NCERT Solutions - Exercise 9.3: Differential Equations

Q1: Which of the following differential equation hasas one of its particular solution?
A.                                            
B. 
C.                                          
D. 
Ans: The given equation of curve is y = x.
Differentiating with respect to x, we get:
Again, differentiating with respect to x, we get:
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.

Hence, the correct answer is C.

Q2: 
Ans: The given differential equation is:

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

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

Q3: 
Ans: The given differential equation is:

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

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

Q4: 
Ans: The given differential equation is:

Now, integrating both sides, we get:

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

Q5: 
Ans: The given differential equation is:

Integrating both sides of this equation, we get:


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

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

Q6: 
Ans: The given differential equation is:

Integrating both sides of this equation, we get:

Let (e^x   e^-x) = t.
Differentiating both sides with respect to x, we get:

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

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

Q7: 
Ans: The given differential equation is:

Integrating both sides of this equation, we get:

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

Q8: 
Ans: 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.

Q9: 
Ans: The given differential equation is:

Integrating both sides, we get:

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

Q10: 
Ans: 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.

Q11: 
Ans: 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.

Q12: 
Ans: 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  + C = 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:


Q13: 
Ans: 

Integrating both sides, we get:

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:

Therefore, equation (1) becomes:

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


Q14: 
Ans: 

Integrating both sides, we get:

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


Q15: 
Ans: 

Integrating both sides, we get:

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

Q16: Find the equation of a curve passing through the point (0, 0) and whose differential equation is.
Ans: 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

Q17: For the differential equation find the solution curve passing through the point (1, -1).
Ans: 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.

Q18: 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.
Ans: Let x 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.

Q19: 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).
Ans: 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:

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.

Q20: 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.
Ans: Let the rate of change of the volume of the balloon be k (where k is a constant).

Integrating both sides, we get:

⇒ 4π × 3³ = 3 (k × 0 C)
⇒ 108π = 3C
⇒ C = 36π
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.

Q21: 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 (log­_e 2 = 0.6931).
Ans: - 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.

Integrating both sides, we get:

It is given that when t = 0, p = 100.
⇒ 100 = e^k ... (2)
Now, if t = 10, then p = 2 × 100 = 200.
Therefore, equation (1) becomes:

Hence, the value of r is 6.93%.

Q22: 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.
Ans: 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 = e^C ... (2)
At t = 10, equation (1) becomes:

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

Q23: 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?
Ans: 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 t = t₁
From equation (4), we get:

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

Q24: The general solution of the differential equation
A.                                                
B. 
C.                                              
D. 
Ans: 

Integrating both sides, we get:

Hence, the correct answer is A.