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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, â€“1).

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

Integrating both sides, we get:

Now, the curve passes through point (1, â€“1).

Substituting C = â€“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, â€“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, â€“2).

âˆ´ (â€“2)2 â€“ 02 = 2C

â‡’ 2C = 4

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

y2 â€“ 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 (â€“4, â€“3). Find the equation of the curve given that it passes through (â€“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 (â€“4, â€“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 (â€“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:

â‡’ 4Ï€ Ã— 3= 3 (k Ã— 0 C)

â‡’ 108Ï€ = 3C

â‡’ C = 36Ï€

At = 3, r = 6:

â‡’ 4Ï€ Ã— 63 = 3 (k Ã— 3 + C)

â‡’ 864Ï€ = 3 (3k + 36Ï€)

â‡’ 3k = â€“288Ï€ â€“ 36Ï€ = 252Ï€

â‡’ k = 84Ï€

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 (logÂ­e 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.

â‡’ 100 = ek â€¦ (2)

Now, if t = 10, then p = 2 Ã— 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.

â‡’ 1000 = eC â€¦ (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.

â‡’ 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.

â‡’ 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 = â€“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(Î»x, Î»y) = Î»n F(xy) for any non-zero constant (Î»).

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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