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# NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

## JEE : NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

The document NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Question 41:

ANSWER : - The given differential equation is:

Integrating both sides of this equation, we get:

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

Question 42:

ANSWER : - The given differential equation is:

Integrating both sides, we get:

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

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

Question 43:

ANSWER : - The given differential equation is:

Integrating both sides, we get:

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

Question 44:

ANSWER : - The given differential equation is:

Integrating both sides, we get:

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

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

Question 45:

ANSWER : - The given differential equation is:

Integrating both sides, we get:

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

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

Question 46:

ANSWER : - The given differential equation is:

Integrating both sides, we get:

Comparing the coefficients of x2 and x, we get:

A  + B = 2

B  + C = 1

A  + = 0

Solving these equations, we get:

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

Therefore, equation (1) becomes:

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

Question 47:

Integrating both sides, we get:

Comparing the coefficients of x2x, and constant, we get:

Solving these equations, we get

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

Therefore, equation (1) becomes:

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

Question 48:

Integrating both sides, we get:

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

Question 49:

Integrating both sides, we get:

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

y = sec x

Question 50: 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:

Integrating both sides, we get:

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

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

Substituting in equation (2), we get:

Hence, the required equation of the curve is

Question 51: For the differential equation find the solution curve passing through the point (1, –1).

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

Integrating both sides, we get:

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 52: 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 and y be the x-coordinate and y-coordinate of the curve respectively.

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:

Integrating both sides, 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 53: At any point (xy) 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 : - It is given that (xy) is the point of contact of the curve and its tangent.

The slope (m1) of the line segment joining (xy) 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:

Integrating both sides, we get:

This is the general equation of the curve.

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

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

y + 3 = (x + 4)2

This is the required equation of the curve.

Question 54: 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 seconds.

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

Integrating both sides, we get:

⇒ 4π × 3= 3 (k × 0 C)

⇒ 108π = 3C

⇒ C = 36π

At = 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 55: 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).

ANSWER : - Let pt, 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.

Integrating both sides, we get:

It is given that when t = 0, p = 100.

⇒ 100 = ek … (2)

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

Therefore, equation (1) becomes:

Hence, the value of r is 6.93%.

Question 56: 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.

Integrating both sides, we get:

Now, when t = 0, p = 1000.

⇒ 1000 = eC … (2)

At t = 10, equation (1) becomes:

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

Question 57: 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.

Integrating both sides, we get:

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.

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

Therefore, equation (2) becomes:

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

⇒ y = 2y0 at tt1

From equation (4), we get:

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

Question 58: The general solution of the differential equation

A.

B.

C.

D.

Integrating both sides, we get:

Hence, the correct answer is A.

Question 59:

ANSWER : - The given differential equation i.e., (x2   xydy = (x2   y2dx can be written as:

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

To solve it, we make the substitution as: vx

Differentiating both sides with respect to x, we get:

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 60:

ANSWER : - The given differential equation is:

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as: vx

Differentiating both sides with respect to x, we get:

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 61:

ANSWER : - The given differential equation is:

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 62:

ANSWER : - The given differential equation is:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 63:

ANSWER : - The given differential equation is:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Integrating both sides, we get:

This is the required solution for the given differential equation.

Question 64:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 65:

ANSWER : - The given differential equation is:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 66:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 67:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Integrating both sides, we get:

Therefore, equation (1) becomes:

This is the required solution of the given differential equation.

Question 68:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vy

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Question 69:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Integrating both sides, we get:

Now, y = 1 at x = 1.

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

This is the required solution of the given differential equation.

Question 70:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Integrating both sides, we get:

Now, y = 1 at x = 1.

Substituting in equation (2), we get:

This is the required solution of the given differential equation.

Question 71:

Therefore, the given differential equation is a homogeneous equation.

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

vx

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

Integrating both sides, we get:

Now, .

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

This is the required solution of the given differential equation.

Question 72:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Integrating both sides, we get:

This is the required solution of the given differential equation.

Now, y = 0 at x = 1.

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

This is the required solution of the given differential equation.

Question 73:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Integrating both sides, we get:

Now, y = 2 at x = 1.

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

This is the required solution of the given differential equation.

Question 74: A homogeneous differential equation of the form can be solved by making the substitution

A. yvx

B. vyx

C. vy

D. xv

ANSWER : - For solving the homogeneous equation of the form, we need to make the substitution as xvy.Hence, the correct answer is C.

Question 75: Which of the following is a homogeneous differential equation?

A.

B.

C.

D.

ANSWER : - Function F(xy) is said to be the homogenous function of degree n, if

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

Consider the equation given in alternativeD:

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

Question 76:

ANSWER : - The given differential equation is

This is in the form of

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

Therefore, equation (1) becomes:

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

Question 77:

ANSWER : - The given differential equation is

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

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

Question 78:

ANSWER : - The given differential equation is:

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

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

Question 79:

ANSWER : - The given differential equation is:

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

Question 80:

ANSWER : - The given differential equation is:

This equation is in the form of:

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

Therefore, equation (1) becomes:

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## Mathematics (Maths) Class 12

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