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

Mathematics (Maths) Class 12

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

ANSWER : - The given differential equation is:

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

Integrating both sides of this equation, we get:

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

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

Question 42: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

Integrating both sides, we get:

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

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

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

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

 

Question 43: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

Integrating both sides, we get:

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

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

 Question 44: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

Integrating both sides, we get:

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

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

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

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

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

Question 45: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

Integrating both sides, we get:

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

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

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

Substituting the values of NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

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

Question 46: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

Integrating both sides, we get:

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

Comparing the coefficients of x2 and x, we get:

A  + B = 2

B  + C = 1

A  + = 0

Solving these equations, we get:

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

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

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

Therefore, equation (1) becomes:

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

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

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

 

 

Question 47: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Integrating both sides, we get:

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

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

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

Solving these equations, we get NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

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

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

Therefore, equation (1) becomes:

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

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

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

 

Question 48: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Integrating both sides, we get:

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

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

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

Question 49: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Integrating both sides, we get:

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

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 isNCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev.

ANSWER : - The differential equation of the curve is:

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

Integrating both sides, we get:

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

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

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

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

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

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

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

Substituting NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (2), we get:

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

Hence, the required equation of the curve isNCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

Question 51: For the differential equation NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevfind the solution curve passing through the point (1, –1).

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

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

Integrating both sides, we get:

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

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

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

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

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

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

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

According to the given information, we get:

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

Integrating both sides, we get:

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

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

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

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

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

According to the given information:

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

Integrating both sides, we get:

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

This is the general equation of the curve.

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

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

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).

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

Integrating both sides, we get:

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

⇒ 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:

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

Thus, the radius of the balloon after t seconds isNCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev.

 

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.

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

Integrating both sides, we get:

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

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:

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

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 yearsNCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev.

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.

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

Integrating both sides, we get:

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

Now, when t = 0, p = 1000.

⇒ 1000 = eC … (2)

At t = 10, equation (1) becomes:

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

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.

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

Integrating both sides, we get:

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

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

⇒ log y0 = C

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

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

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

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

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

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

Therefore, equation (2) becomes:

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

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:

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

Hence, in NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevhours the number of bacteria increases from 100000 to 200000.

Question 58: The general solution of the differential equation NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

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

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

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

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

ANSWER : -

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

Integrating both sides, we get:

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

Hence, the correct answer is A.

 

Question 59: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

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

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

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

Substituting the values of v and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

Question 60: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

 

Question 61: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

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

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

 

 Question 62: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

 

Question 63: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution for the given differential equation.

 

Question 64: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

 

ANSWER : -

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Substituting the values of and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

 

Question 65: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

 

ANSWER : - The given differential equation is:

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

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

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

This is the required solution of the given differential equation.

 

Question 66: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

 

ANSWER : -

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

 

Question 67: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

 

ANSWER : -

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

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

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

Therefore, equation (1) becomes:

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

This is the required solution of the given differential equation.

 

Question 68: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

 

ANSWER : -

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

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vy

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

Substituting the values of x and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

 

Question 69: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

Now, y = 1 at x = 1.

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

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

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

This is the required solution of the given differential equation.

 

Question 70: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as: vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

Now, y = 1 at x = 1.

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

Substituting NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (2), we get:

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

This is the required solution of the given differential equation.

 

Question 71: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Therefore, the given differential equation is a homogeneous equation.

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

vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

Now, NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev.

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

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

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

This is the required solution of the given differential equation.

 

Question 72: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Substituting the values of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

This is the required solution of the given differential equation.

Now, y = 0 at x = 1.

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

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

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

This is the required solution of the given differential equation.

 

Question 73: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : -

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

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

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

Substituting the value of y and NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevin equation (1), we get:

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

Integrating both sides, we get:

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

Now, y = 2 at x = 1.

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

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

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

This is the required solution of the given differential equation.

Question 74: A homogeneous differential equation of the form NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRevcan be solved by making the substitution

A. yvx 

B. vyx

C. vy 

D. xv

ANSWER : - For solving the homogeneous equation of the formNCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev, 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. NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev                   

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

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

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

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:

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

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

Question 76: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

This is in the form of NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

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

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

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

Therefore, equation (1) becomes:

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

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

 

Question 77: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

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

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

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

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

 

Question 78: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

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

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

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

Question 79: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

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

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

Question 80: NCERT Solutions (Part - 2) - Differential Equations JEE Notes | EduRev

ANSWER : - The given differential equation is:

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

This equation is in the form of:

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

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

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

Therefore, equation (1) becomes:

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

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