JEE Exam  >  JEE Notes  >  Mathematics (Maths) Class 12  >  NCERT Solutions - Exercise 9.3: Differential Equations

NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations

Q1: Which of the following differential equation hasNCERT Solutions Class 12 Maths Chapter 9 - Differential Equationsas one of its particular solution?
A. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations                                           
B. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
C. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations                                         
D. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given equation of curve is y = x.
Differentiating with respect to x, we get: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Again, differentiating with respect to x, we get: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, on substituting the values of y, NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct.
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Hence, the correct answer is C.

Q2: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, integrating both sides of this equation, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q3: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, integrating both sides of this equation, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q4: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q5: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides of this equation, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting these values in equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q6: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides of this equation, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Let (ex   e–x) = t.
Differentiating both sides with respect to x, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting this value in equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q7: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides of this equation, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q8: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting this value in equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q9: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q10: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting this value in equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q11: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting the values of NCERT Solutions Class 12 Maths Chapter 9 - Differential Equationsin equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the required general solution of the given differential equation.

Q12: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: The given differential equation is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Comparing the coefficients of x2 and x, we get:
A  + B = 2
B  + C = 1
A  + = 0
Solving these equations, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting the values of A, B, and C in equation (2), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Therefore, equation (1) becomes:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting C = 1 in equation (3), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations

Q13: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: 
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Comparing the coefficients of x2, x, and constant, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Solving these equations, we get NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting the values of A, B, and C in equation (2), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Therefore, equation (1) becomes:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting the value of kin equation (3), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations

Q14: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: 
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting C = 1 in equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations

Q15: NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: 
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
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 isNCERT Solutions Class 12 Maths Chapter 9 - Differential Equations.
Ans: The differential equation of the curve is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting this value in equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, the curve passes through point (0, 0).
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting NCERT Solutions Class 12 Maths Chapter 9 - Differential Equationsin equation (2), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Hence, the required equation of the curve isNCERT Solutions Class 12 Maths Chapter 9 - Differential Equations

Q17: For the differential equation NCERT Solutions Class 12 Maths Chapter 9 - Differential Equationsfind the solution curve passing through the point (1, –1).
Ans: The differential equation of the given curve is:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, the curve passes through point (1, –1).
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting C = –2 in equation (1), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
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 NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations 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 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 Class 12 Maths Chapter 9 - Differential Equations
According to the given information, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, the curve passes through point (0, –2).
∴ (–2)2 – 02 = 2C
⇒ 2C = 4
Substituting 2C = 4 in equation (1), we get:
y2x2 = 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 NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
We know that the slope of the tangent to the curve is given by the relation,
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
According to the given information:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
This is the general equation of the curve.
It is given that it passes through point (–2, 1).
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
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 seconds.
Ans: Let the rate of change of the volume of the balloon be k (where k is a constant).
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
⇒ 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 Class 12 Maths Chapter 9 - Differential Equations
Thus, the radius of the balloon after t seconds isNCERT Solutions Class 12 Maths Chapter 9 - Differential Equations.

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 (log­e 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.
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
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 Class 12 Maths Chapter 9 - Differential Equations
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 yearsNCERT Solutions Class 12 Maths Chapter 9 - Differential Equations.
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.
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Now, when t = 0, p = 1000.
⇒ 1000 = eC … (2)
At t = 10, equation (1) becomes:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
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.
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
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 Class 12 Maths Chapter 9 - Differential Equations
Also, it is given that the number of bacteria increases by 10% in 2 hours.
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Substituting this value in equation (2), we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Therefore, equation (2) becomes:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
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:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Hence, in NCERT Solutions Class 12 Maths Chapter 9 - Differential Equationshours the number of bacteria increases from 100000 to 200000.

Q24: The general solution of the differential equation NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
A. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations                                               
B. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
C. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations                                             
D. NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Ans: 
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Integrating both sides, we get:
NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations
Hence, the correct answer is A.

The document NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations is a part of the JEE Course Mathematics (Maths) Class 12.
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FAQs on NCERT Solutions Class 12 Maths Chapter 9 - Differential Equations

1. What is a differential equation?
Ans. A differential equation is an equation that relates a function with one or more of its derivatives. It describes how a function changes as a function of one or more independent variables.
2. What is the order of a differential equation?
Ans. The order of a differential equation is the highest order derivative present in the equation. For example, if the equation includes the second derivative of a function, it is a second-order differential equation.
3. How can differential equations be classified?
Ans. Differential equations can be classified based on various factors such as order, linearity, and the number of independent variables. They can be categorized into linear or nonlinear, ordinary or partial, and first, second, or higher order differential equations.
4. What is the general solution of a differential equation?
Ans. The general solution of a differential equation includes all possible solutions within a family of functions that satisfy the given differential equation. It typically contains arbitrary constants that can be determined by applying initial or boundary conditions.
5. How are differential equations used in real-world applications?
Ans. Differential equations are used in various fields such as physics, engineering, biology, and economics to model and analyze dynamic systems. They help in predicting the behavior of systems over time and are essential in understanding natural phenomena.
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