥 𝑑𝑥

(1 − 2

𝑦+2)𝑑𝑦 = (1 +

2

𝑥) 𝑑𝑥

Integrating both sides weget

Y –2 log (y + 2 ) = x +2 log x + C

The curve is pasing through the the point (1,-1)

- 1 –2 log 1 = 1 + 2 log 1 + C => C = -2

The equation of the line is 𝑦 − 𝑥 = 2 𝑙𝑜𝑔[x(y+2)] – 2

6. Solve the differential equation (𝑥2 − 1)𝑑𝑦

𝑑𝑥 + 2𝑥𝑦 =

2

𝑥2−1

Solution

(𝑥2 − 1)𝑑𝑦

𝑑𝑥 + 2𝑥𝑦 =

2

𝑥2 − 1

𝑑𝑦

𝑑𝑥+ (

2𝑥

𝑥2−1) . 𝑦 =

2

(𝑥2−1)2

I.F. = 𝑒∫2𝑥

𝑥2−1𝑑𝑥

= 𝑒𝑙𝑜𝑔(𝑥2−1) = (𝑥2 − 1)

𝑑

𝑑𝑥(𝑦 (𝑥2 − 1)) =

2

𝑥2−1

Integrating both sides w.r.t x we get

𝑦 (𝑥2 − 1) = ∫2

𝑥2−1 𝑑𝑥

𝑦 (𝑥2 − 1) = 𝑙𝑜𝑔 (𝑥−1

𝑥+1) + 𝐶

-

PRACTICE PROBLEMS

LEVEL I

1. Find the order and degree of the following differential equation.

(𝑑3𝑦

𝑑𝑥3)2

− 𝑥 (𝑑𝑦

𝑑𝑥)3

= 0

2. Show that 𝑦 = 𝐴𝑥 + 𝐵

𝑥 is a solution of the differential equation.

𝑥2 𝑑2𝑦

𝑑𝑥2 + 𝑥𝑑𝑦

𝑑𝑥− 𝑦 = 0

3. Form the differential equation corresponding to 𝑦2 − 2𝑎𝑦 + 𝑥2 = 𝑎2 by eliminating a.

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4. Form the differential equation representing the family of curves 𝑦 = 𝑎 𝑠𝑖𝑛 (𝑥 + 𝑏),

where 𝑎, 𝑏 are arbitrary constants.

5. Solve the differential equation.

𝑠𝑒𝑐2𝑥 𝑡𝑎𝑛 𝑦 𝑑𝑥 + 𝑠𝑒𝑐2𝑦 𝑡𝑎𝑛 𝑥 𝑑𝑦 = 0

6. Solve the differential equation.

𝑑𝑦

𝑑𝑥=

1+ 𝑦2

1+ 𝑥2

LEVEL II

Solve the following differential equations.

1. 𝑑𝑦

𝑑𝑥+

1

𝑥 . 𝑦 = 2𝑥2

2. (𝑥2 + 𝑥𝑦)𝑑𝑦 = (𝑥2 + 𝑦2)𝑑𝑥

3. 𝑑𝑦

𝑑𝑥+

4𝑥

𝑥2+1 𝑦 +

1

(𝑥2+1)2 = 0

4. 𝑠𝑖𝑛 𝑥 𝑑𝑦

𝑑𝑥+ 𝑐𝑜𝑠 𝑥. 𝑦 = 𝑐𝑜𝑠 𝑥. 𝑠𝑖𝑛2𝑥

5. 3𝑒𝑥 𝑡𝑎𝑛 𝑦 𝑑𝑥 + (1 − 𝑒2)𝑠𝑒𝑐2 𝑦 𝑑𝑦 = 0, 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑦 = 𝜋

4, 𝑤ℎ𝑒𝑛 𝑥 = 1.

6. (𝑥3 + 𝑦3)𝑑𝑦 − 𝑥2𝑦 𝑑𝑥 = 0

7. 𝑐𝑜𝑠2𝑥 𝑑𝑦

𝑑𝑥+ 𝑦 = 𝑡𝑎𝑛 𝑥

8. 𝑑𝑦

𝑑𝑥+ 𝑦 = 𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥

9. 𝑥 𝑙𝑜𝑔 𝑥 𝑑𝑦

𝑑𝑥+ 𝑦 =

2

𝑥 𝑙𝑜𝑔 𝑥

LEVEL III

Solve the following differential equations

1. y – x 𝑑𝑦

𝑑𝑥 = a (𝑦2 + 𝑥2 𝑑𝑦

𝑑𝑥)

2. ( 1 + sin2x ) dy + ( 1 + y2 ) cos x dx = 0, given that x = 𝜋

2 , y = 0

3. ( x3 + x2 + x +1) 𝑑𝑦

𝑑𝑥 = 2x2 + x

class xii chapter 9 – differential equations maths question...

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