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# NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

## JEE : NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev

The document NCERT Solutions (Part - 3) - Differential Equations JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Question 81:

ANSWER : - The given differential equation is:

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

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

Question 82:

ANSWER : - The given differential equation is:

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

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

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

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

Question 83:

This equation is a linear differential equation of the form:

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

Question 84:

This equation is a linear differential equation of the form:

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

Question 85:

This is a linear differential equation of the form:

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

Question 86:

This is a linear differential equation of the form:

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

Question 87:

This is a linear differential equation of the form:

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

Question 88:

ANSWER : - The given differential equation is

This is a linear equation of the form:

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

Now,

Therefore,

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

Hence, the required solution of the given differential equation is

Question 89:

This is a linear differential equation of the form:

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

Now, y = 0 at x = 1.

Therefore,

Substituting  in equation (1), we get:

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

Question 90:

ANSWER : - The given differential equation is

This is a linear differential equation of the form:

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

Now,

Therefore, we get:

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

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

Question 91: Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (xy) is equal to the sum of the coordinates of the point.

ANSWER : - Let F (xy) be the curve passing through the origin.

At point (xy), the slope of the curve will be

According to the given information:

This is a linear differential equation of the form:

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

Substituting in equation (1), we get:

The curve passes through the origin.

Therefore, equation (2) becomes:

1 = C

⇒ C = 1

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

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

Question 92: 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 (xy) be the curve and let (xy) be a point on the curve. The slope of the tangent to the curve at (xy) is

According to the given information:

This is a linear differential equation of the form:

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

Therefore, equation (1) becomes:

The curve passes through point (0, 2).

Therefore, equation (2) becomes:

0 2 – 4 = Ce0

⇒ – 2 = C

⇒ C = – 2

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

This is the required equation of the curve.

Question 93: The integrating factor of the differential equation is

A. ex

B. ey

C.

D. x

ANSWER : - The given differential equation is:

This is a linear differential equation of the form:

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

Hence, the correct answer is C.

Question 94: The integrating factor of the differential equation  is

A.

B.

C.

D.

ANSWER : - The given differential equation is:

This is a linear differential equation of the form:

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

Hence, the correct answer is D.

Question 95: 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:

The highest order derivative present in the differential equation is. Thus, its order is one. The highest power raised to is three. Hence, its degree is three.

(iii) The differential equation is given as:

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

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

Question 96: 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)

Differentiating both sides with respect to x, we get:

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

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)

Differentiating both sides with respect to x, we get:

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

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)

Differentiating both sides with respect to x, we get:

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

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

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

(iv)

Differentiating both sides with respect to x, we get:

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

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

Question 97: Form the differential equation representing the family of curves given by where a is an arbitrary constant.

Differentiating with respect to x, we get:

From equation (1), we get:

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

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

Question 98: Prove that is the general solution of differential equation, where c is a parameter.

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

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

Integrating both sides, we get:

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

Therefore, equation (2) becomes:

Hence, the given result is proved.

Question 99: 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 (aa) and radius (a) which touches the coordinate axes is:

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

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

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

Question 100: Find the general solution of the differential equation

Integrating both sides, we get:

Question 101: Show that the general solution of the differential equation is given by (x    1) = (1 – – y – 2xy), where is parameter

Integrating both sides, we get:

Hence, the given result is proved.

Question 102: Find the equation of the curve passing through the point whose differential equation is,

ANSWER : - The differential equation of the given curve is:

Integrating both sides, we get:

The curve passes through point

On substituting in equation (1), we get:

Hence, the required equation of the curve is

Question 103: Find the particular solution of the differential equation , given that y = 1 when x = 0

Integrating both sides, we get:

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

Now, y = 1 at x = 0.

Therefore, equation (2) becomes:

Substituting in equation (2), we get:

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

Question 104: Solve the differential equation

Differentiating it with respect to y, we get:

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

Integrating both sides, we get:

Question 105: Find a particular solution of the differential equation, given that = – 1, when x = 0 (Hint: put x – yt)

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

Integrating both sides, we get:

Now, y = –1 at = 0.

Therefore, equation (3) becomes:

log 1 = 0 – 1 C

⇒ C = 1

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

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

Question 106: Solve the differential equation

This equation is a linear differential equation of the form

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

Question 107: Find a particular solution of the differential equation , given that y = 0 when

ANSWER : - The given differential equation is:

This equation is a linear differential equation of the form

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

Now,

Therefore, equation (1) becomes:

Substituting in equation (1), we get:

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

Question 108: Find a particular solution of the differential equation, given that y = 0 when x = 0

Integrating both sides, we get:

Substituting this value in equation (1), we get:

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

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

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

Question 109: 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.

Integrating both sides, we get:

log 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 = 25000.

Therefore, we get:

In the year 2009, t = 10 years.

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

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

Question 110: The general solution of the differential equation is

A. xy = C

B. = Cy2

C. = Cx

D. y = Cx2

ANSWER : - The given differential equation is:

Integrating both sides, we get:

Hence, the correct answer is C.

Question 111: 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,

Hence, the correct answer is C.

Question 112: The general solution of the differential equation  is

A. xeyx2 = C
B. xey + y2 = C
C. yexx2 = C
Dyey + x2 = C

ANSWER : - The given differential equation is:

This is a linear differential equation of the form

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

Hence, the correct answer is C.

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## Mathematics (Maths) Class 12

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