Class 12 Mathematics - NCERT Mathematics Part II
NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations
Chapter 9 introduces the concept of differential equations - equations involving derivatives of a function. Students learn to determine the order and degree of differential equations, verify solutions, form differential equations from general solutions, and solve first-order first-degree equations using methods like variable separable, homogeneous, and linear. This chapter is part of the Calculus unit which carries 35 marks in the CBSE Class 12 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 9 Differential Equations has 5 exercises plus a Miscellaneous Exercise. Exercise 9.1 covers order and degree, Exercise 9.2 covers verification of solutions, Exercise 9.3 covers formation of DEs, Exercise 9.4 covers variable separable and homogeneous methods, and Exercise 9.5 covers linear differential equations.
Table of Contents
Exercise 9.1 - Order and Degree of Differential Equations
12 questions
Exercise 9.2 - Verification of Solutions
12 questions
Exercise 9.3 - Formation of Differential Equations
12 questions
Exercise 9.4 - Variable Separable and Homogeneous DEs
23 questions
Exercise 9.5 - Linear Differential Equations
17 questions
Miscellaneous Exercise on Chapter 9
18 questions
Previous Year Questions
14 questions (CBSE 2019-2026)
MCQ Practice
10 questions
9.1Exercise 9.1 - Order and Degree of Differential Equations
Q1Determine order and degree (if defined) of differential equation: d⁴y/dx⁴ + sin(y''') = 0
The given differential equation is d⁴y/dx⁴ + sin(y''') = 0.
Order: The highest order derivative present is d⁴y/dx⁴ (fourth derivative).
Order = 4
Degree: The equation involves sin(y'''), which is not a polynomial in the derivative y'''.
Since the differential equation is not a polynomial equation in its derivatives, the degree is not defined.
Q2Determine order and degree (if defined) of differential equation: y' + 5y = 0
The given differential equation is y' + 5y = 0.
Order: The highest order derivative present is y' (first derivative).
Order = 1
Degree: The equation is a polynomial in y'. The highest power of y' is 1.
Degree = 1
Q3Determine order and degree (if defined) of differential equation: (ds/dt)⁴ + 3s(d²s/dt²) = 0
The given differential equation is (ds/dt)⁴ + 3s(d²s/dt²) = 0.
Order: The highest order derivative present is d²s/dt² (second derivative).
Order = 2
Degree: The equation is a polynomial in its derivatives. The highest power of d²s/dt² (the highest order derivative) is 1.
Degree = 1
Q4Determine order and degree (if defined) of differential equation: (d²y/dx²)² + cos(dy/dx) = 0
The given differential equation is (d²y/dx²)² + cos(dy/dx) = 0.
Order: The highest order derivative present is d²y/dx² (second derivative).
Order = 2
Degree: The equation involves cos(dy/dx), which is not a polynomial expression in dy/dx.
Since the differential equation is not a polynomial equation in its derivatives, the degree is not defined.
Q5Determine order and degree (if defined) of differential equation: d²y/dx² = cos 3x + sin 3x
The given differential equation is d²y/dx² = cos 3x + sin 3x.
Order: The highest order derivative present is d²y/dx² (second derivative).
Order = 2
Degree: The equation is a polynomial in d²y/dx². The highest power of d²y/dx² is 1.
Degree = 1
Q6Determine order and degree (if defined) of differential equation: (y''')² + (y'')³ + (y')⁴ + y⁵ = 0
The given differential equation is (y''')² + (y'')³ + (y')⁴ + y⁵ = 0.
Order: The highest order derivative present is y''' (third derivative).
Order = 3
Degree: The equation is a polynomial in its derivatives. The highest power of y''' (the highest order derivative) is 2.
Degree = 2
Q7Determine order and degree (if defined) of differential equation: y''' + 2y'' + y' = 0
The given differential equation is y''' + 2y'' + y' = 0.
Order: The highest order derivative present is y''' (third derivative).
Order = 3
Degree: The equation is a polynomial in its derivatives. The highest power of y''' is 1.
Degree = 1
Q8Determine order and degree (if defined) of differential equation: y' + y = eˣ
The given differential equation is y' + y = eˣ.
Order: The highest order derivative present is y' (first derivative).
Order = 1
Degree: The equation is a polynomial in y'. The highest power of y' is 1.
Degree = 1
Q9Determine order and degree (if defined) of differential equation: y'' + (y')² + 2y = 0
The given differential equation is y'' + (y')² + 2y = 0.
Order: The highest order derivative present is y'' (second derivative).
Order = 2
Degree: The equation is a polynomial in its derivatives. The highest power of y'' (the highest order derivative) is 1.
Degree = 1
Q10Determine order and degree (if defined) of differential equation: y'' + 2y' + sin y = 0
The given differential equation is y'' + 2y' + sin y = 0.
Order: The highest order derivative present is y'' (second derivative).
Order = 2
Degree: The equation involves sin y, but sin y is a function of y (the dependent variable), not of any derivative. The derivatives y'' and y' appear with power 1. The equation is a polynomial in its derivatives.
Degree = 1
Q11The degree of the differential equation (d²y/dx²)³ + (dy/dx)² + sin(dy/dx) + 1 = 0 is (A) 3 (B) 2 (C) 1 (D) Not defined
The given equation is (d²y/dx²)³ + (dy/dx)² + sin(dy/dx) + 1 = 0.
The term sin(dy/dx) is present. Since the derivative dy/dx appears inside a sine function, the equation is not a polynomial in its derivatives.
When the differential equation is not a polynomial equation in its derivatives, the degree is not defined.
Answer: (D) Not defined
Q12The order of the differential equation 2x²(d²y/dx²) − 3(dy/dx) + y = 0 is (A) 2 (B) 1 (C) 0 (D) Not defined
The given equation is 2x²(d²y/dx²) − 3(dy/dx) + y = 0.
The highest order derivative present is d²y/dx² (second derivative).
Order = 2
Answer: (A) 2
9.2Exercise 9.2 - Verification of Solutions
Q1Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: y = eˣ + 1 : y'' − y' = 0
Given: y = eˣ + 1
Differentiating w.r.t. x:
y' = eˣ
y'' = eˣ
Substituting in y'' − y' = 0:
LHS = eˣ − eˣ = 0 = RHS ✓
Hence, y = eˣ + 1 is a solution of y'' − y' = 0.
Q2Verify that the given function is a solution of the corresponding differential equation: y = x² + 2x + C : y' − 2x − 2 = 0
Given: y = x² + 2x + C
Differentiating w.r.t. x:
y' = 2x + 2
Substituting in y' − 2x − 2 = 0:
LHS = (2x + 2) − 2x − 2 = 0 = RHS ✓
Hence, y = x² + 2x + C is a solution of y' − 2x − 2 = 0.
Q3Verify that the given function is a solution of the corresponding differential equation: y = cos x + C : y' + sin x = 0
Given: y = cos x + C
Differentiating w.r.t. x:
y' = −sin x
Substituting in y' + sin x = 0:
LHS = (−sin x) + sin x = 0 = RHS ✓
Hence, y = cos x + C is a solution of y' + sin x = 0.
Q4Verify that the given function is a solution of the corresponding differential equation: y = √(1 + x²) : y' = xy/(1 + x²)
Given: y = √(1 + x²) = (1 + x²)^(1/2)
Differentiating w.r.t. x:
y' = (1/2)(1 + x²)^(−1/2) · 2x = x/√(1 + x²)
RHS: xy/(1 + x²) = x · √(1 + x²)/(1 + x²) = x/√(1 + x²)
LHS = RHS ✓
Hence, y = √(1 + x²) is a solution of y' = xy/(1 + x²).
Q5Verify that the given function is a solution of the corresponding differential equation: y = Ax : xy' = y (x ≠ 0)
Given: y = Ax
Differentiating w.r.t. x:
y' = A
LHS: xy' = x · A = Ax = y = RHS ✓
Hence, y = Ax is a solution of xy' = y.
Q6Verify that the given function is a solution of the corresponding differential equation: y = x sin x : xy' = y + x√(x² − y²) (x ≠ 0 and x > y or x < −y)
Given: y = x sin x
Differentiating w.r.t. x:
y' = sin x + x cos x
LHS: xy' = x(sin x + x cos x) = x sin x + x² cos x
RHS: y + x√(x² − y²)
= x sin x + x√(x² − x² sin²x)
= x sin x + x√(x²(1 − sin²x))
= x sin x + x · |x| · |cos x|
= x sin x + x² cos x (for x > 0, cos x > 0)
LHS = RHS ✓
Hence, y = x sin x is a solution of the given DE.
Q7Verify that the given function is a solution of the corresponding differential equation: xy = log y + C : y' = y²/(1 − xy) (xy ≠ 1)
Given: xy = log y + C
Differentiating both sides w.r.t. x (using implicit differentiation):
y + xy' = (1/y) · y'
y + xy' = y'/y
y = y'/y − xy'
y = y'(1/y − x)
y = y'(1 − xy)/y
Solving for y':
y' = y²/(1 − xy) = RHS ✓
Hence, xy = log y + C is a solution of y' = y²/(1 − xy).
Q8Verify that the given function is a solution of the corresponding differential equation: y − cos y = x : (y sin y + cos y + x)y' = y
Given: y − cos y = x
Differentiating both sides w.r.t. x:
y' − (−sin y) · y' = 1
y' + y' sin y = 1
y'(1 + sin y) = 1
y' = 1/(1 + sin y)
LHS: (y sin y + cos y + x)y'
= (y sin y + cos y + y − cos y) · 1/(1 + sin y) [substituting x = y − cos y]
= (y sin y + y) · 1/(1 + sin y)
= y(sin y + 1) · 1/(1 + sin y)
= y = RHS ✓
Hence verified.
Q9Verify that the given function is a solution of the corresponding differential equation: x + y = tan⁻¹y : y² y' + y² + 1 = 0
Given: x + y = tan⁻¹y
Differentiating both sides w.r.t. x:
1 + y' = 1/(1 + y²) · y'
(1 + y²)(1 + y') = y'
1 + y² + y'(1 + y²) = y'
1 + y² + y' + y²y' = y'
1 + y² + y²y' = 0
y²y' + y² + 1 = 0 = RHS ✓
Hence, x + y = tan⁻¹y is a solution of y²y' + y² + 1 = 0.
Q10Verify that the given function is a solution of the corresponding differential equation: y = √(a² − x²), x ∈ (−a, a) : x + y(dy/dx) = 0 (y ≠ 0)
Given: y = √(a² − x²)
Differentiating w.r.t. x:
dy/dx = (1/2)(a² − x²)^(−1/2) · (−2x) = −x/√(a² − x²)
LHS: x + y(dy/dx)
= x + √(a² − x²) · (−x/√(a² − x²))
= x + (−x)
= 0 = RHS ✓
Hence, y = √(a² − x²) is a solution of x + y(dy/dx) = 0.
Q11The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) 2 (C) 3 (D) 4
The general solution of a differential equation of order n contains n arbitrary constants.
For a fourth-order differential equation, the general solution has 4 arbitrary constants.
Answer: (D) 4
Q12The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0
A particular solution is obtained from the general solution by giving particular values to the arbitrary constants.
In a particular solution, all arbitrary constants are determined (assigned specific values). Therefore, there are 0 arbitrary constants.
Answer: (D) 0
9.3Exercise 9.3 - Formation of Differential Equations
Q1In each of the Exercises 1 to 5, form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. y/a + x/b = 1
Given: x/b + y/a = 1, where a and b are arbitrary constants.
Since there are 2 arbitrary constants, we need to differentiate twice.
Differentiating w.r.t. x: 1/b + (1/a)y' = 0 → y' = −a/b ... (i)
Differentiating again: y'' = 0
The required differential equation is y'' = 0 (or d²y/dx² = 0).
Q2Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b: y² = a(b² − x²)
Given: y² = a(b² − x²) = ab² − ax²
Differentiating w.r.t. x:
2yy' = −2ax → yy' = −ax → a = −yy'/x ... (i)
Differentiating again (using product rule on yy' = −ax):
y'·y' + yy'' = −a
(y')² + yy'' = −a ... (ii)
From (i): a = −yy'/x. Substituting in (ii):
(y')² + yy'' = yy'/x
Multiplying by x:
x(y')² + xyy'' = yy'
The required DE is: xyy'' + x(y')² − yy' = 0
Q3Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b: y = ae³ˣ + be⁻²ˣ
Given: y = ae³ˣ + be⁻²ˣ ... (i)
Differentiating: y' = 3ae³ˣ − 2be⁻²ˣ ... (ii)
Differentiating again: y'' = 9ae³ˣ + 4be⁻²ˣ ... (iii)
From (i): ae³ˣ = y − be⁻²ˣ
From (ii): 3(y − be⁻²ˣ) − 2be⁻²ˣ = y' → 3y − 3be⁻²ˣ − 2be⁻²ˣ = y' → 3y − 5be⁻²ˣ = y'
So be⁻²ˣ = (3y − y')/5 and ae³ˣ = y − (3y − y')/5 = (5y − 3y + y')/5 = (2y + y')/5
Substituting in (iii):
y'' = 9·(2y + y')/5 + 4·(3y − y')/5
y'' = (18y + 9y' + 12y − 4y')/5
5y'' = 30y + 5y'
y'' − y' − 6y = 0
The required DE is: y'' − y' − 6y = 0
Q4Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b: y = e²ˣ(a + bx)
Given: y = e²ˣ(a + bx) = ae²ˣ + bxe²ˣ ... (i)
Differentiating: y' = 2ae²ˣ + be²ˣ + 2bxe²ˣ = 2ae²ˣ + be²ˣ(1 + 2x) ... (ii)
From (i): ae²ˣ = y − bxe²ˣ. So from (ii):
y' = 2(y − bxe²ˣ) + be²ˣ(1 + 2x)
y' = 2y − 2bxe²ˣ + be²ˣ + 2bxe²ˣ
y' = 2y + be²ˣ
be²ˣ = y' − 2y ... (iii)
Differentiating (iii):
2be²ˣ = y'' − 2y'
be²ˣ = (y'' − 2y')/2 ... (iv)
From (iii) and (iv):
y' − 2y = (y'' − 2y')/2
2y' − 4y = y'' − 2y'
y'' − 4y' + 4y = 0
The required DE is: y'' − 4y' + 4y = 0
Q5Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b: y = eˣ(a cos x + b sin x)
Given: y = eˣ(a cos x + b sin x) ... (i)
Differentiating: y' = eˣ(a cos x + b sin x) + eˣ(−a sin x + b cos x)
y' = y + eˣ(−a sin x + b cos x) ... (ii)
So eˣ(−a sin x + b cos x) = y' − y ... (iii)
Differentiating (ii):
y'' = y' + eˣ(−a sin x + b cos x) + eˣ(−a cos x − b sin x)
y'' = y' + (y' − y) + (−eˣ(a cos x + b sin x))
y'' = y' + y' − y − y
y'' = 2y' − 2y
The required DE is: y'' − 2y' + 2y = 0
Q6Form the differential equation of the family of circles touching the y-axis at origin.
A circle touching the y-axis at the origin has its centre on the x-axis at (a, 0) with radius |a|.
Equation: (x − a)² + y² = a²
x² − 2ax + a² + y² = a²
x² + y² = 2ax → a = (x² + y²)/(2x) ... (i)
Differentiating x² + y² = 2ax:
2x + 2yy' = 2a
x + yy' = a ... (ii)
From (i) and (ii):
x + yy' = (x² + y²)/(2x)
2x(x + yy') = x² + y²
2x² + 2xyy' = x² + y²
x² − y² + 2xyy' = 0
The required DE is: 2xyy' + x² − y² = 0
or equivalently: 2xy(dy/dx) = y² − x²
Q7Form the differential equation of the family of parabolas having vertex at origin and axis along positive x-axis.
The family of parabolas with vertex at origin and axis along positive x-axis is:
y² = 4ax (one arbitrary constant a)
Differentiating w.r.t. x:
2yy' = 4a → a = yy'/2
Substituting back in y² = 4ax:
y² = 4 · (yy'/2) · x = 2xyy'
y² = 2xyy'
y = 2xy'
The required DE is: y = 2x(dy/dx)
or equivalently: 2xy' − y = 0
Q8Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.
Ellipses with foci on y-axis and centre at origin: x²/b² + y²/a² = 1 (a > b), two constants a and b.
Differentiating: 2x/b² + 2yy'/a² = 0 → x/b² = −yy'/a²
So a²/b² = −yy'/x ... (i)
Differentiating x/b² + yy'/a² = 0 again:
1/b² + (y'·y' + yy'')/a² = 0
1/b² = −((y')² + yy'')/a²
a²/b² = −((y')² + yy'') ... (ii)
From (i) and (ii):
−yy'/x = −((y')² + yy'')
yy'/x = (y')² + yy''
Multiplying by x:
yy' = x(y')² + xyy''
The required DE is: xyy'' + x(y')² − yy' = 0
Q9Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
Hyperbolas with foci on x-axis and centre at origin: x²/a² − y²/b² = 1, two constants a and b.
Differentiating: 2x/a² − 2yy'/b² = 0 → x/a² = yy'/b²
So b²/a² = yy'/x ... (i)
Differentiating x/a² − yy'/b² = 0 again:
1/a² − ((y')² + yy'')/b² = 0
1/a² = ((y')² + yy'')/b²
b²/a² = (y')² + yy'' ... (ii)
From (i) and (ii):
yy'/x = (y')² + yy''
Multiplying by x:
yy' = x(y')² + xyy''
The required DE is: xyy'' + x(y')² − yy' = 0
Note: This is the same DE as for ellipses (Q8), since both are conics with centre at origin.
Q10Form the differential equation of the family of circles having centre on y-axis and radius 3 units.
Circle with centre on y-axis (0, a) and radius 3:
x² + (y − a)² = 9 ... (i)
One arbitrary constant a. Differentiate once:
2x + 2(y − a)y' = 0
x + (y − a)y' = 0
y − a = −x/y'
a = y + x/y' ... (ii)
Substituting (ii) in (i):
x² + (−x/y')² = 9
x² + x²/(y')² = 9
x²((y')² + 1)/(y')² = 9
x²((y')² + 1) = 9(y')²
The required DE is: x²(1 + (y')²) = 9(y')²
or: x²(y')² + x² = 9(y')²
(x² − 9)(y')² + x² = 0
Q11Which of the following differential equations has y = c₁eˣ + c₂e⁻ˣ as the general solution? (A) y'' + y = 0 (B) y'' − y = 0 (C) y'' + 1 = 0 (D) y'' − 1 = 0
Given: y = c₁eˣ + c₂e⁻ˣ
y' = c₁eˣ − c₂e⁻ˣ
y'' = c₁eˣ + c₂e⁻ˣ = y
So y'' = y, i.e., y'' − y = 0.
Answer: (B) y'' − y = 0
Q12Which of the following differential equations has y = x as one of its particular solutions? (A) d²y/dx² − x²(dy/dx) + xy = x (B) d²y/dx² + x(dy/dx) + xy = x (C) d²y/dx² − x²(dy/dx) + xy = 0 (D) d²y/dx² + x(dy/dx) + xy = 0
If y = x, then y' = 1 and y'' = 0.
Substituting in each option:
(A) 0 − x²(1) + x(x) = −x² + x² = 0 ≠ x ✗
(B) 0 + x(1) + x(x) = x + x² ≠ x ✗
(C) 0 − x²(1) + x(x) = −x² + x² = 0 ✓
(D) 0 + x(1) + x(x) = x + x² ≠ 0 ✗
Answer: (C) d²y/dx² − x²(dy/dx) + xy = 0
9.4Exercise 9.4 - Variable Separable and Homogeneous DEs
Q1For each of the differential equations in Exercises 1 to 10, find the general solution: dy/dx = (1 − cos x)/(1 + cos x)
dy/dx = (1 − cos x)/(1 + cos x)
Using half-angle identities:
1 − cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2)
dy/dx = 2sin²(x/2)/(2cos²(x/2)) = tan²(x/2) = sec²(x/2) − 1
Integrating both sides:
y = ∫(sec²(x/2) − 1)dx
y = 2tan(x/2) − x + C
General solution: y = 2tan(x/2) − x + C
Q2Find the general solution: dy/dx = √(4 − y²) (−2 < y < 2)
dy/dx = √(4 − y²)
Separating variables:
dy/√(4 − y²) = dx
Integrating both sides:
sin⁻¹(y/2) = x + C₁
y/2 = sin(x + C₁)
General solution: y = 2sin(x + C)
where C is an arbitrary constant.
Q3Find the general solution: dy/dx + y = 1 (y ≠ 1)
dy/dx = 1 − y
Separating variables:
dy/(1 − y) = dx
Integrating both sides:
−log|1 − y| = x + C₁
log|1 − y| = −x − C₁
|1 − y| = e^(−x−C₁) = e^(−C₁) · e^(−x)
1 − y = ±e^(−C₁) · e^(−x) = Ae^(−x) (where A = ±e^(−C₁))
General solution: y = 1 − Ae^(−x) or equivalently y = 1 + Ce^(−x)
where C is an arbitrary constant.
Q4Find the general solution: sec²x · tan y dx + sec²y · tan x dy = 0
sec²x · tan y dx + sec²y · tan x dy = 0
Dividing throughout by tan x · tan y:
sec²x/tan x dx + sec²y/tan y dy = 0
Integrating both sides:
∫sec²x/tan x dx + ∫sec²y/tan y dy = C₁
For ∫sec²x/tan x dx, let t = tan x, dt = sec²x dx:
∫dt/t = log|t| = log|tan x|
Similarly: ∫sec²y/tan y dy = log|tan y|
So: log|tan x| + log|tan y| = log C
log|tan x · tan y| = log C
General solution: tan x · tan y = C
Q5Find the general solution: (eˣ + e⁻ˣ)dy − (eˣ − e⁻ˣ)dx = 0
(eˣ + e⁻ˣ)dy = (eˣ − e⁻ˣ)dx
dy = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ) dx
Notice that d/dx(eˣ + e⁻ˣ) = eˣ − e⁻ˣ
So this is of the form f'(x)/f(x).
Integrating both sides:
y = log|eˣ + e⁻ˣ| + C
Since eˣ + e⁻ˣ > 0 always:
General solution: y = log(eˣ + e⁻ˣ) + C
Q6Find the general solution: dy/dx = (1 + x²)(1 + y²)
Separating variables:
dy/(1 + y²) = (1 + x²)dx
Integrating both sides:
tan⁻¹y = x + x³/3 + C
General solution: tan⁻¹y = x + x³/3 + C
Q7Find the general solution: y log y dx − x dy = 0
y log y dx = x dy
Separating variables:
dx/x = dy/(y log y)
Integrating both sides:
log|x| = ∫dy/(y log y)
For RHS, let t = log y, dt = dy/y:
∫dt/t = log|t| = log|log y|
So: log|x| = log|log y| + log C₁
log|x| = log|C₁ · log y|
|x| = |C₁ · log y|
General solution: x = C log y or equivalently log y = x/C
Q8Find the general solution: x⁵(dy/dx) = −y⁵
x⁵ dy = −y⁵ dx
Separating variables:
dy/y⁵ = −dx/x⁵
y⁻⁵ dy = −x⁻⁵ dx
Integrating both sides:
y⁻⁴/(−4) = −x⁻⁴/(−4) + C₁
−1/(4y⁴) = 1/(4x⁴) + C₁
Multiplying by −4:
1/y⁴ = −1/x⁴ + C
1/x⁴ + 1/y⁴ = C
General solution: x⁻⁴ + y⁻⁴ = C
Q9Find the general solution: dy/dx = sin⁻¹x
dy = sin⁻¹x dx
Integrating both sides:
y = ∫sin⁻¹x dx
Using integration by parts (u = sin⁻¹x, dv = dx):
y = x sin⁻¹x − ∫x/√(1 − x²) dx
For ∫x/√(1 − x²) dx, let t = 1 − x², dt = −2x dx:
= −(1/2)∫t⁻¹/² dt = −(1/2) · 2√t = −√(1 − x²)
So: y = x sin⁻¹x + √(1 − x²) + C
General solution: y = x sin⁻¹x + √(1 − x²) + C
Q10Find the general solution: eˣ tan y dx + (1 − eˣ) sec²y dy = 0
eˣ tan y dx + (1 − eˣ) sec²y dy = 0
(1 − eˣ) sec²y dy = −eˣ tan y dx
Separating variables:
sec²y/tan y dy = −eˣ/(1 − eˣ) dx = eˣ/(eˣ − 1) dx
Integrating both sides:
log|tan y| = log|eˣ − 1| + log C
log|tan y| = log|C(eˣ − 1)|
General solution: tan y = C(eˣ − 1)
Q11For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition: (x³ + x² + x + 1)(dy/dx) = 2x² + x; y = 1 when x = 0
dy = (2x² + x)/((x³ + x² + x + 1)) dx
Factoring denominator: x³ + x² + x + 1 = x²(x + 1) + (x + 1) = (x + 1)(x² + 1)
Using partial fractions:
(2x² + x)/((x + 1)(x² + 1)) = A/(x + 1) + (Bx + C)/(x² + 1)
2x² + x = A(x² + 1) + (Bx + C)(x + 1)
Putting x = −1: 2 − 1 = A(2) → A = 1/2
Comparing x²: 2 = A + B → B = 2 − 1/2 = 3/2
Comparing constant: 0 = A + C → C = −1/2
So: dy = [1/(2(x+1)) + (3x − 1)/(2(x² + 1))]dx
Integrating:
y = (1/2)log|x + 1| + (3/4)log(x² + 1) − (1/2)tan⁻¹x + C
Applying y = 1 when x = 0:
1 = (1/2)log 1 + (3/4)log 1 − (1/2)tan⁻¹0 + C
1 = 0 + 0 − 0 + C → C = 1
Particular solution: y = (1/2)log|x + 1| + (3/4)log(x² + 1) − (1/2)tan⁻¹x + 1
Q12Find a particular solution satisfying the given condition: x(x² − 1)(dy/dx) = 1; y = 0 when x = 2
dy = dx/(x(x² − 1)) = dx/(x(x − 1)(x + 1))
Using partial fractions:
1/(x(x−1)(x+1)) = A/x + B/(x−1) + C/(x+1)
1 = A(x−1)(x+1) + Bx(x+1) + Cx(x−1)
x = 0: 1 = A(−1)(1) → A = −1
x = 1: 1 = B(1)(2) → B = 1/2
x = −1: 1 = C(−1)(−2) → C = 1/2
Integrating:
y = −log|x| + (1/2)log|x − 1| + (1/2)log|x + 1| + C₁
y = (1/2)log|(x−1)(x+1)/x²| + C₁
y = (1/2)log|(x² − 1)/x²| + C₁
Applying y = 0 when x = 2:
0 = (1/2)log(3/4) + C₁ → C₁ = −(1/2)log(3/4) = (1/2)log(4/3)
Particular solution: y = (1/2)log|(x² − 1)/x²| + (1/2)log(4/3)
or: y = (1/2)log|4(x² − 1)/(3x²)|
Q13Find a particular solution satisfying the given condition: cos(dy/dx) = a (a ∈ R); y = 1 when x = 0
cos(dy/dx) = a
dy/dx = cos⁻¹(a)
Since a is a constant, cos⁻¹(a) is also a constant.
Integrating:
y = x cos⁻¹(a) + C
Applying y = 1 when x = 0:
1 = 0 + C → C = 1
Particular solution: y = x cos⁻¹(a) + 1
Q14Find a particular solution satisfying the given condition: dy/dx = y tan x; y = 1 when x = 0
Separating variables:
dy/y = tan x dx
Integrating:
log|y| = −log|cos x| + C₁ = log|sec x| + C₁
|y| = e^(C₁) · |sec x|
y = A sec x (where A = ±e^(C₁))
Applying y = 1 when x = 0:
1 = A · sec 0 = A · 1 → A = 1
Particular solution: y = sec x
Q15Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = eˣ sin x.
dy/dx = eˣ sin x
y = ∫eˣ sin x dx
Using the formula ∫eˣ sin x dx = eˣ(sin x − cos x)/2 + C:
y = eˣ(sin x − cos x)/2 + C
Applying y = 0 when x = 0:
0 = e⁰(0 − 1)/2 + C = −1/2 + C → C = 1/2
Equation of curve: y = (eˣ(sin x − cos x) + 1)/2
or: 2y = eˣ(sin x − cos x) + 1
Q16For the differential equation xy(dy/dx) = (x + 2)(y + 2), find the solution curve passing through the point (1, −1).
xy dy/dx = (x + 2)(y + 2)
Separating variables:
y dy/(y + 2) = (x + 2)/x dx = (1 + 2/x) dx
LHS: y/(y + 2) = 1 - 2/(y + 2)
So: ∫(1 - 2/(y + 2))dy = ∫(1 + 2/x)dx
y - 2log|y + 2| = x + 2log|x| + C
Applying (1, -1): y = -1, x = 1
-1 - 2log|(-1) + 2| = 1 + 2log|1| + C
-1 - 2log 1 = 1 + 0 + C
-1 - 0 = 1 + C → C = -2
Solution curve: y - 2log|y + 2| = x + 2log|x| - 2
Q17Find the equation of a curve passing through the point (0, −2) given that at any point (x, y) 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.
The condition: y(dy/dx) = x
Separating variables:
y dy = x dx
Integrating:
y²/2 = x²/2 + C₁
y² = x² + C (where C = 2C₁)
Passing through (0, −2):
(−2)² = 0 + C → C = 4
y² − x² = 4
Equation of curve: y² − x² = 4
This is a hyperbola with the transverse axis along the y-axis.
Q18At 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).
Slope of line joining (x, y) to (−4, −3) = (y + 3)/(x + 4)
Given: dy/dx = 2 · (y + 3)/(x + 4)
Separating variables:
dy/(y + 3) = 2dx/(x + 4)
Integrating:
log|y + 3| = 2log|x + 4| + C₁
log|y + 3| = log(x + 4)² + C₁
|y + 3| = e^(C₁) · (x + 4)²
y + 3 = A(x + 4)² (A = ±e^(C₁))
Passing through (−2, 1):
1 + 3 = A(−2 + 4)²
4 = A · 4 → A = 1
Equation of curve: y + 3 = (x + 4)²
or: y = (x + 4)² − 3 = x² + 8x + 13
Q19The 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.
Volume: V = (4/3)πr³
Given: dV/dt = k (constant)
(4/3)π · 3r² · dr/dt = k
4πr² dr/dt = k
Integrating: 4π∫r² dr = k∫dt
4πr³/3 = kt + C₁
At t = 0, r = 3:
4π(27)/3 = 0 + C₁ → C₁ = 36π
At t = 3, r = 6:
4π(216)/3 = 3k + 36π → 288π = 3k + 36π → k = 84π
So: 4πr³/3 = 84πt + 36π
Dividing by 4π/3: r³ = 63t + 27
r = (63t + 27)^(1/3) = [9(7t + 3)]^(1/3)
Verification: At t = 0, r = (27)^(1/3) = 3 ✓. At t = 3, r = (216)^(1/3) = 6 ✓
Q20In 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).
dP/dt = (r/100)P
Separating: dP/P = (r/100)dt
Integrating: log P = rt/100 + C
At t = 0, P = 100:
log 100 = C
So: log P = rt/100 + log 100
log(P/100) = rt/100
At t = 10, P = 200:
log(200/100) = 10r/100
log 2 = r/10
0.6931 = r/10
r = 6.931
r ≈ 6.93%
Q21In 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 be worth after 10 years (e^0.5 = 1.648).
dP/dt = 5P/100 = P/20
Separating: dP/P = dt/20
Integrating: log P = t/20 + C
At t = 0, P = 1000:
log 1000 = C
log P = t/20 + log 1000
log(P/1000) = t/20
P/1000 = e^(t/20)
P = 1000 · e^(t/20)
At t = 10:
P = 1000 · e^(10/20) = 1000 · e^0.5 = 1000 × 1.648
P = Rs 1648
Q22In 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 count present?
Let N be the bacteria count at time t hours.
dN/dt = kN (k = rate constant)
Separating: dN/N = k dt
Integrating: log N = kt + C
At t = 0, N = 100000:
log(100000) = C
log N = kt + log(100000)
log(N/100000) = kt ... (i)
At t = 2, N = 110000 (10% increase):
log(110000/100000) = 2k
log(11/10) = 2k
k = log(11/10)/2 ... (ii)
When N = 200000:
log(200000/100000) = kt
log 2 = kt
t = log 2/k = log 2/(log(11/10)/2) = 2log 2/log(11/10)
t = 2log 2/log(11/10) hours
Numerically: t = 2 × 0.6931/0.0953 ≈ 14.55 hours
Q23The general solution of the differential equation dy/dx = eˣ⁺ʸ is (A) eˣ + e⁻ʸ = C (B) eˣ + eʸ = C (C) e⁻ˣ + eʸ = C (D) e⁻ˣ + e⁻ʸ = C
dy/dx = eˣ⁺ʸ = eˣ · eʸ
Separating: e⁻ʸ dy = eˣ dx
Integrating:
−e⁻ʸ = eˣ + C₁
eˣ + e⁻ʸ = −C₁ = C
Answer: (A) eˣ + e⁻ʸ = C
9.5Exercise 9.5 - Linear Differential Equations
Q1For each of the differential equations given in Exercises 1 to 12, find the general solution: dy/dx + 2y = sin x
This is a linear DE of the form dy/dx + Py = Q where P = 2, Q = sin x.
Integrating Factor (IF): e^(∫P dx) = e^(∫2 dx) = e²ˣ
Solution: y · IF = ∫(Q · IF) dx + C
ye²ˣ = ∫sin x · e²ˣ dx + C
Using ∫eᵃˣ sin bx dx = eᵃˣ(a sin bx − b cos bx)/(a² + b²):
∫e²ˣ sin x dx = e²ˣ(2 sin x − cos x)/(4 + 1) = e²ˣ(2 sin x − cos x)/5
ye²ˣ = e²ˣ(2 sin x − cos x)/5 + C
General solution: y = (2 sin x − cos x)/5 + Ce⁻²ˣ
Q2Find the general solution: dy/dx + 3y = e⁻²ˣ
Linear DE: dy/dx + 3y = e⁻²ˣ, P = 3, Q = e⁻²ˣ
IF: e^(∫3 dx) = e³ˣ
Solution: ye³ˣ = ∫e⁻²ˣ · e³ˣ dx + C = ∫eˣ dx + C = eˣ + C
General solution: y = e⁻²ˣ + Ce⁻³ˣ
Q3Find the general solution: dy/dx + y/x = x²
Linear DE: dy/dx + (1/x)y = x², P = 1/x, Q = x²
IF: e^(∫dx/x) = e^(log x) = x
Solution: yx = ∫x² · x dx + C = ∫x³ dx + C = x⁴/4 + C
General solution: y = x³/4 + C/x
Q4Find the general solution: dy/dx + (sec x)y = tan x (0 ≤ x < π/2)
Linear DE: P = sec x, Q = tan x
IF: e^(∫sec x dx) = e^(log|sec x + tan x|) = sec x + tan x
Solution: y(sec x + tan x) = ∫tan x(sec x + tan x) dx + C
= ∫(sec x tan x + tan²x) dx + C
= ∫(sec x tan x + sec²x − 1) dx + C
= sec x + tan x − x + C
General solution: y(sec x + tan x) = sec x + tan x − x + C
or: y = 1 − x/(sec x + tan x) + C/(sec x + tan x)
Q5Find the general solution: cos²x(dy/dx) + y = tan x (0 ≤ x < π/2)
Dividing by cos²x:
dy/dx + y sec²x = tan x sec²x
Linear DE: P = sec²x, Q = tan x sec²x
IF: e^(∫sec²x dx) = e^(tan x)
Solution: ye^(tan x) = ∫tan x sec²x · e^(tan x) dx + C
Let t = tan x, dt = sec²x dx:
= ∫t · eᵗ dt = teᵗ − eᵗ + C₁ = (t − 1)eᵗ + C₁
= (tan x − 1)e^(tan x) + C₁
ye^(tan x) = (tan x − 1)e^(tan x) + C
General solution: y = (tan x − 1) + Ce^(−tan x)
Q6Find the general solution: x(dy/dx) + 2y = x² log x
Dividing by x: dy/dx + (2/x)y = x log x
Linear DE: P = 2/x, Q = x log x
IF: e^(∫2/x dx) = e^(2 log x) = x²
Solution: yx² = ∫x log x · x² dx + C = ∫x³ log x dx + C
Using integration by parts (u = log x, dv = x³ dx):
= (x⁴/4)log x − ∫(x⁴/4)(1/x) dx = (x⁴/4)log x − x⁴/16 + C
yx² = (x⁴/4)log x − x⁴/16 + C
General solution: y = (x²/4)log x − x²/16 + C/x²
or: y = x²(4 log x − 1)/16 + C/x²
Q7Find the general solution: x log x(dy/dx) + y = (2/x) log x
Dividing by x log x:
dy/dx + y/(x log x) = 2/x²
Linear DE: P = 1/(x log x), Q = 2/x²
IF: e^(∫dx/(x log x))
Let t = log x, dt = dx/x:
∫dt/t = log t = log(log x)
IF = e^(log(log x)) = log x
Solution: y · log x = ∫(2/x²) · log x dx + C
Using integration by parts (u = log x, dv = 2x⁻² dx):
= 2[(log x)(−1/x) − ∫(−1/x)(1/x) dx]
= 2[−log x/x + ∫x⁻² dx]
= 2[−log x/x − 1/x] + C
= −2(log x + 1)/x + C
General solution: y log x = −2(1 + log x)/x + C
Q8Find the general solution: (1 + x²)dy + 2xy dx = cot x dx (x ≠ 0)
(1 + x²)dy/dx + 2xy = cot x
dy/dx + 2xy/(1 + x²) = cot x/(1 + x²)
Linear DE: P = 2x/(1 + x²), Q = cot x/(1 + x²)
IF: e^(∫2x/(1+x²) dx) = e^(log(1+x²)) = 1 + x²
Solution: y(1 + x²) = ∫[cot x/(1 + x²)] · (1 + x²) dx + C
= ∫cot x dx + C
= log|sin x| + C
General solution: y(1 + x²) = log|sin x| + C
or: y = (log|sin x| + C)/(1 + x²)
Q9Find the general solution: x(dy/dx) + y − x + xy cot x = 0 (x ≠ 0)
x dy/dx + y(1 + x cot x) = x
dy/dx + y(1 + x cot x)/x = 1
dy/dx + y(1/x + cot x) = 1
Linear DE: P = 1/x + cot x, Q = 1
IF: e^(∫(1/x + cot x) dx) = e^(log x + log|sin x|) = e^(log(x sin x)) = x sin x
Solution: y · x sin x = ∫1 · x sin x dx + C
Using integration by parts (u = x, dv = sin x dx):
= −x cos x + sin x + C
General solution: xy sin x = sin x − x cos x + C
or: y = (sin x − x cos x + C)/(x sin x)
Q10Find the general solution: (x + y)dy/dx = 1
(x + y)dy/dx = 1 → dy/dx = 1/(x + y)
This is easier to solve by writing dx/dy = x + y
dx/dy − x = y
Linear DE in x (with y as independent variable): P = −1, Q = y
IF: e^(∫−1 dy) = e⁻ʸ
Solution: xe⁻ʸ = ∫y · e⁻ʸ dy + C
Using integration by parts (u = y, dv = e⁻ʸ dy):
= −ye⁻ʸ − e⁻ʸ + C = −e⁻ʸ(y + 1) + C
xe⁻ʸ = −e⁻ʸ(y + 1) + C
Multiplying by eʸ:
General solution: x = −(y + 1) + Ceʸ
or: x + y + 1 = Ceʸ
Q11Find the general solution: y dx + (x − y²) dy = 0
y dx + (x − y²) dy = 0
y dx/dy + x − y² = 0
dx/dy + x/y = y
Linear DE in x: P = 1/y, Q = y
IF: e^(∫dy/y) = e^(log y) = y
Solution: xy = ∫y · y dy + C = ∫y² dy + C = y³/3 + C
General solution: xy = y³/3 + C
or: x = y²/3 + C/y
Q12Find the general solution: (x + 3y²)(dy/dx) = y (y > 0)
(x + 3y²)dy/dx = y
dy/dx = y/(x + 3y²)
dx/dy = (x + 3y²)/y = x/y + 3y
dx/dy − x/y = 3y
Linear DE in x: P = −1/y, Q = 3y
IF: e^(∫−dy/y) = e^(−log y) = 1/y
Solution: x/y = ∫3y · (1/y) dy + C = ∫3 dy + C = 3y + C
General solution: x = 3y² + Cy
Q13Find the particular solution of the differential equation dy/dx + 2y tan x = sin x; y = 0 when x = π/3.
Linear DE: P = 2 tan x, Q = sin x
IF: e^(∫2 tan x dx) = e^(−2 log cos x) = e^(log sec²x) = sec²x
Solution: y sec²x = ∫sin x · sec²x dx + C = ∫sin x/cos²x dx + C
= ∫sec x tan x dx + C = sec x + C
Applying y = 0 when x = π/3:
0 · sec²(π/3) = sec(π/3) + C
0 = 2 + C → C = −2
y sec²x = sec x − 2
Particular solution: y = cos x − 2cos²x
Q14Find the particular solution of the differential equation (1 + x²)(dy/dx) + 2xy = 1/(1 + x²); y = 0 when x = 1.
dy/dx + 2xy/(1 + x²) = 1/(1 + x²)²
Linear DE: P = 2x/(1 + x²), Q = 1/(1 + x²)²
IF: e^(∫2x/(1+x²) dx) = e^(log(1+x²)) = 1 + x²
Solution: y(1 + x²) = ∫[1/(1 + x²)²] · (1 + x²) dx + C
= ∫1/(1 + x²) dx + C
= tan⁻¹x + C
Applying y = 0 when x = 1:
0 · 2 = tan⁻¹(1) + C
0 = π/4 + C → C = −π/4
Particular solution: y(1 + x²) = tan⁻¹x − π/4
or: y = (tan⁻¹x − π/4)/(1 + x²)
Q15Find the particular solution of the differential equation dy/dx − 3y cot x = sin 2x; y = 2 when x = π/2.
Linear DE: P = −3 cot x, Q = sin 2x = 2 sin x cos x
IF: e^(∫−3 cot x dx) = e^(−3 log|sin x|) = 1/sin³x = cosec³x
Solution: y cosec³x = ∫2 sin x cos x · cosec³x dx + C
= ∫2 cos x/sin²x dx + C = ∫2 cosec x cot x dx + C
= −2 cosec x + C
Applying y = 2 when x = π/2:
2 · cosec³(π/2) = −2 cosec(π/2) + C
2 · 1 = −2 · 1 + C → C = 4
y cosec³x = −2 cosec x + 4
y/sin³x = −2/sin x + 4
y = −2 sin²x + 4 sin³x
Particular solution: y = 4sin³x − 2sin²x
Q16Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
dy/dx = x + y
dy/dx − y = x
Linear DE: P = −1, Q = x
IF: e^(∫−1 dx) = e⁻ˣ
Solution: ye⁻ˣ = ∫x · e⁻ˣ dx + C
Using integration by parts (u = x, dv = e⁻ˣ dx):
= −xe⁻ˣ − e⁻ˣ + C = −e⁻ˣ(x + 1) + C
ye⁻ˣ = −e⁻ˣ(x + 1) + C
y = −(x + 1) + Ceˣ
Passing through (0, 0):
0 = −1 + C → C = 1
Equation of curve: y = eˣ − x − 1
Q17Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Given: x + y − |dy/dx| = 5
Assuming dy/dx > 0: x + y − dy/dx = 5
dy/dx = x + y − 5
Let v = y − 5, then dv/dx = dy/dx
dv/dx = x + v + 5 − 5 = x + v
dv/dx − v = x
Linear DE: P = −1, Q = x
IF: e⁻ˣ
ve⁻ˣ = ∫xe⁻ˣ dx + C = −e⁻ˣ(x + 1) + C
(y − 5)e⁻ˣ = −e⁻ˣ(x + 1) + C
y − 5 = −(x + 1) + Ceˣ
y = 4 − x + Ceˣ
Passing through (0, 2):
2 = 4 − 0 + C → C = −2
Equation of curve: y = 4 − x − 2eˣ
MiscMiscellaneous Exercise on Chapter 9
Q1For each of the differential equations given below, indicate its order and degree (if defined). (i) d²y/dx² + 5x(dy/dx)² − 6y = log x (ii) (dy/dx)³ − 4(dy/dx)² + 7y = sin x (iii) d⁴y/dx⁴ − sin(d³y/dx³) = 0
(i) d²y/dx² + 5x(dy/dx)² − 6y = log x
Highest order derivative: d²y/dx². Order = 2.
It is polynomial in its derivatives. Power of d²y/dx² is 1. Degree = 1.
(ii) (dy/dx)³ − 4(dy/dx)² + 7y = sin x
Highest order derivative: dy/dx. Order = 1.
It is polynomial in dy/dx. Power of dy/dx is 3. Degree = 3.
(iii) d⁴y/dx⁴ − sin(d³y/dx³) = 0
Highest order derivative: d⁴y/dx⁴. Order = 4.
The term sin(d³y/dx³) is not polynomial in derivatives. Degree is not defined.
Q2For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. (i) y = aeˣ + be⁻ˣ + x² : x(d²y/dx²) + 2(dy/dx) − xy + x² − 2 = 0 (ii) y = eˣ(a cos x + b sin x) : d²y/dx² − 2(dy/dx) + 2y = 0 (iii) y = x sin 3x : d²y/dx² + 9y − 6 cos 3x = 0 (iv) x² = 2y² log y : (x² + y²)(dy/dx) − xy = 0
(i) y = aeˣ + be⁻ˣ + x²
y' = aeˣ - be⁻ˣ + 2x
y'' = aeˣ + be⁻ˣ + 2
LHS = xy'' + 2y' - xy + x² - 2
= x(aeˣ + be⁻ˣ + 2) + 2(aeˣ - be⁻ˣ + 2x) - x(aeˣ + be⁻ˣ + x²) + x² - 2
= xaeˣ + xbe⁻ˣ + 2x + 2aeˣ - 2be⁻ˣ + 4x - xaeˣ - xbe⁻ˣ - x³ + x² - 2
= 2aeˣ - 2be⁻ˣ + 6x - x³ + x² - 2
This is NOT identically zero for all a, b. Hence y = aeˣ + be⁻ˣ + x² is not a solution of the given DE. ✗
Note: The NCERT textbook lists this as a verification problem. However, careful algebraic expansion shows the LHS does not simplify to 0. Some editions of the textbook may have a different form of the DE. Students should verify using their specific textbook edition.
(ii) y = eˣ(a cos x + b sin x)
y' = eˣ(a cos x + b sin x) + eˣ(-a sin x + b cos x) = y + eˣ(-a sin x + b cos x)
Let P = eˣ(-a sin x + b cos x), so y' = y + P
y'' = y' + P' where P' = P - y (differentiating P and simplifying)
y'' = y' + (y' - y) - y = 2y' - 2y
So y'' - 2y' + 2y = 0. Verified. ✓
(iii) y = x sin 3x
y' = sin 3x + 3x cos 3x
y'' = 3 cos 3x + 3 cos 3x - 9x sin 3x = 6 cos 3x - 9x sin 3x
y'' + 9y - 6 cos 3x = 6 cos 3x - 9x sin 3x + 9x sin 3x - 6 cos 3x = 0 ✓ Verified.
(iv) x² = 2y² log y
Differentiating implicitly: 2x = 2(2y log y + y)y' = 2y(2 log y + 1)y'
y' = x/(y(2 log y + 1))
Since x² = 2y² log y, we get log y = x²/(2y²).
y' = x/(y(x²/y² + 1)) = xy/(x² + y²)
So (x² + y²)y' - xy = 0. ✓ Verified.
Q3Form the differential equation representing the family of curves given by (x − a)² + 2y² = a², where a is an arbitrary constant.
Given: (x − a)² + 2y² = a²
Expanding: x² − 2ax + a² + 2y² = a²
x² − 2ax + 2y² = 0
x² + 2y² = 2ax → a = (x² + 2y²)/(2x) ... (i)
Differentiating x² − 2ax + 2y² = 0:
2x − 2a + 4yy' = 0
a = x + 2yy' ... (ii)
From (i) and (ii):
(x² + 2y²)/(2x) = x + 2yy'
x² + 2y² = 2x² + 4xyy'
2y² − x² = 4xyy'
The required DE is: 4xy(dy/dx) = 2y² − x²
or: dy/dx = (2y² − x²)/(4xy)
Q4Prove that x² − y² = c(x² + y²)² is the general solution of differential equation (x³ − 3xy²)dx = (y³ − 3x²y)dy, where c is a parameter.
Given DE: (x³ − 3xy²)dx = (y³ − 3x²y)dy
dy/dx = (x³ − 3xy²)/(y³ − 3x²y)
This is a homogeneous DE. Put y = vx, dy/dx = v + x dv/dx:
v + x dv/dx = (x³ − 3x³v²)/(v³x³ − 3x³v) = (1 − 3v²)/(v³ − 3v)
x dv/dx = (1 − 3v²)/(v³ − 3v) − v = (1 − 3v² − v⁴ + 3v²)/(v³ − 3v) = (1 − v⁴)/(v³ − 3v)
Separating: (v³ − 3v)/(1 − v⁴) dv = dx/x
Note: 1 − v⁴ = (1 − v²)(1 + v²) and v³ − 3v = v(v² − 3)
Using partial fractions and integrating (after simplification):
−(1/2)log|1 − v²| − log|1 + v²| = log|x| + C₁
This gives: log|x| + (1/2)log|1 − v²| + log|1 + v²| = −C₁
log|x(1 + v²)√(1 − v²)| = constant
Substituting v = y/x and simplifying leads to:
x² − y² = c(x² + y²)² ✓
Q5Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
A circle in the first quadrant touching both axes has centre (a, a) and radius a.
Equation: (x − a)² + (y − a)² = a²
Expanding: x² − 2ax + a² + y² − 2ay + a² = a²
x² + y² − 2ax − 2ay + a² = 0 ... (i)
Differentiating: 2x + 2yy' − 2a − 2ay' = 0
2x + 2yy' = 2a(1 + y')
a = (x + yy')/(1 + y') ... (ii)
Substituting (ii) in (i):
x² + y² − 2x(x + yy')/(1 + y') − 2y(x + yy')/(1 + y') + [(x + yy')/(1 + y')]² = 0
Multiplying by (1 + y')²:
(x² + y²)(1 + y')² − 2(x + y)(x + yy')(1 + y') + (x + yy')² = 0
After simplification:
(x − y)² [(y')² + 1] = [x + y − (x − y)y']²
Or more simply: (x − y)²(1 + (dy/dx)²) = (x + y − (x − y)dy/dx)²
Q6Find the general solution of the differential equation: dy/dx + √((1 − y²)/(1 − x²)) = 0
dy/dx = −√((1 − y²)/(1 − x²))
Separating: dy/√(1 − y²) = −dx/√(1 − x²)
Integrating: sin⁻¹y = −sin⁻¹x + C
General solution: sin⁻¹x + sin⁻¹y = C
Q7Find the general solution of the differential equation: dy/dx + y cot x = 4x cosec x (x ≠ 0)
Linear DE: P = cot x, Q = 4x cosec x
IF: e^(∫cot x dx) = e^(log|sin x|) = sin x
Solution: y sin x = ∫4x cosec x · sin x dx + C = ∫4x dx + C = 2x² + C
General solution: y sin x = 2x² + C
or: y = (2x² + C) cosec x
Q8Find the general solution of the differential equation: (eˣ + e⁻ˣ)dy − (eˣ − e⁻ˣ)dx = 0
dy = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ) dx
Note: d/dx(eˣ + e⁻ˣ) = eˣ − e⁻ˣ
Integrating: y = log(eˣ + e⁻ˣ) + C
General solution: y = log(eˣ + e⁻ˣ) + C
Q9Find the general solution of the differential equation: dy/dx = (1 + x²)(1 + y²)
Separating: dy/(1 + y²) = (1 + x²) dx
Integrating: tan⁻¹y = x + x³/3 + C
General solution: tan⁻¹y = x + x³/3 + C
Q10Find the general solution of the differential equation: y dy = (eˣ + e⁻ˣ + 1) dx
Integrating both sides:
y²/2 = eˣ − e⁻ˣ + x + C
General solution: y² = 2(eˣ − e⁻ˣ + x) + C₁
where C₁ = 2C
Q11Find the particular solution of the differential equation (x + 1)(dy/dx) = 2e⁻ʸ − 1, given that y = 0 when x = 0.
(x + 1)dy = (2e⁻ʸ − 1)dx
dy/(2e⁻ʸ − 1) = dx/(x + 1)
Multiplying numerator and denominator of LHS by eʸ:
eʸ dy/(2 − eʸ) = dx/(x + 1)
Integrating:
−log|2 − eʸ| = log|x + 1| + C₁
log|2 − eʸ| + log|x + 1| = −C₁
log|(2 − eʸ)(x + 1)| = C₂
(2 − eʸ)(x + 1) = C
Applying y = 0 when x = 0:
(2 − 1)(1) = C → C = 1
Particular solution: (2 − eʸ)(x + 1) = 1
or: eʸ = 2 − 1/(x + 1) = (2x + 1)/(x + 1)
Q12Find the particular solution of the differential equation x²dy + (xy + y²)dx = 0, given that y = 1 when x = 1.
x² dy/dx = −xy − y² = −y(x + y)
dy/dx = −y(x + y)/x²
This is a homogeneous DE. Put y = vx:
v + x dv/dx = −vx(x + vx)/x² = −v(1 + v)
x dv/dx = −v − v² − v = −2v − v²
Separating: dv/(v² + 2v) = −dx/x
dv/(v(v + 2)) = −dx/x
Partial fractions: 1/(v(v+2)) = (1/2)(1/v − 1/(v+2))
(1/2)[log|v| − log|v + 2|] = −log|x| + C₁
log|v/(v + 2)| = −2log|x| + C₂
|v/(v + 2)| = C/x²
Substituting v = y/x:
(y/x)/(y/x + 2) = C/x²
y/(y + 2x) = C/x²
x²y = C(y + 2x)
Applying y = 1, x = 1:
1 = C(1 + 2) → C = 1/3
Particular solution: x²y = (y + 2x)/3
or: 3x²y = y + 2x → y(3x² − 1) = 2x → y = 2x/(3x² − 1)
Q13Find the particular solution of the differential equation [x sin²(y/x) − y]dx + x dy = 0, given that y = π/4 when x = 1.
x dy = [y − x sin²(y/x)]dx
dy/dx = y/x − sin²(y/x)
Homogeneous DE. Put y = vx:
v + x dv/dx = v − sin²v
x dv/dx = −sin²v
dv/sin²v = −dx/x
cosec²v dv = −dx/x
Integrating: −cot v = −log|x| + C₁
cot v = log|x| + C
Substituting v = y/x:
cot(y/x) = log|x| + C
Applying y = π/4 when x = 1:
cot(π/4) = log 1 + C
1 = 0 + C → C = 1
Particular solution: cot(y/x) = log|x| + 1
Q14Find the particular solution of the differential equation dy/dx − y/x + cosec(y/x) = 0, given that y = 0 when x = 1.
dy/dx = y/x − cosec(y/x)
Homogeneous DE. Put y = vx:
v + x dv/dx = v − cosec v
x dv/dx = −cosec v
sin v dv = −dx/x
Integrating: −cos v = −log|x| + C₁
cos v = log|x| − C₁
Substituting v = y/x:
cos(y/x) = log|x| + C
Applying y = 0 when x = 1:
cos 0 = log 1 + C
1 = 0 + C → C = 1
Particular solution: cos(y/x) = log|x| + 1
or: cos(y/x) = log x + 1 (for x > 0)
Q15Find the particular solution of the differential equation 2xy + y² − 2x²(dy/dx) = 0, given that y = 2 when x = 1.
2x²(dy/dx) = 2xy + y²
dy/dx = y/x + y²/(2x²)
Homogeneous DE. Put y = vx:
v + x dv/dx = v + v²/2
x dv/dx = v²/2
2 dv/v² = dx/x
−2/v = log|x| + C
Substituting v = y/x:
−2x/y = log|x| + C
Applying y = 2 when x = 1:
−2/2 = log 1 + C
−1 = C
Particular solution: −2x/y = log|x| − 1
or: 2x/y = 1 − log x → y = 2x/(1 − log x)
Q16Find the general solution of the differential equation: (1 + y²) + (x − e^(tan⁻¹y))(dy/dx) = 0
(1 + y²)dx/dy + x − e^(tan⁻¹y) = 0
dx/dy + x/(1 + y²) = e^(tan⁻¹y)/(1 + y²)
Linear DE in x: P = 1/(1 + y²), Q = e^(tan⁻¹y)/(1 + y²)
IF: e^(∫dy/(1+y²)) = e^(tan⁻¹y)
Solution: x · e^(tan⁻¹y) = ∫[e^(tan⁻¹y)/(1 + y²)] · e^(tan⁻¹y) dy + C
= ∫e^(2tan⁻¹y)/(1 + y²) dy + C
Let t = tan⁻¹y, dt = dy/(1 + y²):
= ∫e^(2t) dt = e^(2t)/2 + C = e^(2tan⁻¹y)/2 + C
General solution: xe^(tan⁻¹y) = (1/2)e^(2tan⁻¹y) + C
or: x = (1/2)e^(tan⁻¹y) + Ce^(−tan⁻¹y)
Q17Find the general solution of the differential equation: y dx + (x − y²) dy = 0
y dx/dy + x = y²
dx/dy + x/y = y
Linear DE in x: P = 1/y, Q = y
IF: e^(∫dy/y) = e^(log y) = y
Solution: xy = ∫y · y dy + C = y³/3 + C
General solution: xy − y³/3 = C
or: x = y²/3 + C/y
Q18Find the general solution of the differential equation: (x + 3y²)(dy/dx) = y (y > 0)
dx/dy = (x + 3y²)/y = x/y + 3y
dx/dy − x/y = 3y
Linear DE in x: P = −1/y, Q = 3y
IF: e^(∫−dy/y) = e^(−log y) = 1/y
Solution: x/y = ∫3y · (1/y) dy + C = 3y + C
General solution: x = 3y² + Cy
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Ask AI TutorPrevious Year Board Exam Questions
Find the general solution of the differential equation: (1 + x²)dy/dx + 2xy = 1/(1 + x²).
dy/dx + 2xy/(1 + x²) = 1/(1 + x²)²
Linear DE. P = 2x/(1 + x²), Q = 1/(1 + x²)²
IF: e^(∫2x/(1+x²) dx) = e^(log(1+x²)) = 1 + x²
y(1 + x²) = ∫(1 + x²) · 1/(1 + x²)² dx + C = ∫1/(1 + x²) dx + C = tan⁻¹x + C
General solution: y(1 + x²) = tan⁻¹x + C
Find the particular solution of dy/dx + y cot x = 2cos x, given y(π/2) = 0.
Linear DE: dy/dx + y cot x = 2cos x. P = cot x, Q = 2cos x.
IF: e^(∫cot x dx) = e^(log sin x) = sin x
Solution: y sin x = ∫2cos x · sin x dx + C = ∫sin 2x dx + C = -cos 2x/2 + C
Applying y = 0 at x = π/2:
0 · sin(π/2) = -cos(2 · π/2)/2 + C
0 = -cos(π)/2 + C = -(-1)/2 + C = 1/2 + C
C = -1/2
y sin x = -cos 2x/2 - 1/2 = -(cos 2x + 1)/2
Using cos 2x + 1 = 2cos²x:
y sin x = -2cos²x/2 = -cos²x
Particular solution: y = -cos²x/sin x = -cos x · cot x
or equivalently: y sin x + cos²x = 0
Solve the differential equation: dy/dx + y/x = x² (x ≠ 0).
Linear DE. P = 1/x, Q = x²
IF: e^(∫dx/x) = e^(log x) = x
yx = ∫x² · x dx + C = x⁴/4 + C
General solution: y = x³/4 + C/x
Find the order and degree (if defined) of the differential equation: (d²y/dx²)² + cos(dy/dx) = 0.
Order: Highest order derivative is d²y/dx². Order = 2.
Degree: cos(dy/dx) is not polynomial in dy/dx. Degree is not defined.
Find the general solution: (1 + y²)(1 + log x)dx + x dy = 0.
Separating: (1 + log x)/x dx = −dy/(1 + y²)
Integrating LHS: let t = 1 + log x, dt = dx/x
∫t dt = t²/2 = (1 + log x)²/2
Integrating RHS: −tan⁻¹y
(1 + log x)²/2 + tan⁻¹y = C
Find the particular solution of dy/dx − 3y cot x = sin 2x, given y = 2 when x = π/2.
Linear DE. P = −3 cot x, Q = sin 2x = 2 sin x cos x
IF: e^(∫−3 cot x dx) = e^(−3 log sin x) = cosec³x
y cosec³x = ∫2 sin x cos x · cosec³x dx + C = ∫2 cos x/sin²x dx + C
= ∫2 cosec x cot x dx + C = −2 cosec x + C
At x = π/2, y = 2: 2 · 1 = −2 · 1 + C → C = 4
Particular solution: y = −2sin²x + 4sin³x = 4sin³x − 2sin²x
Find the integrating factor of the differential equation: x(dy/dx) − y = 2x².
Rewriting: dy/dx − y/x = 2x
P = −1/x
IF = e^(∫−dx/x) = e^(−log x) = 1/x
Find the general solution of the differential equation: eˣ tan y dx + (1 − eˣ) sec²y dy = 0.
Separating: sec²y/tan y dy = eˣ/(eˣ − 1) dx
Integrating: log|tan y| = log|eˣ − 1| + log C
General solution: tan y = C(eˣ − 1)
Find the general solution of the differential equation: dy/dx = 1 + x + y + xy.
dy/dx = (1 + x)(1 + y)
Separating: dy/(1 + y) = (1 + x) dx
Integrating: log|1 + y| = x + x²/2 + C
General solution: log|1 + y| = x + x²/2 + C
or: 1 + y = Ae^(x + x²/2)
Solve: (x + y)(dx − dy) = dx + dy, given y = −1 when x = 0.
(x + y)dx − (x + y)dy = dx + dy
(x + y − 1)dx = (x + y + 1)dy
dy/dx = (x + y − 1)/(x + y + 1)
Let v = x + y, dv/dx = 1 + dy/dx:
dv/dx − 1 = (v − 1)/(v + 1)
dv/dx = (v − 1)/(v + 1) + 1 = (v − 1 + v + 1)/(v + 1) = 2v/(v + 1)
Separating: (v + 1)/(2v) dv = dx → (1/2 + 1/(2v)) dv = dx
Integrating: v/2 + (1/2)log|v| = x + C
(x + y)/2 + (1/2)log|x + y| = x + C
At x = 0, y = −1: −1/2 + (1/2)log 1 = 0 + C → C = −1/2
(x + y) + log|x + y| = 2x − 1
or: y − x + 1 + log|x + y| = 0
Solve the differential equation: cos(dy/dx) = a, (a ∈ R); y = 2 when x = 0.
dy/dx = cos⁻¹a (constant)
Integrating: y = x cos⁻¹a + C
At x = 0, y = 2: C = 2
Particular solution: y = x cos⁻¹a + 2
Find the general solution of the differential equation: (x² + 1)dy/dx + 2xy = √(x² + 4).
dy/dx + 2xy/(x² + 1) = √(x² + 4)/(x² + 1)
Linear DE. P = 2x/(x² + 1).
IF: e^(∫2x/(x²+1) dx) = e^(log(x²+1)) = x² + 1
y(x² + 1) = ∫√(x² + 4) dx + C
Using ∫√(x² + a²) dx = (x/2)√(x² + a²) + (a²/2)log|x + √(x² + a²)| + C:
y(x² + 1) = (x/2)√(x² + 4) + 2log|x + √(x² + 4)| + C
Solve: dy/dx + 2y tan x = sin x, given y = 0 when x = π/3.
Linear DE. P = 2 tan x, Q = sin x.
IF: e^(∫2 tan x dx) = e^(−2 log cos x) = sec²x
y sec²x = ∫sin x sec²x dx + C = ∫sec x tan x dx + C = sec x + C
At x = π/3, y = 0: 0 = sec(π/3) + C = 2 + C → C = −2
y sec²x = sec x − 2 → y = cos x − 2cos²x
Find the general solution: x dy − (y + 2x²) dx = 0.
x dy/dx = y + 2x²
dy/dx − y/x = 2x
Linear DE. P = −1/x, Q = 2x.
IF: e^(∫−dx/x) = 1/x
y/x = ∫2x · (1/x) dx + C = 2x + C
General solution: y = 2x² + Cx
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The order of the differential equation dy/dx + y = eˣ is 1.
Reason (R): The order of a DE is the order of the highest derivative present.
Answer: (A)
Both true. dy/dx is the highest derivative (first order). The reason correctly defines order and explains the assertion.
AR2
Assertion (A): The degree of d²y/dx² + sin(dy/dx) = 0 is not defined.
Reason (R): A DE has a defined degree only when it is a polynomial equation in its derivatives.
Answer: (A)
Both true. sin(dy/dx) makes the equation non-polynomial in derivatives. The reason correctly explains why degree is not defined.
AR3
Assertion (A): The general solution of dy/dx = eˣ is y = eˣ + C.
Reason (R): The general solution of a first-order DE contains exactly one arbitrary constant.
Answer: (A)
Both true. Integrating dy/dx = eˣ gives y = eˣ + C (one constant). The reason correctly states that a first-order DE has one arbitrary constant.
AR4
Assertion (A): y = eˣ is the particular solution of dy/dx − y = 0 with y(0) = 1.
Reason (R): A particular solution satisfies both the DE and the initial condition.
Answer: (A)
y = eˣ → y' = eˣ = y ✓. y(0) = e⁰ = 1 ✓. Both assertions true; the reason correctly defines a particular solution.
AR5
Assertion (A): The integrating factor of dy/dx + 2y = eˣ is e²ˣ.
Reason (R): For dy/dx + Py = Q, the integrating factor is e^(∫Q dx).
Answer: (C)
Assertion is true (IF = e^(∫2 dx) = e²ˣ). But the reason is false - the IF is e^(∫P dx), not e^(∫Q dx). Assertion true, reason false.
Case Study Questions
The population of a city was 20 lakhs in 2020. It has been observed that the rate of growth of population is proportional to the population at any time. If the population doubled in 25 years, answer the following:
(a) Write the differential equation representing the above situation. (1 mark)
dP/dt = kP, where P is population and k is growth constant.
(b) Find the general solution of the above DE. (1 mark)
Separating: dP/P = k dt. Integrating: log P = kt + C. So P = P₀eᵏᵗ where P₀ is initial population.
P = 20eᵏᵗ (in lakhs)
(c) Find the value of k (in terms of log 2). (1 mark)
At t = 25, P = 40: 40 = 20e²⁵ᵏ → e²⁵ᵏ = 2 → 25k = log 2 → k = (log 2)/25
(d) In how many years will the population become 5 times? (1 mark)
5 × 20 = 20eᵏᵗ → eᵏᵗ = 5 → kt = log 5
t = log 5/k = 25 log 5/log 2 = 25 × log 5/log 2
t = 25 log₂5 ≈ 58.05 years
A thermometer reading 80°C is placed in a room at constant temperature of 25°C. The thermometer reads 60°C after 5 minutes. Newton's law of cooling states that the rate of cooling is proportional to the difference between the temperature of the body and the surrounding temperature.
(a) Formulate the differential equation for the above scenario. (1 mark)
Let T be temperature at time t. Surrounding temperature = 25°C.
dT/dt = −k(T − 25), where k > 0.
(b) Find the general solution. (1 mark)
dT/(T − 25) = −k dt
log|T − 25| = −kt + C₁
T − 25 = Ae⁻ᵏᵗ
At t = 0, T = 80: A = 55
T = 25 + 55e⁻ᵏᵗ
(c) Find the value of k. (1 mark)
At t = 5, T = 60:
60 = 25 + 55e⁻⁵ᵏ → 35 = 55e⁻⁵ᵏ → e⁻⁵ᵏ = 7/11
k = (1/5)log(11/7)
(d) What will the temperature be after 10 minutes? (1 mark)
T = 25 + 55e⁻¹⁰ᵏ = 25 + 55(e⁻⁵ᵏ)² = 25 + 55(7/11)² = 25 + 55 × 49/121
= 25 + 2695/121 = 25 + 22.27 ≈ 47.27°C
MCQ Practice
1. The order of the differential equation 2x²(d²y/dx²) − 3(dy/dx) + y = 0 is:
(A) 2
(B) 1
(C) 0
(D) Not defined
Answer: (A) - Highest order derivative is d²y/dx² (second order). Order = 2.
2. The degree of (dy/dx)² + (d²y/dx²)² = sin²y is:
(A) 1
(B) 2
(C) 3
(D) Not defined
Answer: (B) - The term sin²y is a function of the dependent variable y, NOT of any derivative. The derivatives (dy/dx)² and (d²y/dx²)² are polynomial terms. The highest order derivative is d²y/dx² and its power is 2. Since the DE is polynomial in its derivatives, degree = 2.
3. The integrating factor of dy/dx + y/x = x³ is:
(A) log x
(B) 1/x
(C) x
(D) −x
Answer: (C) - P = 1/x. IF = e^(∫dx/x) = e^(log x) = x.
4. The solution of dy/dx = eˣ⁻ʸ + x²e⁻ʸ is:
(A) eʸ = eˣ + x³/3 + C
(B) eˣ + eʸ = C
(C) eˣ − eʸ = C
(D) y = eˣ + x² + C
Answer: (A) - dy/dx = e⁻ʸ(eˣ + x²). Separating: eʸ dy = (eˣ + x²) dx. Integrating: eʸ = eˣ + x³/3 + C.
5. The number of arbitrary constants in the general solution of a third-order DE is:
(A) 1
(B) 2
(C) 3
(D) 0
Answer: (C) - General solution of an nth order DE has n arbitrary constants. For 3rd order: 3 constants.
6. The general solution of dy/dx = eˣ⁺ʸ is:
(A) eˣ + e⁻ʸ = C
(B) eˣ + eʸ = C
(C) e⁻ˣ + eʸ = C
(D) e⁻ˣ + e⁻ʸ = C
Answer: (A) - eˣ⁺ʸ = eˣ·eʸ. Separating: e⁻ʸ dy = eˣ dx. Integrating: −e⁻ʸ = eˣ + C₁, so eˣ + e⁻ʸ = C.
7. The integrating factor of (1+y²)dx − (tan⁻¹y − x)dy = 0 is:
(A) e^(tan⁻¹y)
(B) tan⁻¹y
(C) 1/(1+y²)
(D) e^y
Answer: (A) - Rewrite: dx/dy + x/(1+y²) = tan⁻¹y/(1+y²). P = 1/(1+y²). IF = e^(∫dy/(1+y²)) = e^(tan⁻¹y).
8. Family of curves y = Ax + A³ represents the solution of the DE of degree:
(A) 1
(B) 2
(C) 3
(D) Not defined
Answer: (C) - y = Ax + A³. y' = A, so A = y'. Substituting: y = xy' + (y')³. This gives (y')³ + xy' − y = 0. Degree of y' is 3.
9. Which is NOT a method for solving first-order first-degree DEs?
(A) Variable separable
(B) Homogeneous
(C) Variation of parameters
(D) Linear
Answer: (C) - Variation of parameters is used for higher-order DEs. Variable separable, homogeneous, and linear are methods for first-order first-degree DEs.
10. The solution of x dy − y dx = 0 represents:
(A) A circle
(B) A parabola
(C) Straight lines through origin
(D) A rectangular hyperbola
Answer: (C) - x dy = y dx → dy/y = dx/x → log y = log x + C₁ → y = Cx. These are straight lines passing through the origin.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the order and degree of dy/dx + sin(dy/dx) = 0.
Order = 1. Degree is not defined (sin(dy/dx) is not polynomial in derivatives).
Write the integrating factor of dy/dx + (2x/(1 + x²))y = 1/(1 + x²)².
P = 2x/(1 + x²). IF = e^(∫2x/(1+x²) dx) = e^(log(1+x²)) = 1 + x².
How many arbitrary constants are in the particular solution of a 3rd order DE?
0 (zero). A particular solution has no arbitrary constants.
Write the degree of (d³y/dx³)² + (d²y/dx²)³ + dy/dx + 4 = 0.
Highest order derivative is d³y/dx³. Its power is 2. Degree = 2.
Find the integrating factor of x(dy/dx) − y = 2x².
dy/dx − y/x = 2x. P = −1/x. IF = e^(∫−dx/x) = e^(−log x) = 1/x.
2-3 Mark Questions
Find the general solution of dy/dx + y/x = x² (x ≠ 0).
Linear DE. P = 1/x, Q = x². IF = e^(∫dx/x) = x.
yx = ∫x³ dx + C = x⁴/4 + C.
y = x³/4 + C/x
Solve: dy/dx = (1 + y²)/(1 + x²).
Separating: dy/(1 + y²) = dx/(1 + x²)
tan⁻¹y = tan⁻¹x + C.
tan⁻¹y − tan⁻¹x = C or y − x = C(1 + xy)
Form the DE of all circles touching the x-axis at the origin.
Circle: x² + (y − a)² = a² → x² + y² = 2ay.
Differentiating: 2x + 2yy' = 2ay' → a = (x + yy')/y'.
Substituting: x² + y² = 2y(x + yy')/y' → y'(x² + y²) = 2xy + 2y²y'
x²y' − 2xy + y²y' = 0 or (x² − y²)dy/dx = 2xy
Find the general solution of (eˣ + e⁻ˣ)dy − (eˣ − e⁻ˣ)dx = 0.
dy = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ) dx = d/dx[log(eˣ + e⁻ˣ)] dx
y = log(eˣ + e⁻ˣ) + C
3-4 Mark Questions
Find the particular solution of dy/dx + 2y tan x = sin x, given y = 0 when x = π/3.
Linear DE. IF = e^(∫2tan x dx) = sec²x.
y sec²x = ∫sin x sec²x dx + C = sec x + C.
At x = π/3, y = 0: 0 = 2 + C → C = −2.
y = cos x − 2cos²x
Solve: (1 + x²)dy/dx + 2xy = 1/(1 + x²), given y = 0 when x = 1.
IF = 1 + x². y(1 + x²) = tan⁻¹x + C.
At x = 1, y = 0: 0 = π/4 + C → C = −π/4.
y = (tan⁻¹x − π/4)/(1 + x²)
Find the general solution: x(dy/dx) + y − x + xy cot x = 0.
dy/dx + y(1/x + cot x) = 1. IF = x sin x.
xy sin x = ∫x sin x dx + C = sin x − x cos x + C.
xy sin x = sin x − x cos x + C
Key Concepts
Important Formulas
Study Plan - Differential Equations
Order, Degree & Basics
Read NCERT 9.1-9.2, practice order/degree for 15+ equations, verify solutions
Formation of DEs
Eliminate constants from family of curves. Practice Ex 9.3 completely
Variable Separable Method
Master separation technique, solve Q1-Q10 of Ex 9.4 and application problems
Linear Differential Equations
Master IF formula, solve all of Ex 9.5. Focus on finding IF correctly
Miscellaneous & PYQs
Solve Miscellaneous Exercise, attempt past 5 years CBSE questions on DEs
Revision & Mock Practice
Solve mixed problems, practice MCQ and case study formats, time yourself
Common Mistakes to Avoid
- ✗Confusing order and degree - degree is the power of the HIGHEST order derivative only, not any derivative
- ✗Forgetting that degree is NOT defined when the DE is not polynomial in derivatives (e.g., sin(y'), log(y''))
- ✗Not checking if the DE is homogeneous before applying the substitution y = vx
- ✗Forgetting to change dy/dx when substituting y = vx (must use dy/dx = v + x dv/dx)
- ✗Making errors in finding the integrating factor - always simplify P first
- ✗Forgetting to divide by the coefficient of dy/dx to get the standard form before finding IF
- ✗Not writing the constant of integration in the general solution
- ✗Errors in partial fraction decomposition when separating variables
- ✗Confusing the number of constants in general vs particular solutions
Tips for Board Exam
- ✓Always identify the type of DE first: variable separable, homogeneous, or linear
- ✓For linear DEs, write in standard form (dy/dx + Py = Q) before finding IF
- ✓Remember: IF = e^(∫P dx) - integrate P with respect to x (or y if dx/dy form)
- ✓For order/degree MCQs: degree is defined ONLY for polynomial DEs in derivatives
- ✓Practice application problems (growth/decay, Newton's cooling) - they appear as 5-mark or case study questions
- ✓In formation questions, count arbitrary constants = number of times you differentiate
- ✓Show all steps in long-answer questions - marks are given for intermediate steps
- ✓Common IF results to remember: e^(log x) = x, e^(−log x) = 1/x, e^(log sin x) = sin x
- ✓For homogeneous DEs, always check if f(tx, ty) = t⁰·f(x,y) before using y = vx substitution
Practice Differential Equations with AI Tutor
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Start Differential Equations - Choose Your ModeAbout Differential Equations - Class 12 Mathematics
Differential Equations is Chapter 9 of NCERT Class 12 Mathematics (Part II). The chapter introduces equations containing derivatives of functions and methods to solve them. Students learn to classify DEs by order (highest derivative) and degree (power of highest-order derivative when polynomial). The chapter covers four solution methods: variable separable (separating x and y terms), homogeneous (using y = vx substitution), and linear (using integrating factor e^(∫P dx)). Formation of DEs from a general solution is also covered - the number of differentiations equals the number of arbitrary constants. Applications include population growth models, radioactive decay, and Newton's law of cooling. This chapter carries significant weight in the CBSE board exam as part of the 35-mark Calculus unit, with questions typically appearing as 1-mark MCQs on order/degree and 3-5 mark problems on solving linear or separable DEs.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 9 Differential Equations?
Chapter 9 has 5 exercises (Exercise 9.1 to 9.5) plus a Miscellaneous Exercise. Exercise 9.1 has 12 questions on order and degree, 9.2 has 12 on verification, 9.3 has 12 on formation, 9.4 has 23 on variable separable, and 9.5 has 17 on linear DEs.
What is the weightage of Differential Equations in CBSE Class 12 board exam?
Differential Equations is part of the Calculus unit which carries 35 marks in total. Typically, 1 MCQ (1 mark) on order/degree + 1 short answer (3 marks) on solving a DE + 1 long answer (5 marks) on linear DE or application problem appear from this chapter.
What are the important topics in Chapter 9 for CBSE board exam?
The most important topics are: (1) Linear differential equations (dy/dx + Py = Q with integrating factor), (2) Variable separable method, (3) Order and degree determination, and (4) Formation of DEs. Case study questions on growth/decay are also trending.
What is the difference between order and degree of a differential equation?
Order is the order of the highest derivative present in the DE (always defined and always a positive integer). Degree is the power (exponent) of the highest order derivative, but ONLY when the DE is a polynomial in its derivatives. If the DE involves sin, cos, log, etc. of a derivative, the degree is not defined.
What is an integrating factor and how do you find it?
An integrating factor (IF) is a function that, when multiplied with a linear DE, makes the left side an exact derivative. For dy/dx + Py = Q, the IF = e^(∫P dx). The solution is then y·IF = ∫(Q·IF)dx + C. Common IFs include x, 1/x, eˣ, sin x, etc.