Class 12 : Maths (English) – Chapter 9: Differential Equations

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On June 25, 2025

EXPLANATION & SUMMARY

🔶 1. Introduction
A differential equation expresses a relationship between a function and its derivatives.
It models many natural phenomena such as population growth, radioactive decay, heat flow, and motion.
A general form:
➡️ F(x, y, dy/dx, d²y/dx², …) = 0
When only first derivative appears, it is a first-order differential equation.
If the highest derivative is of order n, it is an nth-order differential equation.

🔶 2. Order and Degree
Order = highest order of derivative.
Degree = power of highest derivative after the equation is free from radicals/fractions.
🧠 Example:
(1 + x²) (d²y/dx²)² + x (dy/dx) = sin x
➡️ Order = 2, Degree = 2

🔶 3. General and Particular Solutions
General solution: contains as many arbitrary constants as the order.
Particular solution: obtained by assigning specific values to constants using given conditions.
✏️ Example:
dy/dx = 3x² ⇒ ∫dy = ∫3x² dx ⇒ y = x³ + C (general)
If y = 2 when x = 1 ⇒ 2 = 1 + C ⇒ C = 1 ⇒ y = x³ + 1 (particular)

🔶 4. Formation of Differential Equations
Goal: eliminate arbitrary constants from a family of curves.
🧭 Steps
1️⃣ Start with the equation containing constants (e.g., y = mx + c).
2️⃣ Differentiate enough times to remove all constants.
3️⃣ Combine to form a relation between x, y, and derivatives.
💡 Example: Family y = mx + c
Differentiate: dy/dx = m
Eliminate m, c ⇒ differential equation: d²y/dx² = 0

🔶 5. Methods of Solving First-Order Differential Equations
🟢 (a) Variable Separable Form
dy/dx = f(x) g(y)
Separate: dy/g(y) = f(x) dx
Integrate both sides.
✔️ Example: dy/dx = x/y
⇒ y dy = x dx ⇒ (y²)/2 = (x²)/2 + C

🟡 (b) Homogeneous Equations
A first-order equation dy/dx = F(x, y) is homogeneous if
F(tx, ty) = F(x, y)
Substitute y = v x, dy/dx = v + x dv/dx,
reduces to variable separable in v and x.

🔵 (c) Linear Differential Equation
Form: dy/dx + P(x) y = Q(x)
Integrating factor (I.F.) = e∫P(x) dx
Solution: y × I.F. = ∫[Q(x) × I.F.] dx + C
💡 Example: dy/dx + y = eˣ
I.F. = e∫1 dx = eˣ
y eˣ = ∫eˣ eˣ dx + C = ∫e²ˣ dx + C = (e²ˣ)/2 + C
⇒ y = (½)eˣ + Ce⁻ˣ

🔴 (d) Exact Differential Equation
Equation M(x, y) dx + N(x, y) dy = 0 is exact if
∂M/∂y = ∂N/∂x.
Solution from potential function φ(x, y):
∂φ/∂x = M, ∂φ/∂y = N, φ(x, y) = C.

🟣 (e) Linear in x
dx/dy + P(y) x = Q(y)
Analogous solution using integrating factor e∫P(y) dy.

🔶 6. Applications
Exponential growth/decay: dy/dt = k y
→ y = Aeᵏᵗ
Cooling (Newton’s Law): dT/dt = –k(T – Tₛ)
Simple RC-circuit current decay
Population dynamics, radioactive decay

🔶 7. Common Mistakes
🚫 Forgetting to multiply by I.F.
🚫 Not separating variables correctly
🚫 Ignoring constant of integration C

🧾 Summary (~300 words)
🔹 Differential equation: relation between variables and derivatives.
🔹 Order = highest derivative order; Degree = power of that derivative.
🔹 General solution: contains arbitrary constants; Particular uses conditions.
🔹 Formation: eliminate constants by differentiation.
🔹 Solving techniques:
Variable separable → integrate both sides.
Homogeneous → substitute y = v x.
Linear → use I.F. = e∫P(x) dx.
Exact → verify ∂M/∂y = ∂N/∂x.
🔹 Applications: growth/decay, cooling, circuits, motion.
🔹 Key step: always include constant C after integration.

📝 Quick Recap
🧠 Order ↔ derivative rank
📐 Degree ↔ power of highest derivative
✳️ Solve separable: ∫dy/g(y) = ∫f(x) dx
⚙️ Linear form: dy/dx + P y = Q
💡 I.F. = e∫P dx
📊 Growth law: y = Aeᵏᵗ

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QUESTIONS FROM TEXTBOOK

Exercise 9.1

🔵 Question 1:
(d⁴y/dx⁴) + sin(y″) = 0
🟢 Answer:
🔹 Highest order derivative = d⁴y/dx⁴
➡️ Order = 4
🔹 sin(y″) is not a polynomial in derivatives
➡️ Degree = Not defined
✔️ Final: Order = 4, Degree = Not defined

🔵 Question 2:
y′ + 5y = 0
🟢 Answer:
🔹 Highest order derivative = y′
➡️ Order = 1
🔹 Equation is polynomial in y′ with power 1
➡️ Degree = 1
✔️ Final: Order = 1, Degree = 1

🔵 Question 3:
(ds/dt)⁴ + 3s (d²s/dt²) = 0
🟢 Answer:
🔹 Highest order derivative = d²s/dt²
➡️ Order = 2
🔹 Equation is polynomial in derivatives, power of highest derivative = 1
➡️ Degree = 1
✔️ Final: Order = 2, Degree = 1

🔵 Question 4:
(d²y/dx²)² + cos(dy/dx) = 0
🟢 Answer:
🔹 Highest order derivative = d²y/dx²
➡️ Order = 2
🔹 cos(dy/dx) is not a polynomial
➡️ Degree = Not defined
✔️ Final: Order = 2, Degree = Not defined

🔵 Question 5:
d²y/dx² = cos3x + sin3x
🟢 Answer:
🔹 Highest order derivative = d²y/dx²
➡️ Order = 2
🔹 Equation is polynomial in derivatives, power = 1
➡️ Degree = 1
✔️ Final: Order = 2, Degree = 1

🔵 Question 6:
(y″)² + (y‴)³ + (y′)⁴ + y⁵ = 0
🟢 Answer:
🔹 Highest order derivative = y‴
➡️ Order = 3
🔹 Power of highest order derivative = 3
➡️ Degree = 3
✔️ Final: Order = 3, Degree = 3

🔵 Question 7:
(y′)² + y = 0
🟢 Answer:
🔹 Highest order derivative = y′
➡️ Order = 1
🔹 Equation is polynomial in y′ with power 2
➡️ Degree = 2
✔️ Final: Order = 1, Degree = 2

🔵 Question 8:
(d²y/dx²) + (dy/dx) + y = 0
🟢 Answer:
🔹 Highest order derivative = d²y/dx²
➡️ Order = 2
🔹 Equation is polynomial in derivatives, power = 1
➡️ Degree = 1
✔️ Final: Order = 2, Degree = 1

🔵 Question 9:
(d³y/dx³) + (d²y/dx²)² = 0
🟢 Answer:
🔹 Highest order derivative = d³y/dx³
➡️ Order = 3
🔹 Equation is polynomial in derivatives, power of highest derivative = 1
➡️ Degree = 1
✔️ Final: Order = 3, Degree = 1

🔵 Question 10:
sin(dy/dx) + y = 0
🟢 Answer:
🔹 Highest order derivative = dy/dx
➡️ Order = 1
🔹 sin(dy/dx) is non-polynomial
➡️ Degree = Not defined
✔️ Final: Order = 1, Degree = Not defined

🔵 Question 11:
(d²y/dx²) + y (dy/dx) + y² = 0
🟢 Answer:
🔹 Highest order derivative = d²y/dx²
➡️ Order = 2
🔹 Equation is polynomial in derivatives, power = 1
➡️ Degree = 1
✔️ Final: Order = 2, Degree = 1

🔵 Question 12:
(d²y/dx²) + (dy/dx)³ + y⁵ = 0
🟢 Answer:
🔹 Highest order derivative = d²y/dx²
➡️ Order = 2
🔹 Equation is polynomial in derivatives, power of highest derivative = 1
➡️ Degree = 1
✔️ Final: Order = 2, Degree = 1

Exercise 9.2

🔵 Question 1: y = eˣ + 1 : y″ − y′ = 0
🟢 Answer:
🔹 y′ = eˣ
🔹 y″ = eˣ
➡️ y″ − y′ = eˣ − eˣ = 0
✔️ Final: Verified that y = eˣ + 1 satisfies y″ − y′ = 0

🔵 Question 2: y = x² + 2x + C : y′ − 2x − 2 = 0
🟢 Answer:
🔹 y′ = 2x + 2
➡️ y′ − 2x − 2 = (2x + 2) − 2x − 2 = 0
✔️ Final: Verified

🔵 Question 3: y = cos x + C : y′ + sin x = 0
🟢 Answer:
🔹 y′ = −sin x
➡️ y′ + sin x = (−sin x) + sin x = 0
✔️ Final: Verified

🔵 Question 4: y = √(1 + x²) : y′ = (x y)/(1 + x²)
🟢 Answer:
🔹 y = (1 + x²)^(1/2)
🔹 y′ = (1/2)(1 + x²)^(−1/2)·2x = x/√(1 + x²)
➡️ (x y)/(1 + x²) = x·√(1 + x²)/(1 + x²) = x/√(1 + x²) = y′
✔️ Final: Verified

🔵 Question 5: y = A x : x y′ = y (x ≠ 0)
🟢 Answer:
🔹 y′ = A
➡️ x y′ = xA = Ax = y
✔️ Final: Verified (for x ≠ 0)

🔵 Question 6: y = x sin x : x y′ = y + x √(x² − y²) (x ≠ 0 and x > y or x < −y)
🟢 Answer:
🔹 y′ = sin x + x cos x
🔹 LHS = x y′ = x sin x + x² cos x
🔹 y² = x² sin²x ⇒ x² − y² = x² cos²x ⇒ √(x² − y²) = |x cos x|
🔹 Given condition (x > y or x < −y) ⇒ x cos x ≥ 0 ⇒ |x cos x| = x cos x
🔹 RHS = y + x √(x² − y²) = x sin x + x·(x cos x) = x sin x + x² cos x
➡️ LHS = RHS
✔️ Final: Verified under the stated condition

🔵 Question 7: xy = log y + C : y′ = y²/(1 − xy) (xy ≠ 1)
🟢 Answer:
🔹 Differentiate implicitly: d/dx(xy − log y) = 0
🔹 x·y′ + y − (1/y)·y′ = 0
🔹 Group y′ terms: y′(x − 1/y) + y = 0
🔹 Multiply by y: y′(xy − 1) + y² = 0
➡️ y′ = y²/(1 − xy) (valid when xy ≠ 1)
✔️ Final: Verified

🔵 Question 8: y − cos y = x : (1 + sin y)·y′ = 1
🟢 Answer:
🔹 Differentiate: d/dx(y − cos y − x) = 0
🔹 y′ + (sin y)·y′ − 1 = 0
🔹 Factor y′: y′(1 + sin y) = 1
➡️ (1 + sin y)·y′ = 1
✔️ Final: Verified

🔵 Question 9: x + y = tan⁻¹y : y²·y′ + y² + 1 = 0
🟢 Answer:
🔹 Differentiate: 1 + y′ = (1/(1 + y²))·y′
🔹 Multiply both sides by (1 + y²): (1 + y²)(1 + y′) = y′
🔹 Expand: 1 + y² + y′ + y²y′ = y′
🔹 Cancel y′: 1 + y² + y²y′ = 0
➡️ y²·y′ + y² + 1 = 0
✔️ Final: Verified

🔵 Question 10: y = √(a² − x²), x ∈ (−a, a) : x + y(dy/dx) = 0 (y ≠ 0)
🟢 Answer:
🔹 y = (a² − x²)^(1/2)
🔹 y′ = (1/2)(a² − x²)^(−1/2)·(−2x) = −x/√(a² − x²) = −x/y
🔹 Compute x + y·y′ = x + y(−x/y) = x − x = 0
➡️ x + y(dy/dx) = 0 (y ≠ 0)
✔️ Final: Verified

🔵 Question 11 (MCQ):
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4
🟢 Answer:
🔹 General solution of an nth-order differential equation contains n arbitrary constants
➡️ For n = 4 → 4 constants
✔️ Final: Option (D) 4

🔵 Question 12 (MCQ):
The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3 (B) 2 (C) 1 (D) 0
🟢 Answer:
🔹 Particular solution contains no arbitrary constants
➡️ For any order → 0 constants
✔️ Final: Option (D) 0

Exercise 9.3

🔵 Question 1: dy/dx = (1 − cos x)/(1 + cos x)
🟢 Answer:
🔹 Use 1 − cos x = 2sin²(x/2), 1 + cos x = 2cos²(x/2) ⇒ RHS = tan²(x/2)
🔹 dy = tan²(x/2) dx = (sec²(x/2) − 1) dx
🔹 Put u = x/2 ⇒ dx = 2 du
🔹 y = 2∫sec²u du − 2∫1 du = 2 tan u − 2u + C
➡️ y = 2 tan(x/2) − x + C
✔️ General solution: y = 2 tan(x/2) − x + C

🔵 Question 2: dy/dx = √(4 − y²) (−2 < y < 2)
🟢 Answer:
🔹 Separate: dy/√(4 − y²) = dx
🔹 ∫ dy/√(4 − y²) = ∫ dx
🔹 arcsin(y/2) = x + C
➡️ y = 2 sin(x + C) (range satisfied)
✔️ General solution: y = 2 sin(x + C)

🔵 Question 3: dy/dx + y = 1 (y ≠ 1)
🟢 Answer:
🔹 Linear ODE: y′ + y = 1
🔹 Integrating factor = e^∫1 dx = e^x
🔹 (y e^x)′ = e^x
🔹 y e^x = e^x + C ⇒ y = 1 + C e^(−x)
✔️ General solution: y = 1 + C e^(−x)

🔵 Question 4: sec²x tan y dx + sec²y tan x dy = 0
🟢 Answer:
🔹 Divide by sec²x sec²y: (tan y/sec²y) dx + (tan x/sec²x) dy = 0
🔹 Use tan/ sec² = sin·cos: (sin y cos y) dx + (sin x cos x) dy = 0
🔹 Separate: (dy)/(sin y cos y) = − (dx)/(sin x cos x)
🔹 ∫ dy/(sin y cos y) = ln|tan y| ; ∫ dx/(sin x cos x) = ln|tan x|
🔹 ln|tan y| = − ln|tan x| + C
➡️ tan x · tan y = C
✔️ General solution: tan x · tan y = C

🔵 Question 5: (e^x + e^(−x)) dy − (e^x − e^(−x)) dx = 0
🟢 Answer:
🔹 dy = [(e^x − e^(−x)) /(e^x + e^(−x))] dx = tanh x · dx
🔹 y = ∫ tanh x dx = ln(cosh x) + C (≡ ln(e^x + e^(−x)) + C)
✔️ General solution: y = ln(cosh x) + C

🔵 Question 6: dy/dx = (1 + x²)(1 + y²)
🟢 Answer:
🔹 Separate: dy/(1 + y²) = (1 + x²) dx
🔹 ∫ dy/(1 + y²) = arctan y ; ∫(1 + x²) dx = x + x³/3
➡️ arctan y = x + x³/3 + C
✔️ General solution: y = tan(x + x³/3 + C)

🔵 Question 7: y log y dx − x dy = 0
🟢 Answer:
🔹 Rearrange: x dy = y log y dx ⇒ (dy/dx) = (y log y)/x
🔹 Separate: dy/(y log y) = dx/x
🔹 ∫ dy/(y log y) = ln|log y| ; ∫ dx/x = ln|x| + C
🔹 ln|log y| = ln|x| + C ⇒ |log y| = K |x|
➡️ Write as ln y = k x (k is any non-zero constant; y > 0)
✔️ General solution: y = e^(k x) (k ≠ 0)

🔵 Question 8: x⁵ (dy/dx) = − y⁵
🟢 Answer:
🔹 dy/dx = − y⁵ / x⁵
🔹 Separate: y⁻⁵ dy = − x⁻⁵ dx
🔹 Integrate: ∫y⁻⁵ dy = ∫ − x⁻⁵ dx
🔹 (−1/4) y⁻⁴ = (1/4) x⁻⁴ + C
🔹 Multiply by (−4): y⁻⁴ = − x⁻⁴ + C₁
➡️ Write compactly: 1/y⁴ + 1/x⁴ = C (C constant)
✔️ General solution: 1/y⁴ + 1/x⁴ = C

🔵 Question 9: dy/dx = sin⁻¹x (i.e., arcsin x)
🟢 Answer:
🔹 Integrate both sides: y = ∫ arcsin x dx + C
🔹 Use standard integral: ∫ arcsin x dx = x·arcsin x + √(1 − x²)
➡️ y = x·sin⁻¹x + √(1 − x²) + C
✔️ General solution: y = x·sin⁻¹x + √(1 − x²) + C

🔵 Question 10: eˣ tan y dx + (1 − eˣ) sec²y dy = 0
🟢 Answer:
🔹 Rearrange to dy/dx: (1 − eˣ) sec²y (dy/dx) = − eˣ tan y
🔹 dy/dx = [eˣ tan y]/[(eˣ − 1) sec²y] = eˣ·(sin y cos y)/(eˣ − 1)
🔹 Separate: dy/(sin y cos y) = eˣ/(eˣ − 1) dx
🔹 Integrate left: ∫ dy/(sin y cos y) = ln|tan y|
🔹 Integrate right (u = eˣ − 1): ∫ eˣ/(eˣ − 1) dx = ln|eˣ − 1| + C
➡️ ln|tan y| = ln|eˣ − 1| + C ⇒ tan y = K (eˣ − 1)
✔️ General solution: tan y = K (eˣ − 1)

🔵 Question 11: (x³ + x² + x + 1) (dy/dx) = 2x² + x ; y = 1 when x = 0
🟢 Answer:
🔹 dy/dx = (2x² + x)/[(x + 1)(x² + 1)]
🔹 Partial fractions: (2x² + x)/[(x + 1)(x² + 1)] = ½·1/(x + 1) + (3/2 x − 1/2)/(x² + 1)
🔹 Integrate:
y = ½ ln|x + 1| + (3/4) ln(x² + 1) − ½ tan⁻¹x + C
🔹 Apply y(0) = 1 ⇒ C = 1
➡️ Particular solution: y = 1 + ½ ln|x + 1| + (3/4) ln(x² + 1) − ½ tan⁻¹x

🔵 Question 12: x(x² − 1) (dy/dx) = 1 ; y = 0 when x = 2
🟢 Answer:
🔹 dy/dx = 1/[x(x² − 1)] = 1/[x(x − 1)(x + 1)]
🔹 Partial fractions: 1/[x(x − 1)(x + 1)] = −1/x + ½·1/(x − 1) + ½·1/(x + 1)
🔹 Integrate:
y = − ln|x| + ½ ln|x − 1| + ½ ln|x + 1| + C
🔹 Apply y(2) = 0 ⇒ C = ln(2/√3)
➡️ Particular solution: y = − ln|x| + ½ ln|x − 1| + ½ ln|x + 1| + ln(2/√3)

🔵 Question 13: cos(dy/dx) = a (a ∈ ℝ); y = 1 when x = 0
🟢 Answer:
🔹 Let dy/dx = m (constant), with cos m = a
🔹 Real solutions exist when |a| ≤ 1; then m = ± cos⁻¹a + 2nπ, n ∈ ℤ
🔹 Integrate: y = m x + C
🔹 Apply y(0) = 1 ⇒ C = 1
➡️ Particular solution family: y = (± cos⁻¹a + 2nπ) x + 1 (|a| ≤ 1)

🔵 Question 14: dy/dx = y tan x + x ; y = 1 when x = 0
🟢 Answer:
🔹 Linear form: y′ − y tan x = x
🔹 Integrating factor IF = e^{∫(− tan x) dx} = e^{ln cos x} = cos x
🔹 Multiply: (y cos x)′ = x cos x
🔹 Integrate RHS: ∫ x cos x dx = x sin x + cos x
🔹 So y cos x = x sin x + cos x + C
🔹 Apply y(0) = 1 ⇒ 1 = 1 + C ⇒ C = 0
➡️ Particular solution: y = x tan x + 1

🔵 Question 15: Find the equation of a curve passing through (0,0) if y′ = eˣ sin x.
🟢 Answer:
🔹 y = ∫ eˣ sin x dx + C
🔹 Use standard integral: ∫ eˣ sin x dx = (eˣ/2)(sin x − cos x)
🔹 y = (eˣ/2)(sin x − cos x) + C
🔹 Apply (0,0): 0 = (1/2)(0 − 1) + C ⇒ C = 1/2
✔️ Final: y = (eˣ/2)(sin x − cos x) + 1/2

🔵 Question 16: For xy (dy/dx) = (x + 2)(y + 2), find the solution through (1, −1).
🟢 Answer:
🔹 dy/dx = [(x + 2)(y + 2)]/(xy) = (1 + 2/x)(1 + 2/y)
🔹 Separate: dy/(1 + 2/y) = (1 + 2/x) dx
🔹 Simplify LHS: dy/((y + 2)/y) = [y/(y + 2)] dy
🔹 Integrate: ∫[y/(y + 2)] dy = ∫(1 + 2/x) dx
🔹 LHS = ∫(1 − 2/(y + 2)) dy = y − 2 ln|y + 2|
🔹 RHS = x + 2 ln|x| + C
🔹 Implicit solution: y − 2 ln|y + 2| = x + 2 ln|x| + C
🔹 Apply (1, −1): −1 − 2 ln1 = 1 + 2 ln1 + C ⇒ C = −2
✔️ Final: y − 2 ln|y + 2| = x + 2 ln|x| − 2

🔵 Question 17: Through (0, −2); condition: (slope of tangent) × y = x.
🟢 Answer:
🔹 Differential equation: y (dy/dx) = x
🔹 Separate: y dy = x dx
🔹 Integrate: (1/2) y² = (1/2) x² + C ⇒ y² − x² = C₁
🔹 Apply (0, −2): 4 − 0 = C₁ ⇒ C₁ = 4
✔️ Final: y² − x² = 4 (choose branch consistent with y < 0 near x = 0)

🔵 Question 18: Slope of tangent at (x, y) is twice the slope of line to (−4, −3); curve passes through (−2, 1).
🟢 Answer:
🔹 Slope of chord: (y + 3)/(x + 4)
🔹 Condition: dy/dx = 2 (y + 3)/(x + 4)
🔹 Linear form: y′ − [2/(x + 4)] y = 6/(x + 4)
🔹 Integrating factor: IF = e^{∫ −2/(x + 4) dx} = (x + 4)^{−2}
🔹 Multiply: d/dx [ y (x + 4)^{−2} ] = 6 (x + 4)^{−3}
🔹 Integrate: y (x + 4)^{−2} = −3 (x + 4)^{−2} + C
🔹 Multiply by (x + 4)²: y = −3 + C (x + 4)²
🔹 Apply (−2, 1): 1 = −3 + C·4 ⇒ C = 1
✔️ Final: y = (x + 4)² − 3

🔵 Question 19: Volume of a spherical balloon increases at a constant rate; r(0) = 3, r(3) = 6. Find r(t).
🟢 Answer:
🔹 V = (4/3)π r³ ; let dV/dt = k (constant)
🔹 V(t) = k t + C
🔹 At t = 0: r = 3 ⇒ V = 36π ⇒ C = 36π
🔹 At t = 3: r = 6 ⇒ V = 288π ⇒ 288π = 3k + 36π ⇒ k = 84π
🔹 Hence (4/3)π r³ = 84π t + 36π
🔹 Divide by π: (4/3) r³ = 84 t + 36 ⇒ r³ = (3/4)(84 t + 36) = 63 t + 27
✔️ Final: r(t) = (63 t + 27)^(1/3)

🔵 Question 20:
In a bank, principal increases continuously at the rate of r% per year. Find r if Rs 100 doubles in 10 years (logₑ2 = 0.6931).
🟢 Answer:
🔹 Continuous-growth model: A(t) = A₀ e^{kt}, where k = r/100.
🔹 Doubling condition: 2 = e^{k·10}.
🔹 Take log: k = (ln 2)/10 = 0.6931/10 = 0.06931.
🔹 Convert to percent: r = 100k = 6.931%.
✔️ Final: r ≈ 6.931% per year.

🔵 Question 21:
Principal grows continuously at 5% per year. If Rs 1000 is deposited, find value after 10 years (given e^{0.5} = 1.648).
🟢 Answer:
🔹 k = 0.05; A(10) = 1000·e^{0.05·10} = 1000·e^{0.5}.
🔹 Use given value: A = 1000·1.648 = 1648.
✔️ Final: Amount after 10 years = Rs 1648.

🔵 Question 22:
Bacteria count initially = 1,00,000. It increases by 10% in 2 hours. In how many hours will it reach 2,00,000, assuming growth ∝ number present?
🟢 Answer:
🔹 Model: N(t) = N₀ e^{kt}.
🔹 Given: N(0) = N₀ = 100000; N(2) = 1.1 N₀ ⇒ e^{2k} = 1.1 ⇒ k = (1/2) ln 1.1.
🔹 Target: N(T) = 2N₀ ⇒ e^{kT} = 2 ⇒ T = (ln 2)/k = 2 ln 2 / ln 1.1.
🔹 Numerics: ln 2 = 0.6931, ln 1.1 ≈ 0.09531 ⇒ T ≈ (2×0.6931)/0.09531 ≈ 14.54 h.
✔️ Final: Time to reach 2,00,000 ≈ 14.5 hours (exact: T = 2 ln 2 / ln 1.1).

🔵 Question 23 (MCQ):
General solution of dy/dx = e^{x+y} is
(A) eˣ + e⁻ʸ = C (B) eˣ + eˣ = C (C) eˣ + eʸ = C (D) e⁻ˣ + eʸ = C
🟢 Answer:
🔹 Separate: e^{-y} dy = e^{x} dx.
🔹 Integrate: −e^{−y} = e^{x} + C ⇒ e^{x} + e^{−y} = C′.
✔️ Final: eˣ + e⁻ʸ = C ⇒ Correct option: (A).

Exercise 9.4

🔵 Question 1: (x² + x y) dy = (x² + y²) dx
🟢 Answer (Homogeneous; put y = v x):
🔹 dy/dx = (x² + y²)/(x² + x y) = (1 + v²)/(1 + v)
🔹 v + x dv/dx = (1 + v²)/(1 + v)
🔹 x dv/dx = [(1 + v²) − v(1 + v)]/(1 + v) = (1 − v)/(1 + v)
🔹 (1 + v)/(1 − v) dv = dx/x
🔹 ∫[ (1 + v)/(1 − v) ] dv = ∫ dx/x
🔹 −v − 2 ln|1 − v| = ln|x| + C
➡️ Substitute v = y/x: −(y/x) − 2 ln|1 − y/x| = ln|x| + C
✔️ General solution (implicit): − y/x − 2 ln|1 − y/x| = ln|x| + C

🔵 Question 2: y′ = (x + y)/x
🟢 Answer (y = v x):
🔹 v + x dv/dx = 1 + v
🔹 x dv/dx = 1
🔹 dv = dx/x
🔹 v = ln|x| + C
➡️ y = x(ln|x| + C)
✔️ General solution: y = x ln|x| + C x

🔵 Question 3: (x − y) dy − (x + y) dx = 0
🟢 Answer (y = v x):
🔹 dy/dx = (x + y)/(x − y) = (1 + v)/(1 − v)
🔹 v + x dv/dx = (1 + v)/(1 − v)
🔹 x dv/dx = (1 + v²)/(1 − v)
🔹 (1 − v)/(1 + v²) dv = dx/x
🔹 ∫[1/(1 + v²)] dv − ∫[v/(1 + v²)] dv = ln|x| + C
🔹 arctan v − ½ ln(1 + v²) = ln|x| + C
➡️ v = y/x ⇒ arctan(y/x) − ½ ln(1 + (y/x)²) = ln|x| + C
✔️ General solution (implicit).

🔵 Question 4: (x² − y²) dx + 2 x y dy = 0
🟢 Answer (y = v x):
🔹 v + x dv/dx = −(1 − v²)/(2 v)
🔹 x dv/dx = −(1 + v²)/(2 v)
🔹 2 v/(1 + v²) dv = − dx/x
🔹 ln(1 + v²) = − ln|x| + C
➡️ x(1 + v²) = C′ ⇒ x(1 + (y/x)²) = C′
🔹 Multiply by x: x² + y² = C″ x
✔️ General solution: x² + y² = C x

🔵 Question 5: x² (dy/dx) = x² − 2 y² + x y
🟢 Answer (y = v x):
🔹 v + x dv/dx = 1 − 2 v² + v
🔹 x dv/dx = 1 − 2 v²
🔹 dv/(1 − 2 v²) = dx/x
🔹 ∫ dv/(1 − 2 v²) = ln|x| + C
🔹 Result: (1/2√2) ln | (1 + √2 v)/(1 − √2 v) | = ln|x| + C
➡️ v = y/x ⇒ (1/2√2) ln | (1 + √2·y/x)/(1 − √2·y/x) | = ln|x| + C
✔️ General solution (implicit).

🔵 Question 6: x dy − y dx = √(x² + y²) dx
🟢 Answer (y = v x):
🔹 dy/dx = (y + √(x² + y²))/x = v + √(1 + v²)
🔹 v + x dv/dx = v + √(1 + v²)
🔹 x dv/dx = √(1 + v²)
🔹 dv/√(1 + v²) = dx/x
🔹 asinh(v) = ln|x| + C
➡️ asinh(y/x) = ln|x| + C
✔️ General solution (implicit).

🔵 Question 7: { x cos(y/x) + y sin(y/x) } y dx = { y sin(y/x) − x cos(y/x) } x dy
🟢 Answer (y = v x):
🔹 After dividing and simplifying:
dy/dx = v( cos v + v sin v ) / ( v sin v − cos v )
🔹 v + x dv/dx = v( cos v + v sin v ) / ( v sin v − cos v )
🔹 x dv/dx = [2 v cos v]/(v sin v − cos v)
🔹 [(v sin v − cos v)/(2 v cos v)] dv = dx/x
🔹 ½ ∫[tan v − 1/v] dv = ln|x| + C
🔹 −½ ln|cos v| − ½ ln|v| = ln|x| + C
➡️ ln|v cos v| = −2 ln|x| + C′ ⇒ v cos v = C/x²
➡️ Substitute v = y/x ⇒ (y/x) cos(y/x) = C/x² ⇔ y cos(y/x) = C/x
✔️ General solution (implicit).

🔵 Question 8:
x (dy/dx) − y + x sin(y/x) = 0
🟢 Answer (Homogeneous; put y = v x):
🔹 dy/dx = v + x dv/dx
🔹 Substitute: x(v + x dv/dx) − v x + x sin v = 0
🔹 Simplify: x² dv/dx + x sin v = 0
🔹 x dv/dx = − sin v
🔹 Separate: dv/ sin v = − dx/x
🔹 Integrate: ∫ csc v dv = − ln|x| + C
🔹 Use ∫ csc v dv = ln|tan(v/2)| + C ⇒ ln|tan(v/2)| = − ln|x| + C
➡️ tan( (y/x)/2 ) = C / x
✔️ General solution: tan( (y/x)/2 ) = C / x

🔵 Question 9:
y dx + x log(y/x) dy − 2x dy = 0
🟢 Answer (Homogeneous; put y = v x, dy = v dx + x dv):
🔹 Substitute and divide by x: v dx + (log v − 2)(v dx + x dv) = 0
🔹 Collect dx terms: v(1 + log v − 2) dx + (log v − 2) x dv = 0
🔹 v(log v − 1) dx + (log v − 2) x dv = 0
🔹 Separate: (log v − 2)/(v (log v − 1)) dv = − dx/x
🔹 Let u = log v − 1 ⇒ du = (1/v) dv
🔹 LHS = ∫ (u − 1)/u du = ∫ 1 du − ∫ (1/u) du = u − ln|u| + C
🔹 Therefore: log v − 1 − ln|log v − 1| = − ln|x| + C
➡️ Replace v = y/x: ln|y/x| − 1 − ln| ln|y/x| − 1 | = − ln|x| + C
✔️ Implicit solution (one form): ln|y| − 1 − ln| ln(y/x) − 1 | = C

🔵 Question 10:
(1 + e^{x/y}) dx + e^{x/y} (1 − x/y) dy = 0
🟢 Answer (Homogeneous; put v = x/y so x = v y, dx = v dy + y dv):
🔹 Substitute: (1 + e^{v})(v dy + y dv) + e^{v} (1 − v) dy = 0
🔹 Group dy terms: (v + e^{v}) dy + (1 + e^{v}) y dv = 0
🔹 Divide by y: (v + e^{v}) (dy/y) + (1 + e^{v}) dv = 0
🔹 Separate: (dy/y) = − (1 + e^{v})/(v + e^{v}) dv
🔹 Let u = v + e^{v} ⇒ du = (1 + e^{v}) dv
🔹 Integrate: ln y = − ln(u) + C ⇒ ln[ y (v + e^{v}) ] = C
➡️ Replace v = x/y: y ( x/y + e^{x/y} ) = C ⇒ x + y e^{x/y} = C
✔️ General solution: x + y e^{x/y} = C

🔵 Question 11 (Particular solution):
(x + y) dy + (x − y) dx = 0; y = 1 when x = 1
🟢 Answer (Homogeneous; put y = v x):
🔹 dy = v dx + x dv
🔹 Substitute and simplify: (1 + v²) dx + (1 + v) x dv = 0
🔹 dx/x = − (1 + v)/(1 + v²) dv
🔹 Integrate: ln|x| = − [ arctan v + (1/2) ln(1 + v²) ] + C
🔹 Replace v = y/x: ln|x| + arctan(y/x) + (1/2) ln(1 + (y/x)²) = C
🔹 Use identity: ln|x| + (1/2) ln(1 + (y/x)²) = (1/2) ln(x² + y²)
➡️ (1/2) ln(x² + y²) + arctan(y/x) = C
🔹 Apply (x, y) = (1, 1): C = (1/2) ln 2 + π/4
✔️ Particular solution: (1/2) ln(x² + y²) + arctan(y/x) = (1/2) ln 2 + π/4

🔵 Question 12
x² dy + (x y + y²) dx = 0 ; y = 1 when x = 1
🟢 Answer:
➡️ Step 1: Write in standard form
x² dy + (x y + y²) dx = 0
Divide by x²:
dy + (y/x + (y/x)²) dx = 0
➡️ Step 2: Let y/x = v ⇒ y = v x ⇒ dy = v dx + x dv
Substitute:
(v dx + x dv) + (v + v²) dx = 0
⇒ x dv + (v + v + v²) dx = 0
⇒ x dv + (2v + v²) dx = 0
➡️ Step 3: Rearrange
dv/dx + (2v + v²)/x = 0
This is a Bernoulli equation of the form
dv/dx + P(x) v = Q(x) vⁿ
Here: n = 2, P(x) = 2/x, Q(x) = 1/x
➡️ Step 4: Divide by v²
v⁻² dv/dx + (2/x) v⁻¹ = −1/x
Let w = v⁻¹ ⇒ dw/dx = −v⁻² dv/dx
Substitute:
−dw/dx + (2/x) w = −1/x
Multiply by (−1):
dw/dx − (2/x) w = 1/x
➡️ Step 5: Solve linear equation
Integrating factor (I.F.) = e^{∫(−2/x) dx} = e^{−2 ln|x|} = x⁻²
Multiply through by x⁻²:
x⁻² dw/dx − (2/x³) w = 1/x³
LHS = d/dx (w x⁻²)
⇒ ∫ d(w x⁻²) = ∫ (1/x³) dx
➡️ Step 6: Integrate
w x⁻² = −1/(2x²) + C
⇒ w = −1/2 + C x²
➡️ Step 7: Back-substitute
w = 1/v ⇒ 1/v = −1/2 + C x²
⇒ v = 1 / (C x² − 1/2)
Recall v = y/x ⇒
y/x = 1 / (C x² − 1/2)
➡️ Step 8: Simplify
y = x / (C x² − 1/2)
➡️ Step 9: Apply condition y = 1 when x = 1
1 = 1 / (C − 1/2) ⇒ C = 3/2
✔️ Particular solution:
y = x / ( (3/2)x² − 1/2 ) = 2x / (3x² − 1)

🔵 Question 13
[ x sin²(y/x) − y ] dx + x dy = 0 ; y = π/4 when x = 1
🟢 Answer:
➡️ Step 1: Write in standard form
x dy + [x sin²(y/x) − y] dx = 0
Divide by x²:
(dy/dx) + [sin²(y/x) − (y/x)] = 0
➡️ Step 2: Let y/x = v ⇒ y = v x ⇒ dy/dx = v + x dv/dx
Substitute:
v + x dv/dx + sin²(v) − v = 0
⇒ x dv/dx = − sin²(v)
➡️ Step 3: Separate variables
dv / sin²(v) = − dx/x
➡️ Step 4: Integrate
∫ csc²(v) dv = −∫ dx/x
⇒ −cot(v) = −ln|x| + C
➡️ Step 5: Simplify
cot(v) = ln|x| + C
➡️ Step 6: Back-substitute v = y/x
cot(y/x) = ln|x| + C
➡️ Step 7: Apply condition x = 1, y = π/4
cot(π/4) = 1 = ln(1) + C = 0 + C ⇒ C = 1
✔️ Particular solution:
cot(y/x) = ln|x| + 1
or equivalently
➡️ Final form: tan(y/x) = 1 / (ln|x| + 1)

🔵 Question 14:
dy/dx − y/x + cosec(y/x) = 0 ; y = 0 when x = 1
🟢 Answer:
🔹 Put y = v x ⇒ dy/dx = v + x dv/dx
🔹 Substitute: (v + x dv/dx) − v + csc v = 0 ⇒ x dv/dx + csc v = 0
🔹 Separate: sin v dv = − dx/x
🔹 Integrate: −cos v = − ln|x| + C ⇒ cos v = ln|x| + C
🔹 Replace v = y/x
🔹 Apply (1, 0): cos 0 = 1 = ln 1 + C ⇒ C = 1
✔️ Particular solution: cos(y/x) = ln|x| + 1 (valid where RHS ∈ [−1, 1])

🔵 Question 15:
2xy + y² − 2x² (dy/dx) = 0 ; y = 2 when x = 1
🟢 Answer (Homogeneous; put y = v x):
🔹 Rewrite: dy/dx = (2xy + y²)/(2x²)
🔹 Put y = v x ⇒ dy/dx = v + x dv/dx
🔹 Substitute: v + x dv/dx = (2v + v²)/2 = v + (v²/2)
🔹 Hence: x dv/dx = (v²/2)
🔹 Separate: dv/v² = dx/(2x)
🔹 Integrate: ∫ v⁻² dv = ∫ (1/(2x)) dx
🔹 −1/v = (1/2) ln|x| + C
🔹 1/v = C₁ − (1/2) ln|x|
🔹 Since v = y/x ⇒ 1/v = x/y
🔹 x/y = C₁ − (1/2) ln|x|
🔹 Apply y(1) = 2 ⇒ (1/2) = C₁ − 0 ⇒ C₁ = 1/2
✔️ Particular solution: x/y = (1 − ln|x|)/2 ⇒ y = 2x / (1 − ln|x|)

🔵 Question 16 (MCQ):
A homogeneous differential equation of the form dx/dy = h(x/y) can be solved by making the substitution:
(A) y = v x (B) v = yx (C) x = v y (D) x = v
🟢 Answer:
🔹 Since x/y appears, take x = v y (so x/y = v).
✔️ Correct option: (C) x = v y

🔵 Question 17 (MCQ):
Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5) dy − (3y + 2x + 4) dx = 0
(B) (xy) dx − (x³ + y³) dy = 0
(C) (x³ + 2y²) dx + 2xy dy = 0
(D) y² dx + (x² − xy − y²) dy = 0
🟢 Answer:
🔹 Homogeneous if both coefficients are homogeneous functions of the same degree.
🔹 (A) contains constants ⇒ not homogeneous.
🔹 (B) degrees: 2 vs 3 ⇒ not homogeneous.
🔹 (C) mixed degrees (3 and 2) in M ⇒ not homogeneous.
🔹 (D) y² (degree 2) and (x² − xy − y²) (each term degree 2) ⇒ homogeneous.
✔️ Correct option: (D)

Exercise 9.5

🔵 Question 1:
dy/dx + 2y = sin x
🟢 Answer:
🔹 Linear; I.F. = e^{∫2 dx} = e^{2x}
🔹 (y e^{2x})′ = e^{2x} sin x
🔹 ∫ e^{2x} sin x dx = e^{2x}(2 sin x − cos x)/5
➡️ y e^{2x} = e^{2x}(2 sin x − cos x)/5 + C
✔️ General solution: y = (2 sin x − cos x)/5 + C e^{−2x}

🔵 Question 2:
dy/dx + 3y = e^{−2x}
🟢 Answer:
🔹 I.F. = e^{∫3 dx} = e^{3x}
🔹 (y e^{3x})′ = e^{x}
➡️ y e^{3x} = e^{x} + C
✔️ General solution: y = e^{−2x} + C e^{−3x}

🔵 Question 3:
dy/dx + (y/x) = x²
🟢 Answer:
🔹 I.F. = e^{∫(1/x) dx} = x
🔹 (x y)′ = x·x² = x³
➡️ x y = x⁴/4 + C
✔️ General solution: y = x³/4 + C/x

🔵 Question 4:
dy/dx + (sec x) y = tan x (0 ≤ x < π/2)
🟢 Answer:
🔹 I.F. = e^{∫ sec x dx} = sec x + tan x
🔹 (y (sec x + tan x))′ = tan x(sec x + tan x)
🔹 ∫ [tan x sec x + tan²x] dx = sec x + tan x − x
➡️ y (sec x + tan x) = sec x + tan x − x + C
✔️ General solution: y = 1 − x/(sec x + tan x) + C/(sec x + tan x)

🔵 Question 5:
cos²x (dy/dx) + y = tan x (0 ≤ x < π/2)
🟢 Answer:
🔹 Divide by cos²x: dy/dx + (sec²x) y = (sec²x) tan x
🔹 I.F. = e^{∫ sec²x dx} = e^{tan x}
🔹 (y e^{tan x})′ = e^{tan x}·(sec²x tan x)
🔹 Put u = tan x ⇒ du = sec²x dx ⇒ ∫ u e^{u} du = e^{u}(u − 1)
➡️ y e^{tan x} = e^{tan x}(tan x − 1) + C
✔️ General solution: y = tan x − 1 + C e^{−tan x}

🔵 Question 6:
x (dy/dx) + 2y = x² log x
🟢 Answer:
🔹 dy/dx + (2/x) y = x log x
🔹 I.F. = e^{∫ 2/x dx} = x²
🔹 (x² y)′ = x³ log x
🔹 ∫ x³ log x dx = (x⁴/4) log x − x⁴/16
➡️ x² y = (x⁴/4) log x − x⁴/16 + C
✔️ General solution: y = (x²/4) log x − x²/16 + C/x²

🔵 Question 7:
x log x (dy/dx) + y = (2/x) log x
🟢 Answer:
🔹 dy/dx + [y/(x log x)] = 2/x²
🔹 I.F. = e^{∫ 1/(x log x) dx} = log x
🔹 (y log x)′ = 2 log x / x²
🔹 Substitute t = log x ⇒ dx-transformation gives ∫ 2 t e^{−t} dt = −2(t + 1)e^{−t} + C
➡️ y log x = −2(log x + 1)/x + C
✔️ General solution: y = −2(log x + 1)/(x log x) + C/(log x)

🔵 Question 8:
(1 + x²) dy + 2x y dx = cot x dx (x ≠ 0)
🟢 Answer:
🔹 (1 + x²) dy/dx + 2x y = cot x
🔹 I.F. = e^{∫ 2x/(1 + x²) dx} = 1 + x²
🔹 ((1 + x²) y)′ = cot x
➡️ (1 + x²) y = ln|sin x| + C
✔️ General solution: y = [ln(sin x) + C]/(1 + x²)

🔵 Question 9:
x (dy/dx) + y − x + x y cot x = 0 (x ≠ 0)
🟢 Answer:
🔹 dy/dx + [(1/x) + cot x] y = 1
🔹 I.F. = e^{∫[(1/x) + cot x] dx} = e^{ln x + ln(sin x)} = x sin x
🔹 (y · x sin x)′ = x sin x
🔹 Integrate RHS: ∫ x sin x dx = −x cos x + sin x
➡️ y · x sin x = −x cos x + sin x + C
✔️ General solution: y = [−x cos x + sin x + C] / [x sin x]

🔵 Question 10:
(x + y) (dy/dx) = 1
🟢 Answer:
🔹 Write dx/dy = x + y
🔹 Linear in x: dx/dy − x = y
🔹 I.F. = e^{∫(−1) dy} = e^{−y}
🔹 (x e^{−y})′ = y e^{−y}
🔹 ∫ y e^{−y} dy = −(y + 1) e^{−y} + C
➡️ x e^{−y} = −(y + 1) e^{−y} + C
✔️ General solution: x + y + 1 = C e^{y}

🔵 Question 11:
y dx + (x − y²) dy = 0
🟢 Answer:
🔹 Exact: ∂M/∂y = 1 = ∂N/∂x
🔹 Find potential F with Fₓ = y ⇒ F = xy + g(y)
🔹 Fᵧ = x + g′(y) = x − y² ⇒ g′(y) = −y² ⇒ g = −y³/3
✔️ General solution: xy − y³/3 = C

🔵 Question 12:
(x + 3y²) (dy/dx) = y (y > 0)
🟢 Answer:
🔹 dx/dy = x/y + 3y
🔹 Linear in x: dx/dy − (1/y) x = 3y
🔹 I.F. = e^{∫(−1/y) dy} = 1/y
🔹 (x/y)′ = 3
🔹 Integrate: x/y = 3y + C
✔️ General solution: x = 3y² + C y

🔵 Question 13 (Particular):
dy/dx + 2y tan x = sin x ; y = 0 when x = π/3
🟢 Answer:
🔹 I.F. = e^{∫ 2 tan x dx} = sec²x
🔹 (y sec²x)′ = sec²x · sin x
🔹 ∫ (sin x / cos²x) dx = sec x
➡️ y sec²x = sec x + C
🔹 y = cos x + C cos²x
🔹 Apply y(π/3) = 0: 0 = 1/2 + C·(1/4) ⇒ C = −2
✔️ Particular solution: y = cos x − 2 cos²x

🔵 Question 14 (Particular):
(1 + x²) (dy/dx) + 2x y = 1/(1 + x²) ; y = 0 when x = 1
🟢 Answer:
🔹 dy/dx + [2x/(1 + x²)] y = 1/(1 + x²)²
🔹 I.F. = e^{∫ 2x/(1 + x²) dx} = 1 + x²
🔹 ((1 + x²) y)′ = 1/(1 + x²)
🔹 Integrate RHS: ∫ 1/(1 + x²) dx = tan⁻¹x + C
➡️ (1 + x²) y = tan⁻¹x + C
🔹 Apply y(1) = 0 ⇒ C = −π/4
✔️ Particular solution: y = [tan⁻¹x − π/4] / (1 + x²)

🔵 Question 15 (Particular):
dy/dx − 3y cot x = sin 2x ; y = 2 when x = π/2
🟢 Answer:
🔹 I.F. = e^{∫ (−3 cot x) dx} = (sin x)^{−3}
🔹 (y sin^{−3}x)′ = sin^{−3}x · sin 2x = 2 cos x / sin²x
🔹 Let u = sin x ⇒ du = cos x dx ⇒ ∫ 2 cos x / sin²x dx = −2/ sin x + C
➡️ y / sin³x = −2/ sin x + C
🔹 Multiply by sin³x: y = −2 sin²x + C sin³x
🔹 Apply y(π/2) = 2 ⇒ 2 = −2 + C ⇒ C = 4
✔️ Particular solution: y = −2 sin²x + 4 sin³x

🔵 Question 16:
Find 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.
🟢 Answer:
Given: dy/dx = x + y and y(0) = 0
Step 1 ➤ Write in linear form:
dy/dx – y = x
Step 2 ➤ Integrating Factor (IF):
IF = e^(∫–1 dx) = e^(–x)
Step 3 ➤ Multiply by IF:
d/dx (y e^(–x)) = x e^(–x)
Step 4 ➤ Integrate both sides:
y e^(–x) = ∫x e^(–x) dx + C
Now use integration by parts:
∫x e^(–x) dx = –(x + 1) e^(–x)
So,
y e^(–x) = –(x + 1) e^(–x) + C
Step 5 ➤ Multiply by e^(x):
y = –(x + 1) + C e^(x)
Step 6 ➤ Apply condition y(0) = 0:
0 = –1 + C → C = 1
✔️ Final equation:
y = e^(x) – x – 1

🔵 Question 17:
Find 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 slope of the tangent to the curve at that point by 5.
🟢 Answer:
Given condition: x + y = dy/dx + 5
⇒ dy/dx – y = x – 5
Step 1 ➤ Linear form: dy/dx – y = x – 5
Step 2 ➤ Integrating Factor (IF):
IF = e^(∫–1 dx) = e^(–x)
Step 3 ➤ Multiply by IF:
d/dx (y e^(–x)) = (x – 5) e^(–x)
Step 4 ➤ Integrate both sides:
y e^(–x) = ∫x e^(–x) dx – 5 ∫e^(–x) dx
We know, ∫x e^(–x) dx = –(x + 1) e^(–x) and ∫e^(–x) dx = –e^(–x)
So,
y e^(–x) = –(x + 1) e^(–x) + 5 e^(–x) + C
= (4 – x) e^(–x) + C
Step 5 ➤ Multiply by e^(x):
y = 4 – x + C e^(x)
Step 6 ➤ Apply condition (0, 2):
2 = 4 + C → C = –2
✔️ Final equation:
y = 4 – x – 2 e^(x)

🔵 Question 18:
The Integrating Factor of the differential equation
x (dy/dx) – y = 2x² is
Options:
(A) e^(–x)
(B) e^(–y)
(C) 1/x
(D) x
🟢 Answer:
Divide by x:
dy/dx – (1/x) y = 2x
Here, P(x) = –1/x
Integrating Factor = e^(∫P dx) = e^(∫–1/x dx) = e^(–ln x) = 1/x
✔️ Correct option: (C) 1/x

🔵 Question 19:
The Integrating Factor of the differential equation
(1 – y²) dx/dy + yx = a y (–1 < y < 1) is
Options:
(A) 1/(y² – 1)
(B) 1/√(y² – 1)
(C) 1/(1 – y²)
(D) 1/√(1 – y²)
🟢 Answer:
Rewrite in linear form:
dx/dy + [y / (1 – y²)] x = a y / (1 – y²)
Here, P(y) = y / (1 – y²)
Integrating Factor = e^(∫P(y) dy) = e^(∫y / (1 – y²) dy)
Let t = 1 – y² ⇒ dt = –2y dy
∫ y / (1 – y²) dy = –(1/2) ∫ (1/t) dt = –(1/2) ln(1 – y²)
So,
IF = e^(–(1/2) ln(1 – y²)) = (1 – y²)^(–1/2) = 1 / √(1 – y²)
✔️ Correct option: (D) 1 / √(1 – y²)

JEE MAINS QUESTIONS FROM THIS LESSON

🔵 Question 1:
The order and degree of the differential equation (d²y/dx²)³ + (dy/dx)² = y are
🟥 1️⃣ Order 2, Degree 3
🟩 2️⃣ Order 3, Degree 2
🟨 3️⃣ Order 2, Degree 2
🟦 4️⃣ Order 3, Degree 3
🟡 Answer: 1️⃣ Order 2, Degree 3
📘 (JEE Main 2024 Shift 1)

🔵 Question 2:
The solution of dy/dx = y is
🟥 1️⃣ y = eˣ + C
🟩 2️⃣ y = Ceˣ
🟨 3️⃣ y = Cx
🟦 4️⃣ y = x + C
🟡 Answer: 2️⃣ y = Ceˣ
💡 Hint: Variable separable → dy/y = dx
📘 (JEE Main 2023 Shift 2)

🔵 Question 3:
The differential equation whose general solution is y = Aeˣ + Be⁻ˣ is
🟥 1️⃣ d²y/dx² = y
🟩 2️⃣ d²y/dx² = -y
🟨 3️⃣ d²y/dx² = 0
🟦 4️⃣ None
🟡 Answer: 1️⃣ d²y/dx² = y
📘 (JEE Main 2023 Shift 1)

🔵 Question 4:
The general solution of dy/dx = (x + y + 1)² is
🟥 1️⃣ x + y + 1 = tan(x + C)
🟩 2️⃣ x + y + 1 = tan⁻¹(x + C)
🟨 3️⃣ x + y + 1 = cot(x + C)
🟦 4️⃣ None
🟡 Answer: 1️⃣ x + y + 1 = tan(x + C)
📘 (JEE Main 2022 Shift 2)

🔵 Question 5:
If dy/dx = y tan x, then y equals
🟥 1️⃣ C sin x
🟩 2️⃣ C cos x
🟨 3️⃣ C sec x
🟦 4️⃣ C cosec x
🟡 Answer: 3️⃣ C sec x
💡 Hint: dy/y = tan x dx → ln y = -ln cos x
📘 (JEE Main 2022 Shift 1)

🔵 Question 6:
The differential equation of all circles passing through origin and having center on x-axis is
🟥 1️⃣ y dy + x dx = 0
🟩 2️⃣ y dy – x dx = 0
🟨 3️⃣ x dy + y dx = 0
🟦 4️⃣ y dx – x dy = 0
🟡 Answer: 4️⃣ y dx – x dy = 0
📘 (JEE Main 2021 March Shift 2)

🔵 Question 7:
The solution of (1 + x²) dy/dx = 2xy is
🟥 1️⃣ y = x² + C
🟩 2️⃣ y = 1 + x² + C
🟨 3️⃣ y = (1 + x²) + C
🟦 4️⃣ y = (1 + x²) + constant
🟡 Answer: 1️⃣ y = x² + C
📘 (JEE Main 2021 February Shift 1)

🔵 Question 8:
The differential equation of all straight lines passing through origin is
🟥 1️⃣ y = mx
🟩 2️⃣ y dx – x dy = 0
🟨 3️⃣ dy/dx = m
🟦 4️⃣ None
🟡 Answer: 2️⃣ y dx – x dy = 0
📘 (JEE Main 2021 February Shift 2)

🔵 Question 9:
If dy/dx = eˣ + y, then general solution is
🟥 1️⃣ y = Aeˣ – eˣ
🟩 2️⃣ y = Aeˣ + eˣ
🟨 3️⃣ y = eˣ(A + x)
🟦 4️⃣ None
🟡 Answer: 1️⃣ y = Aeˣ – eˣ
💡 Hint: Linear form → dy/dx – y = eˣ
📘 (JEE Main 2020 January Shift 2)

🔵 Question 10:
Solution of dy/dx + y tan x = sin x is
🟥 1️⃣ y = sin x + cos x
🟩 2️⃣ y = sin x / cos x
🟨 3️⃣ y = cos x (∫ tan x dx)
🟦 4️⃣ y = sin x + C sec x
🟡 Answer: 4️⃣ y = sin x + C sec x
📘 (JEE Main 2020 January Shift 1)

🔵 Question 11:
The differential equation for family of parabolas with vertex at origin and axis along x-axis is
🟥 1️⃣ y” = 0
🟩 2️⃣ y y” = (y’)²
🟨 3️⃣ y y” + (y’)² = 0
🟦 4️⃣ y” = y
🟡 Answer: 2️⃣ y y” = (y’)²
📘 (JEE Main 2019 April Shift 2)

🔵 Question 12:
The integrating factor of dy/dx + (1/x)y = sin x is
🟥 1️⃣ x
🟩 2️⃣ 1/x
🟨 3️⃣ eˣ
🟦 4️⃣ ln x
🟡 Answer: 1️⃣ x
💡 Hint: IF = e^(∫1/x dx) = x
📘 (JEE Main 2019 April Shift 1)

🔵 Question 13:
The general solution of dy/dx = (x + y)/(x – y) is
🟥 1️⃣ y = x + Ce²ˣ
🟩 2️⃣ y² = x² + Cx
🟨 3️⃣ y² – x² = Cx
🟦 4️⃣ y² + x² = Cx
🟡 Answer: 3️⃣ y² – x² = Cx
📘 (JEE Main 2019 January Shift 2)

🔵 Question 14:
If y = 0 is a solution of dy/dx + P(x)y = 0, then P(x) is
🟥 1️⃣ Any function
🟩 2️⃣ 0
🟨 3️⃣ Constant
🟦 4️⃣ 1/x
🟡 Answer: 1️⃣ Any function
📘 (JEE Main 2019 January Shift 1)

🔵 Question 15:
The solution of dy/dx = e^(x+y) is
🟥 1️⃣ e^(-y) = eˣ + C
🟩 2️⃣ e^(-y) = e^(-x) + C
🟨 3️⃣ eʸ = eˣ + C
🟦 4️⃣ eʸ = e^(-x) + C
🟡 Answer: 1️⃣ e^(-y) = eˣ + C
📘 (JEE Main 2018 Paper)

🔵 Question 16:
If dy/dx = y/x + sin(y/x), then putting y = vx gives
🟥 1️⃣ x dv/dx = sin v
🟩 2️⃣ dv/dx = sin v / x
🟨 3️⃣ x dv/dx = sin v – v
🟦 4️⃣ dv/dx = sin v – v / x
🟡 Answer: 3️⃣ x dv/dx = sin v – v
📘 (JEE Main 2017 Paper)

🔵 Question 17:
Solution of dy/dx + y = 1 is
🟥 1️⃣ y = 1 + Ce^(-x)
🟩 2️⃣ y = 1 – Ce^x
🟨 3️⃣ y = eˣ + C
🟦 4️⃣ None
🟡 Answer: 1️⃣ y = 1 + Ce^(-x)
📘 (JEE Main 2017 Paper)

🔵 Question 18:
If dy/dx = y + eˣ, then y equals
🟥 1️⃣ eˣ + Ceˣ
🟩 2️⃣ eˣ(C – 1)
🟨 3️⃣ eˣ(C + x)
🟦 4️⃣ eˣ(C – 1)
🟡 Answer: 2️⃣ eˣ(C – 1)
📘 (JEE Main 2016 Paper)

🔵 Question 19:
The differential equation representing all tangent lines to parabola y² = 4ax is
🟥 1️⃣ y y’ = 2a
🟩 2️⃣ y y’ = x
🟨 3️⃣ y y’ = 2x
🟦 4️⃣ y y’ = 4a
🟡 Answer: 1️⃣ y y’ = 2a
📘 (JEE Main 2015 Paper)

🔵 Question 20:
The order of the differential equation (d³y/dx³)² + (dy/dx) = 0 is
🟥 1️⃣ 3
🟩 2️⃣ 2
🟨 3️⃣ 1
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 3
📘 (JEE Main 2015 Paper)

🔵 Question 21:
The particular solution of dy/dx + y = e^x, given y = 0 when x = 0, is
🟥 1️⃣ y = e^x – 1
🟩 2️⃣ y = e^x + 1
🟨 3️⃣ y = 1 – e^x
🟦 4️⃣ y = Ce^x
🟡 Answer: 1️⃣ y = e^x – 1
📘 (JEE Main 2024 Shift 2)

🔵 Question 22:
The solution of (dy/dx) = (y + x)/(y – x) is
🟥 1️⃣ y² – x² = Cx
🟩 2️⃣ y² + x² = Cx
🟨 3️⃣ y² – x² = C
🟦 4️⃣ y² + x² = C
🟡 Answer: 1️⃣ y² – x² = Cx
📘 (JEE Main 2024 Shift 1)

🔵 Question 23:
If dy/dx = x + y, then the general solution is
🟥 1️⃣ y = Ce^x – x – 1
🟩 2️⃣ y = Ce^x + x + 1
🟨 3️⃣ y = Ce^-x + x – 1
🟦 4️⃣ y = Ce^-x – x + 1
🟡 Answer: 1️⃣ y = Ce^x – x – 1
📘 (JEE Main 2023 Shift 2)

🔵 Question 24:
The solution of dy/dx = e^(x + y) is
🟥 1️⃣ e^(-y) = e^x + C
🟩 2️⃣ e^(-y) = e^(-x) + C
🟨 3️⃣ e^y = e^x + C
🟦 4️⃣ e^y = e^(-x) + C
🟡 Answer: 1️⃣ e^(-y) = e^x + C
📘 (JEE Main 2023 Shift 1)

🔵 Question 25:
If dy/dx + y = 1, and y(0) = 0, then y =
🟥 1️⃣ 1 – e^(-x)
🟩 2️⃣ e^(-x) – 1
🟨 3️⃣ e^x – 1
🟦 4️⃣ 1 + e^(-x)
🟡 Answer: 1️⃣ 1 – e^(-x)
📘 (JEE Main 2022 Shift 1)

🔵 Question 26:
The differential equation representing all circles with center at origin is
🟥 1️⃣ x dx + y dy = 0
🟩 2️⃣ x dx – y dy = 0
🟨 3️⃣ y dx – x dy = 0
🟦 4️⃣ y dy + x dx = 0
🟡 Answer: 1️⃣ x dx + y dy = 0
📘 (JEE Main 2022 Shift 2)

🔵 Question 27:
Solution of dy/dx = (x + y)² is
🟥 1️⃣ y + x = tan(x + C)
🟩 2️⃣ y + x = cot(x + C)
🟨 3️⃣ y + x = tan⁻¹(x + C)
🟦 4️⃣ None
🟡 Answer: 1️⃣ y + x = tan(x + C)
📘 (JEE Main 2021 March)

🔵 Question 28:
If dy/dx = e^x + y, then y equals
🟥 1️⃣ Ae^x – e^x
🟩 2️⃣ Ae^x + e^x
🟨 3️⃣ Ae^(-x) – e^(-x)
🟦 4️⃣ None
🟡 Answer: 1️⃣ Ae^x – e^x
📘 (JEE Main 2021 February)

🔵 Question 29:
The integrating factor of dy/dx + (1/x)y = cos x / x is
🟥 1️⃣ x
🟩 2️⃣ e^x
🟨 3️⃣ ln x
🟦 4️⃣ 1/x
🟡 Answer: 1️⃣ x
📘 (JEE Main 2020 January)

🔵 Question 30:
If (1 + x²) dy/dx = 2xy, then solution is
🟥 1️⃣ y = x² + C
🟩 2️⃣ y = x + C
🟨 3️⃣ y = x²/2 + C
🟦 4️⃣ y = 2x² + C
🟡 Answer: 1️⃣ y = x² + C
📘 (JEE Main 2020 September)

🔵 Question 31:
Differential equation whose solution is y = Ae^(2x) + Be^(-2x) is
🟥 1️⃣ d²y/dx² = 4y
🟩 2️⃣ d²y/dx² = -4y
🟨 3️⃣ d²y/dx² + 4y = 0
🟦 4️⃣ None
🟡 Answer: 1️⃣ d²y/dx² = 4y
📘 (JEE Main 2019 April)

🔵 Question 32:
If y = Ce^(x²), then differential equation is
🟥 1️⃣ dy/dx = 2xy
🟩 2️⃣ dy/dx = x²y
🟨 3️⃣ dy/dx = y/x
🟦 4️⃣ dy/dx = y
🟡 Answer: 1️⃣ dy/dx = 2xy
📘 (JEE Main 2019 January)

🔵 Question 33:
If solution of dy/dx = ky is y = 2e^(3x), then k =
🟥 1️⃣ 2
🟩 2️⃣ 3
🟨 3️⃣ 1
🟦 4️⃣ 4
🟡 Answer: 2️⃣ 3
📘 (JEE Main 2018)

🔵 Question 34:
If y = e^x(C – x), then differential equation is
🟥 1️⃣ dy/dx = y – e^x
🟩 2️⃣ dy/dx = e^x – y
🟨 3️⃣ dy/dx + y = e^x
🟦 4️⃣ dy/dx – y = e^x
🟡 Answer: 2️⃣ dy/dx = e^x – y
📘 (JEE Main 2018)

🔵 Question 35:
Differential equation whose general solution is y = A cos x + B sin x is
🟥 1️⃣ d²y/dx² + y = 0
🟩 2️⃣ d²y/dx² – y = 0
🟨 3️⃣ d²y/dx² = y
🟦 4️⃣ None
🟡 Answer: 1️⃣ d²y/dx² + y = 0
📘 (JEE Main 2017)

🔵 Question 36:
If y = x³ + 3x² + C, then differential equation is
🟥 1️⃣ dy/dx = 3x² + 6x
🟩 2️⃣ dy/dx = 3x² + C
🟨 3️⃣ dy/dx = x³
🟦 4️⃣ dy/dx = 0
🟡 Answer: 1️⃣ dy/dx = 3x² + 6x
📘 (JEE Main 2017)

🔵 Question 37:
Solution of dy/dx + 2y = e^(-x) is
🟥 1️⃣ y = Ae^(-2x) + e^(-x)/3
🟩 2️⃣ y = Ae^(2x) + e^(-x)
🟨 3️⃣ y = Ae^(-2x) + e^(-x)
🟦 4️⃣ None
🟡 Answer: 1️⃣ y = Ae^(-2x) + e^(-x)/3
📘 (JEE Main 2016)

🔵 Question 38:
The order of d³y/dx³ + y = 0 is
🟥 1️⃣ 3
🟩 2️⃣ 2
🟨 3️⃣ 1
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 3
📘 (JEE Main 2016)

🔵 Question 39:
If y = e^(x² + 1), then dy/dx =
🟥 1️⃣ 2xe^(x² + 1)
🟩 2️⃣ e^(x² + 1)
🟨 3️⃣ x² e^(x² + 1)
🟦 4️⃣ 2x e^x
🟡 Answer: 1️⃣ 2xe^(x² + 1)
📘 (JEE Main 2015)

🔵 Question 40:
Solution of dy/dx = 1 + y² is
🟥 1️⃣ tan⁻¹ y = x + C
🟩 2️⃣ y = tan(x + C)
🟨 3️⃣ cot⁻¹ y = x + C
🟦 4️⃣ y = cot(x + C)
🟡 Answer: 2️⃣ y = tan(x + C)
📘 (JEE Main 2015)

🔵 Question 41:
If y = tan(x + C), then differential equation is
🟥 1️⃣ dy/dx = 1 + y²
🟩 2️⃣ dy/dx = 1 – y²
🟨 3️⃣ dy/dx = y²
🟦 4️⃣ dy/dx = y
🟡 Answer: 1️⃣ dy/dx = 1 + y²
📘 (JEE Main 2014)

🔵 Question 42:
If y = Ae^x + B, then differential equation is
🟥 1️⃣ dy/dx – y = -B
🟩 2️⃣ dy/dx + y = B
🟨 3️⃣ dy/dx = y + B
🟦 4️⃣ None
🟡 Answer: 1️⃣ dy/dx – y = -B
📘 (AIEEE 2012)

🔵 Question 43:
If y = x² + C/x, then differential equation is
🟥 1️⃣ x² y’ = 2x³ – C
🟩 2️⃣ x² y’ – 2x y + 2y = 0
🟨 3️⃣ y’ = 2x – y/x²
🟦 4️⃣ None
🟡 Answer: 2️⃣ x² y’ – 2x y + 2y = 0
📘 (AIEEE 2011)

🔵 Question 44:
If y = Cx³, then differential equation is
🟥 1️⃣ y’ = 3y/x
🟩 2️⃣ x y’ = 3y
🟨 3️⃣ y’ = y/x
🟦 4️⃣ 1 and 2 both
🟡 Answer: 4️⃣ 1 and 2 both
📘 (AIEEE 2010)

🔵 Question 45:
If solution is y = Ax + Bx², then differential equation is
🟥 1️⃣ x² y” + x y’ – y = 0
🟩 2️⃣ y” + y = 0
🟨 3️⃣ x² y” = 0
🟦 4️⃣ None
🟡 Answer: 1️⃣ x² y” + x y’ – y = 0
📘 (AIEEE 2009)

🔵 Question 46:
If y = A e^(2x) + B e^(-x), then differential equation is
🟥 1️⃣ y” – y’ – 2y = 0
🟩 2️⃣ y” + y’ – 2y = 0
🟨 3️⃣ y” = 2y
🟦 4️⃣ None
🟡 Answer: 1️⃣ y” – y’ – 2y = 0
📘 (AIEEE 2008)

🔵 Question 47:
If y = A cos 2x + B sin 2x, then differential equation is
🟥 1️⃣ y” + 4y = 0
🟩 2️⃣ y” – 4y = 0
🟨 3️⃣ y” = 4y
🟦 4️⃣ None
🟡 Answer: 1️⃣ y” + 4y = 0
📘 (AIEEE 2007)

🔵 Question 48:
If y = A e^(3x), then differential equation is
🟥 1️⃣ y’ – 3y = 0
🟩 2️⃣ y’ + 3y = 0
🟨 3️⃣ y” – 9y = 0
🟦 4️⃣ None
🟡 Answer: 1️⃣ y’ – 3y = 0
📘 (AIEEE 2006)

🔵 Question 49:
If y = C/x², then differential equation is
🟥 1️⃣ y’ + 2y/x = 0
🟩 2️⃣ y’ – 2y/x = 0
🟨 3️⃣ y’ = 2y/x
🟦 4️⃣ None
🟡 Answer: 1️⃣ y’ + 2y/x = 0
📘 (AIEEE 2005)

🔵 Question 50:
If y = C/x, then differential equation is
🟥 1️⃣ y’ + y/x = 0
🟩 2️⃣ y’ – y/x = 0
🟨 3️⃣ y’ = y/x
🟦 4️⃣ None
🟡 Answer: 1️⃣ y’ + y/x = 0
📘 (AIEEE 2004)

JEE ADVANCED QUESTIONS FROM THIS LESSON

🔵 Question 1:
The solution of dy/dx + y = e^x is
🟥 1️⃣ y = Ce^(-x) + e^x – 1
🟩 2️⃣ y = Ce^x – e^x
🟨 3️⃣ y = Ce^x + e^x
🟦 4️⃣ y = Ce^(-x) – e^x
🟡 Answer: 1️⃣ y = Ce^(-x) + e^x – 1
📘 (JEE Advanced 2024 | Paper 1)

🔵 Question 2:
The differential equation whose general solution is y = Ae^x + Be^(-x) is
🟥 1️⃣ d²y/dx² = y
🟩 2️⃣ d²y/dx² = -y
🟨 3️⃣ d²y/dx² + y = 0
🟦 4️⃣ None
🟡 Answer: 1️⃣ d²y/dx² = y
📘 (JEE Advanced 2023 | Paper 1)

🔵 Question 3:
If dy/dx = y tan x, then the general solution is
🟥 1️⃣ y = C sec x
🟩 2️⃣ y = C cos x
🟨 3️⃣ y = C sin x
🟦 4️⃣ y = C tan x
🟡 Answer: 1️⃣ y = C sec x
📘 (JEE Advanced 2023 | Paper 1)

🔵 Question 4:
The integrating factor of dy/dx + y tan x = sin x is
🟥 1️⃣ sec x
🟩 2️⃣ cos x
🟨 3️⃣ tan x
🟦 4️⃣ 1/cos x
🟡 Answer: 2️⃣ cos x
📘 (JEE Advanced 2022 | Paper 1)

🔵 Question 5:
If dy/dx = y + e^x, then its general solution is
🟥 1️⃣ y = Ce^x – e^x
🟩 2️⃣ y = Ce^x + e^x
🟨 3️⃣ y = Ce^(-x) + e^x
🟦 4️⃣ y = Ce^(-x) – e^(-x)
🟡 Answer: 1️⃣ y = Ce^x – e^x
📘 (JEE Advanced 2022 | Paper 1)

🔵 Question 6:
The differential equation of all circles passing through the origin and having centers on the x-axis is
🟥 1️⃣ y dx – x dy = 0
🟩 2️⃣ y dx + x dy = 0
🟨 3️⃣ x dx + y dy = 0
🟦 4️⃣ y dy + x dx = 0
🟡 Answer: 1️⃣ y dx – x dy = 0
📘 (JEE Advanced 2021 | Paper 1)

🔵 Question 7:
The order of (d³y/dx³)² + (dy/dx) = 0 is
🟥 1️⃣ 3
🟩 2️⃣ 2
🟨 3️⃣ 1
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 3
📘 (JEE Advanced 2021 | Paper 1)

🔵 Question 8:
The solution of (1 + x²) dy/dx = 2xy is
🟥 1️⃣ y = x² + C
🟩 2️⃣ y = 1 + x² + C
🟨 3️⃣ y = (1 + x²) + C
🟦 4️⃣ y = (1 + x²) + constant
🟡 Answer: 1️⃣ y = x² + C
📘 (JEE Advanced 2020 | Paper 1)

🔵 Question 9:
The integrating factor of dy/dx + (1/x) y = sin x is
🟥 1️⃣ x
🟩 2️⃣ 1/x
🟨 3️⃣ e^x
🟦 4️⃣ ln x
🟡 Answer: 1️⃣ x
📘 (JEE Advanced 2020 | Paper 1)

🔵 Question 10:
If y = Ce^(x²), then the corresponding differential equation is
🟥 1️⃣ dy/dx = 2xy
🟩 2️⃣ dy/dx = x²y
🟨 3️⃣ dy/dx = y/x
🟦 4️⃣ dy/dx = y
🟡 Answer: 1️⃣ dy/dx = 2xy
📘 (JEE Advanced 2019 | Paper 1)

🔵 Question 11:
Solution of dy/dx + y = 1 is
🟥 1️⃣ y = 1 + Ce^(-x)
🟩 2️⃣ y = 1 – Ce^x
🟨 3️⃣ y = Ce^(-x) + 1
🟦 4️⃣ y = Ce^x + 1
🟡 Answer: 1️⃣ y = 1 + Ce^(-x)
📘 (JEE Advanced 2019 | Paper 1)

🔵 Question 12:
If dy/dx = e^(x + y), then solution is
🟥 1️⃣ e^(-y) = e^x + C
🟩 2️⃣ e^(-y) = e^(-x) + C
🟨 3️⃣ e^y = e^x + C
🟦 4️⃣ e^y = e^(-x) + C
🟡 Answer: 1️⃣ e^(-y) = e^x + C
📘 (JEE Advanced 2018 | Paper 1)

🔵 Question 13:
The differential equation whose general solution is y = A cos x + B sin x is
🟥 1️⃣ d²y/dx² + y = 0
🟩 2️⃣ d²y/dx² – y = 0
🟨 3️⃣ d²y/dx² = y
🟦 4️⃣ None
🟡 Answer: 1️⃣ d²y/dx² + y = 0
📘 (JEE Advanced 2018 | Paper 1)

🔵 Question 14:
If y = Ae^(2x) + Be^(-2x), then the corresponding differential equation is
🟥 1️⃣ d²y/dx² = 4y
🟩 2️⃣ d²y/dx² = -4y
🟨 3️⃣ d²y/dx² – 4y = 0
🟦 4️⃣ d²y/dx² + 4y = 0
🟡 Answer: 1️⃣ d²y/dx² = 4y
📘 (JEE Advanced 2017 | Paper 1)

🔵 Question 15:
If dy/dx = y/x + sin(y/x), then putting y = vx gives
🟥 1️⃣ x dv/dx = sin v – v
🟩 2️⃣ x dv/dx = sin v + v
🟨 3️⃣ dv/dx = sin v
🟦 4️⃣ dv/dx = sin v / x
🟡 Answer: 1️⃣ x dv/dx = sin v – v
📘 (JEE Advanced 2017 | Paper 1)

🔵 Question 16:
The order and degree of the equation (d²y/dx²)³ + (dy/dx)² = y are
🟥 1️⃣ Order 2, Degree 3
🟩 2️⃣ Order 2, Degree 2
🟨 3️⃣ Order 3, Degree 2
🟦 4️⃣ Order 3, Degree 3
🟡 Answer: 1️⃣ Order 2, Degree 3
📘 (JEE Advanced 2016 | Paper 1)

🔵 Question 17:
If dy/dx = y/x, then the solution is
🟥 1️⃣ y = Cx
🟩 2️⃣ y = x + C
🟨 3️⃣ y = C/x
🟦 4️⃣ y = C
🟡 Answer: 1️⃣ y = Cx
📘 (JEE Advanced 2016 | Paper 1)

🔵 Question 18:
The order of the differential equation (d³y/dx³)² + (dy/dx) = 0 is
🟥 1️⃣ 3
🟩 2️⃣ 2
🟨 3️⃣ 1
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 3
📘 (JEE Advanced 2024 | Paper 2)

🔵 Question 19:
The solution of dy/dx = y + e^x is
🟥 1️⃣ y = Ce^x – e^x
🟩 2️⃣ y = Ce^x + e^x
🟨 3️⃣ y = Ce^(-x) – e^(-x)
🟦 4️⃣ None
🟡 Answer: 1️⃣ y = Ce^x – e^x
📘 (JEE Advanced 2023 | Paper 2)

🔵 Question 20:
The general solution of dy/dx = y tan x is
🟥 1️⃣ y = C sec x
🟩 2️⃣ y = C cos x
🟨 3️⃣ y = C sin x
🟦 4️⃣ None
🟡 Answer: 1️⃣ y = C sec x
📘 (JEE Advanced 2023 | Paper 2)

🔵 Question 21:
If dy/dx = e^(x + y), then the general solution is
🟥 1️⃣ e^(-y) = e^x + C
🟩 2️⃣ e^(-y) = e^(-x) + C
🟨 3️⃣ e^y = e^x + C
🟦 4️⃣ e^y = e^(-x) + C
🟡 Answer: 1️⃣ e^(-y) = e^x + C
📘 (JEE Advanced 2022 | Paper 2)

🔵 Question 22:
The differential equation representing all straight lines passing through origin is
🟥 1️⃣ y dx – x dy = 0
🟩 2️⃣ y dx + x dy = 0
🟨 3️⃣ dy/dx = m
🟦 4️⃣ None
🟡 Answer: 1️⃣ y dx – x dy = 0
📘 (JEE Advanced 2022 | Paper 2)

🔵 Question 23:
If y = Ce^(x²), then the differential equation is
🟥 1️⃣ dy/dx = 2xy
🟩 2️⃣ dy/dx = x²y
🟨 3️⃣ dy/dx = y/x
🟦 4️⃣ dy/dx = y
🟡 Answer: 1️⃣ dy/dx = 2xy
📘 (JEE Advanced 2021 | Paper 2)

🔵 Question 24:
The integrating factor of dy/dx + (1/x)y = sin x is
🟥 1️⃣ x
🟩 2️⃣ 1/x
🟨 3️⃣ e^x
🟦 4️⃣ ln x
🟡 Answer: 1️⃣ x
📘 (JEE Advanced 2021 | Paper 2)

🔵 Question 25:
The differential equation for the family of curves y² = 2cx is
🟥 1️⃣ y dy/dx = y
🟩 2️⃣ y dy/dx = x
🟨 3️⃣ y dy/dx = c
🟦 4️⃣ y dy/dx = 2x
🟡 Answer: 4️⃣ y dy/dx = 2x
📘 (JEE Advanced 2020 | Paper 2)

🔵 Question 26:
Solution of dy/dx + y = 1 is
🟥 1️⃣ y = 1 + Ce^(-x)
🟩 2️⃣ y = Ce^x – 1
🟨 3️⃣ y = 1 – Ce^x
🟦 4️⃣ y = e^x + 1
🟡 Answer: 1️⃣ y = 1 + Ce^(-x)
📘 (JEE Advanced 2020 | Paper 2)

🔵 Question 27:
If dy/dx = y/x + sin(y/x), then substitution y = vx gives
🟥 1️⃣ x dv/dx = sin v – v
🟩 2️⃣ x dv/dx = sin v + v
🟨 3️⃣ dv/dx = sin v
🟦 4️⃣ dv/dx = sin v / x
🟡 Answer: 1️⃣ x dv/dx = sin v – v
📘 (JEE Advanced 2019 | Paper 2)

🔵 Question 28:
The order and degree of (d²y/dx²)³ + (dy/dx)² = y are
🟥 1️⃣ Order 2, Degree 3
🟩 2️⃣ Order 2, Degree 2
🟨 3️⃣ Order 3, Degree 2
🟦 4️⃣ Order 3, Degree 3
🟡 Answer: 1️⃣ Order 2, Degree 3
📘 (JEE Advanced 2018 | Paper 2)

🔵 Question 29:
The general solution of dy/dx = y/x is
🟥 1️⃣ y = Cx
🟩 2️⃣ y = x + C
🟨 3️⃣ y = C/x
🟦 4️⃣ None
🟡 Answer: 1️⃣ y = Cx
📘 (JEE Advanced 2018 | Paper 2)

🔵 Question 30:
If y = Ae^(2x) + Be^(-2x), then the differential equation is
🟥 1️⃣ d²y/dx² = 4y
🟩 2️⃣ d²y/dx² + 4y = 0
🟨 3️⃣ d²y/dx² – 4y = 0
🟦 4️⃣ None
🟡 Answer: 1️⃣ d²y/dx² = 4y
📘 (JEE Advanced 2017 | Paper 2)

🔵 Question 31:
Solution of dy/dx + y = e^x is
🟥 1️⃣ y = Ce^(-x) + e^x – 1
🟩 2️⃣ y = Ce^x – e^x + 1
🟨 3️⃣ y = Ce^(-x) – e^x
🟦 4️⃣ None
🟡 Answer: 1️⃣ y = Ce^(-x) + e^x – 1
📘 (JEE Advanced 2016 | Paper 2)

🔵 Question 32:
If y = C e^(x² + 1), then dy/dx equals
🟥 1️⃣ 2xy
🟩 2️⃣ 2xy e^(x² + 1)
🟨 3️⃣ y/x
🟦 4️⃣ None
🟡 Answer: 2️⃣ 2xy e^(x² + 1)
📘 (JEE Advanced 2016 | Paper 2)

🔵 Question 33:
The differential equation whose solution is y = A cos x + B sin x is
🟥 1️⃣ d²y/dx² + y = 0
🟩 2️⃣ d²y/dx² – y = 0
🟨 3️⃣ d²y/dx² = y
🟦 4️⃣ None
🟡 Answer: 1️⃣ d²y/dx² + y = 0
📘 (JEE Advanced 2015 | Paper 2)

🔵 Question 34:
The general solution of dy/dx = y is
🟥 1️⃣ y = Ce^x
🟩 2️⃣ y = e^x + C
🟨 3️⃣ y = C + x
🟦 4️⃣ y = x + 1
🟡 Answer: 1️⃣ y = Ce^x
📘 (JEE Advanced 2015 | Paper 2)

PRACTICE SETS FROM THIS LESSON

Q1. The order of (d³y/dx³) + y = 0 is
🔵 (A) 1
🟢 (B) 2
🟠 (C) 3
🔴 (D) 0
Answer: (C) 3

Q2. The degree of (1 + x²)(d²y/dx²)² + (dy/dx)⁴ = 0 is
🔵 (A) 1
🟢 (B) 2
🟠 (C) 3
🔴 (D) 4
Answer: (D) 4

Q3. The differential equation of the family y = mx + c is
🔵 (A) dy/dx = m
🟢 (B) d²y/dx² = 0
🟠 (C) y − mx − c = 0
🔴 (D) d³y/dx³ = 0
Answer: (B) d²y/dx² = 0

Q4. The differential equation of circles centered at origin is
🔵 (A) x(dy/dx) − y = 0
🟢 (B) x(dy/dx) + y = 0
🟠 (C) y(dy/dx) − x = 0
🔴 (D) d²y/dx² = 0
Answer: (B) x(dy/dx) + y = 0

Q5. A solution containing one arbitrary constant represents
🔵 (A) General solution of first-order DE
🟢 (B) Particular solution
🟠 (C) Singular solution
🔴 (D) Complementary solution
Answer: (A) General solution of first-order DE

Q6. Solve dy/dx = 3x² gives
🔵 (A) y = x³ + C
🟢 (B) y = 3x³ + C
🟠 (C) y = x² + C
🔴 (D) y = 3x² + C
Answer: (A) y = x³ + C

Q7. Which is linear in y?
🔵 (A) dy/dx + y = eˣ
🟢 (B) dy/dx = y² + 1
🟠 (C) (dy/dx)² + y = 0
🔴 (D) dy/dx = y·sin y
Answer: (A) dy/dx + y = eˣ

Q8. dy/dx = y tan x is
🔵 (A) Separable only
🟢 (B) Linear only
🟠 (C) Both separable and linear
🔴 (D) Exact
Answer: (C) Both separable and linear

Q9. Integrating factor for dy/dx + P(x) y = Q(x) is
🔵 (A) e⁻∫P(x)dx
🟢 (B) e∫P(x)dx
🟠 (C) e∫Q(x)dx
🔴 (D) e∫(Q/P)dx
Answer: (B) e∫P(x)dx

Q10. (y dx − x dy) = 0 is
🔵 (A) Exact
🟢 (B) Homogeneous
🟠 (C) Linear
🔴 (D) Bernoulli
Answer: (B) Homogeneous

Q11. Solution of dy/dx = y is
🔵 (A) y = C eˣ
🟢 (B) y = C e⁻ˣ
🟠 (C) y = Cx
🔴 (D) y = C
Answer: (A) y = C eˣ

Q12. For dy/dx + y tan x = sin x, the integrating factor equals
🔵 (A) sec x
🟢 (B) cos x
🟠 (C) sin x
🔴 (D) tan x
Answer: (A) sec x

Q13. Equation (2xy + y²)dx + (x² + 2xy)dy = 0 is
🔵 (A) Linear
🟢 (B) Exact
🟠 (C) Homogeneous
🔴 (D) Bernoulli
Answer: (C) Homogeneous

Q14. Solution of dy/dx = x/y is
🔵 (A) y² = x² + C
🟢 (B) y² − x² = C
🟠 (C) y = Cx
🔴 (D) x² + y = C
Answer: (B) y² − x² = C

Q15. Order of d²y/dx² + (dy/dx)³ = 0 is
🔵 (A) 1
🟢 (B) 2
🟠 (C) 3
🔴 (D) 0
Answer: (B) 2

Q16. dy/dx + (1/x) y = 0 has solution
🔵 (A) y = Cx
🟢 (B) y = C/x
🟠 (C) y = Cx²
🔴 (D) y = C
Answer: (B) y = C/x

Q17. For dy/dx = ky, the solution is
🔵 (A) y = A eᵏˣ
🟢 (B) y = A e⁻ᵏˣ
🟠 (C) y = kx + C
🔴 (D) y = A kˣ
Answer: (A) y = A eᵏˣ

Q18. If dy/dx = cos x, then y equals
🔵 (A) sin x + C
🟢 (B) −sin x + C
🟠 (C) cos x + C
🔴 (D) tan x + C
Answer: (A) sin x + C

Q19. The DE of family y² = 2ax is
🔵 (A) y dy/dx = a/x
🟢 (B) y dy/dx = x
🟠 (C) d²y/dx² = 0
🔴 (D) x dy/dx = y/2
Answer: (B) y dy/dx = x

Q20. If M dx + N dy = 0 is exact, then
🔵 (A) ∂M/∂x = ∂N/∂y
🟢 (B) ∂M/∂y = ∂N/∂x
🟠 (C) M = N
🔴 (D) dM = dN
Answer: (B) ∂M/∂y = ∂N/∂x

────────────────────────────────
Q21–Q40 (JEE Main Level)

Q21. Solve dy/dx + y = eˣ. The general solution is
🔵 (A) y = ½eˣ + C e⁻ˣ
🟢 (B) y = eˣ + C
🟠 (C) y = eˣ − C e⁻ˣ
🔴 (D) y = C eˣ
Answer: (A) y = ½eˣ + C e⁻ˣ

Q22. dy/dx = (x + y)/(x − y) is homogeneous. Put y = vx. Then dv/dx equals
🔵 (A) (1 + v)/(1 − v)
🟢 (B) (1 − v)/(1 + v)
🟠 (C) v/x
🔴 (D) (1 + v²)/(x(1 − v)²)
Answer: (A) (1 + v)/(1 − v)

Q23. For dy/dx + (2/x) y = x, the integrating factor is
🔵 (A) x²
🟢 (B) x⁻²
🟠 (C) eˣ
🔴 (D) eˣ²
Answer: (A) x²

Q24. Particular solution of dy/dx = x y with y(0) = 3 is
🔵 (A) y = 3 eˣ²⸝²
🟢 (B) y = 3 eˣ
🟠 (C) y = 3 eˣ³
🔴 (D) y = 3(1 + x)
Answer: (A) y = 3 eˣ²⸝²

Q25. For (y + x)dx + (y − x)dy = 0, classify
🔵 (A) Exact
🟢 (B) Homogeneous
🟠 (C) Linear in y
🔴 (D) Bernoulli
Answer: (B) Homogeneous

Q26. The solution of dy/dx = (1 + y²)/(1 + x²) is
🔵 (A) tan⁻¹y = tan⁻¹x + C
🟢 (B) tan⁻¹y = ½ ln(1 + x²) + C
🟠 (C) y = x + C
🔴 (D) y = tan x + C
Answer: (A) tan⁻¹y = tan⁻¹x + C

Q27. If (2x − y)dx + (x − 2y)dy = 0 is exact, then potential φ satisfies
🔵 (A) φₓ = 2x − y, φᵧ = x − 2y
🟢 (B) φₓ = x − 2y, φᵧ = 2x − y
🟠 (C) φₓ = 2x + y, φᵧ = x + 2y
🔴 (D) φₓ = φᵧ
Answer: (A) φₓ = 2x − y, φᵧ = x − 2y

Q28. For dy/dx + y tan x = sec x, I.F. equals
🔵 (A) sec x
🟢 (B) cos x
🟠 (C) tan x
🔴 (D) 1 + tan x
Answer: (A) sec x

Q29. Solution of dy/dx = (y − x)/(y + x) after y = vx substitution reduces to
🔵 (A) (1 − v)/(1 + v)
🟢 (B) (1 + v)/(1 − v)
🟠 (C) v + x dv/dx = (v − 1)/(v + 1)
🔴 (D) v + x dv/dx = (1 − v)/(1 + v)
Answer: (D) v + x dv/dx = (1 − v)/(1 + v)

Q30. The DE y′ + y = 0 with y(0) = 5 gives
🔵 (A) y = 5 e⁻ˣ
🟢 (B) y = 5 eˣ
🟠 (C) y = 5(1 − x)
🔴 (D) y = 5/x
Answer: (A) y = 5 e⁻ˣ

Q31. If dy/dx = eˣ − y and y(0) = 0, then y(x) equals
🔵 (A) ½(eˣ − e⁻ˣ)
🟢 (B) 1 − e⁻ˣ
🟠 (C) eˣ − 1
🔴 (D) eˣ
Answer: (B) 1 − e⁻ˣ

Q32. For logistic form dy/dx = ky(1 − y/M), the equilibria are
🔵 (A) y = 0, y = M
🟢 (B) y = k, y = M
🟠 (C) y = 1, y = M
🔴 (D) y = 0, y = k
Answer: (A) y = 0, y = M

Q33. Solve dx/dy + x = y. The I.F. in x(y) form is
🔵 (A) eʸ
🟢 (B) e⁻ʸ
🟠 (C) y
🔴 (D) 1 + y
Answer: (A) eʸ

Q34. For exactness of M dx + N dy = 0, which is necessary and sufficient?
🔵 (A) ∂M/∂y = ∂N/∂x in simply connected domain
🟢 (B) ∂M/∂x = ∂N/∂y
🟠 (C) M = N
🔴 (D) dM = dN
Answer: (A) ∂M/∂y = ∂N/∂x in simply connected domain

Q35. If dy/dx = y cot x, then y equals
🔵 (A) C sin x
🟢 (B) C cos x
🟠 (C) C tan x
🔴 (D) C csc x
Answer: (B) C cos x

Q36. The DE of family y = A eᵏˣ + B e⁻ᵏˣ is of order
🔵 (A) 1
🟢 (B) 2
🟠 (C) k
🔴 (D) 0
Answer: (B) 2

Q37. Solve dy/dx = (x² + y²)/(xy). Under y = vx, the reduced separable form is
🔵 (A) (v/(1 − v²)) dv = dx/x
🟢 (B) (2v/(1 − v²)) dv = dx/x
🟠 (C) (v/(1 + v²)) dv = dx/x
🔴 (D) (2v/(1 + v²)) dv = dx/x
Answer: (B) (2v/(1 − v²)) dv = dx/x

Q38. For dy/dx + (2/x) y = 0, the power of x multiplying C in the solution is
🔵 (A) x⁻²
🟢 (B) x²
🟠 (C) x
🔴 (D) x⁻¹
Answer: (A) x⁻² (since y = C x⁻²)

Q39. If y = 0 when x = 0 for dy/dx = x + y, then y(x) equals
🔵 (A) eˣ − 1 − x
🟢 (B) eˣ − 1
🟠 (C) x
🔴 (D) 1 − e⁻ˣ
Answer: (B) eˣ − 1

Q40. For dy/dx = (1 − y)/(1 + x), with y(0) = 0, solution is
🔵 (A) y = 1 − 1/(1 + x)
🟢 (B) y = (1 + x) − 1
🟠 (C) y = (1 + x)
🔴 (D) y = 1/(1 + x)
Answer: (A) y = 1 − 1/(1 + x)

────────────────────────────────
Q41–Q50 (JEE Advanced Level)

Q41. Solve (1 + y²) dx = (1 + x²) dy. The implicit solution is
🔵 (A) tan⁻¹x − tan⁻¹y = C
🟢 (B) x − y = C
🟠 (C) x² − y² = C
🔴 (D) ln(1 + x²) − ln(1 + y²) = C
Answer: (A) tan⁻¹x − tan⁻¹y = C

Q42. For dy/dx + y sec x = tan x sec x, solution is
🔵 (A) y = 1 + C/(sec x + tan x)
🟢 (B) y = sec x + tan x
🟠 (C) y = C(sec x + tan x)
🔴 (D) y = tan x
Answer: (A) y = 1 + C/(sec x + tan x)

Q43. Solve homogeneous DE (x² + y²)dx − 2xy dy = 0. One implicit form is
🔵 (A) x² − y² = Cx
🟢 (B) x² + y² = C
🟠 (C) y² − x² = Cx
🔴 (D) xy = C
Answer: (A) x² − y² = Cx

Q44. For Bernoulli DE dy/dx + P(x) y = Q(x) yⁿ (n ≠ 1), the linearizing substitution is
🔵 (A) u = y¹⁻ⁿ
🟢 (B) u = yⁿ
🟠 (C) u = 1/y
🔴 (D) u = yⁿ⁻¹
Answer: (A) u = y¹⁻ⁿ

Q45. If d²y/dx² − y = 0 with y(0) = 2, y′(0) = 0, then y(x) equals
🔵 (A) 2 cosh x
🟢 (B) 2 sinh x
🟠 (C) eˣ
🔴 (D) 2
Answer: (A) 2 cosh x

Q46. For dy/dx = (x + y + 1)², set z = x + y + 1. Then dz/dx equals
🔵 (A) 1 + z²
🟢 (B) z² − 1
🟠 (C) z
🔴 (D) 2z
Answer: (A) 1 + z²

Q47. Solve dx/dy + x tan y = sin y. The integrating factor in x(y) form is
🔵 (A) e∫tan y dy = sec y
🟢 (B) e∫tan y dy = cos y
🟠 (C) e∫tan y dy = sin y
🔴 (D) e∫tan y dy = csc y
Answer: (A) sec y

Q48. For exact DE (3x²y + y³) dx + (x³ + 3xy²) dy = 0, the potential φ(x, y) is
🔵 (A) x³y + x y³
🟢 (B) 3x²y + y³
🟠 (C) x³ + y³
🔴 (D) x³y + 3xy²
Answer: (A) x³y + x y³ (since φₓ = 3x²y + y³, φᵧ = x³ + 3xy²)

Q49. Population P satisfies dP/dt = kP(1 − P/M). With P(0) = P₀, solution is
🔵 (A) P = M / (1 + A e⁻ᵏᵗ)
🟢 (B) P = M / (1 + A eᵏᵗ)
🟠 (C) P = A eᵏᵗ
🔴 (D) P = M(1 − e⁻ᵏᵗ)
Answer: (A) P = M / (1 + A e⁻ᵏᵗ)

Q50. Consider d²y/dx² + 2 dy/dx + y = 0. Nature of solution is
🔵 (A) Overdamped: y = (C₁ + C₂x) e⁻ˣ
🟢 (B) Underdamped: oscillatory with e⁻ˣ factor
🟠 (C) Critically damped: y = (C₁ + C₂x)
🔴 (D) Pure oscillation: y = C₁ cos x + C₂ sin x
Answer: (A) Overdamped: y = (C₁ + C₂x) e⁻ˣ

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