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Differential Equations Exercise Solutions | KalviGavani

TN Class 12 Business Maths Chapter 4 exercise solutions — order, degree, variable separable, homogeneous, linear differential equations with worked examples.

Updated 15 June 202613 min read

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These are complete exercise solutions for Samacheer Kalvi 12th Business Maths — Differential Equations (Chapter 4). Solve each question yourself first, then verify below. This chapter is calculation-heavy — practice the methods repeatedly until they become automatic.

Key Concepts (Quick Recap)

  • Differential equation (DE): An equation involving derivatives of a function (e.g., dy/dx, d²y/dx²).
  • Order: The highest order derivative present. Example: d²y/dx² + y = 0 has order 2.
  • Degree: The power of the highest-order derivative after making the DE polynomial in derivatives. Example: (dy/dx)² + y = 0 has degree 2.
  • Variable separable method: Rewrite as f(y) dy = g(x) dx, then integrate both sides.
  • Homogeneous DE: A DE of the form dy/dx = f(x,y) where f is homogeneous (f(tx,ty) = f(x,y)). Substitute y = vx, dy/dx = v + x dv/dx.
  • Linear first-order DE: dy/dx + Py = Q, where P and Q are functions of x (or constants).
  • Integrating factor: IF = e^(∫P dx)
Solutiony · IF = ∫(Q · IF) dx + C
  • Formation of DE: Given a family of curves with n arbitrary constants, differentiate n times to eliminate all constants and obtain the DE.
  • Particular solution: Obtained by substituting given conditions (initial/boundary conditions) into the general solution to find C.

I. Choose the Best Answer (MCQs)

Q1. The order and degree of (d²y/dx²)³ + (dy/dx)² + y = 0 are respectively: (a) 2, 3 ✓ (b) 3, 2 (c) 2, 2 (d) 1, 3

Answer(a) ✓ — Highest-order derivative is d²y/dx² (order 2). Its power is 3 (degree 3). Order = 2, Degree = 3.

Q2. The degree of the DE (d²y/dx²)^(3/2) = (dy/dx) + y is: (a) 3 (b) 3/2 (c) Not defined ✓ (d) 2

Answer(c) ✓ — The derivative d²y/dx² appears with a fractional power (3/2). The degree is not defined because we cannot express the DE as a polynomial in derivatives without raising both sides to powers, which changes the DE.

Q3. The solution of dy/dx = 2x is: (a) y = x² + C ✓ (b) y = 2x + C (c) y = x + C (d) y = 2x² + C

Answer(a) ✓ — ∫dy = ∫2x dx → y = x² + C.

Q4. The integrating factor for dy/dx + 2y = 3 is: (a) e^(2x) ✓ (b) e^x (c) e^(3x) (d) e^(−2x)

Answer(a) ✓ — P = 2, IF = e^(∫2 dx) = e^(2x).

Q5. The DE formed from y = aeˣ + be^(−x) (a, b arbitrary constants) is: (a) d²y/dx² − y = 0 ✓ (b) d²y/dx² + y = 0 (c) dy/dx − y = 0 (d) d²y/dx² = 0

Answer(a) ✓ — dy/dx = aeˣ − be^(−x). d²y/dx² = aeˣ + be^(−x) = y. So d²y/dx² − y = 0.

Q6. The particular solution of dy/dx = 3x² passing through (1,2) is: (a) y = x³ + 1 ✓ (b) y = x³ + 2 (c) y = x³ (d) y = 3x³ + 1

Answer(a) ✓ — y = x³ + C. At (1,2): 2 = 1 + C → C = 1. So y = x³ + 1.

Q7. The DE dy/dx = x/y is: (a) Linear (b) Variable separable ✓ (c) Homogeneous (d) Exact

Answer(b) ✓ — Can be written as y dy = x dx (variables separated). While it is also homogeneous, the most direct method is variable separable.

Q8. If y = e^(2x) is a solution of d²y/dx² − 4y = 0, then another independent solution is: (a) e^(−2x) ✓ (b) e^x (c) e^(4x) (d) 2e^(2x)

Answer(a) ✓ — The characteristic equation is m² − 4 = 0 → m = ±2. Solutions: e^(2x) and e^(−2x).

II. Fill in the Blanks

1. The order of d⁴y/dx⁴ + (d³y/dx³)² + y = 0 is ______. 4

2. The degree of (dy/dx)² + y = 0 is ______. 2

3. The integrating factor of dy/dx + y/x = x² is ______. e^(∫1/x dx) = e^(ln x) = x

4. The general solution of dy/dx = e^x is ______. y = e^x + C

5. A DE of first order and first degree has ______ arbitrary constant(s) in its general solution. 1

6. The DE formed from y = mx + c (where m and c are arbitrary) has order ______. 2

7. The integrating factor for dy/dx + (1/x)y = x is ______. x

III. True or False

1. The order of a DE is always equal to the number of arbitrary constants in its general solution. → True. A first-order DE has one arbitrary constant, a second-order DE has two, etc.

2. The degree of dy/dx + y = 0 is 1. → True. The highest-order derivative dy/dx has power 1, and the equation is already polynomial in derivatives.

3. Every homogeneous DE can be solved by variable separation. → False. Homogeneous DEs require the substitution y = vx (or x = vy), not simple variable separation. They are not directly separable.

4. The integrating factor for dy/dx + Py = Q is always e^(∫P dx). → True. This is the definition of the integrating factor for a linear first-order DE in standard form dy/dx + Py = Q.

5. dy/dx = x + y is a homogeneous DE. → False. A homogeneous DE must be expressible as dy/dx = f(y/x). x + y cannot be written purely as a function of y/x.

6. A linear DE has the form dy/dx + Py = Q where P and Q are functions of x only. → True. This is the standard form of a linear first-order DE. P and Q may be constants or functions of x, but not functions of y.

IV. Match the Following

LeftRight
1. Order of DEc. Highest order derivative
2. Degree of DEd. Power of highest-order derivative (after polynomial form)
3. Variable separablee. f(y) dy = g(x) dx
4. Integrating factora. e^(∫P dx)
5. Homogeneous DE substitutionb. y = vx
Answer1→c, 2→d, 3→e, 4→a, 5→b

V. Answer Briefly

Q1. Find the order and degree of: (d³y/dx³)² + (d²y/dx²)⁴ + dy/dx = x.
Answer

Order = 3 (highest derivative is d³y/dx³) Degree = 2 (power of d³y/dx³ after writing in polynomial form)

Q2. Form the differential equation from y = Ae^(3x) + Be^(−3x).
Answer

dy/dx = 3Ae^(3x) − 3Be^(−3x) d²y/dx² = 9Ae^(3x) + 9Be^(−3x) = 9(Ae^(3x) + Be^(−3x)) = 9y d²y/dx² − 9y = 0

Q3. Solve dy/dx = (x+1)/(y+1).
Answer

This is variable separable: (y+1) dy = (x+1) dx ∫(y+1) dy = ∫(x+1) dx y²/2 + y = x²/2 + x + C y² + 2y = x² + 2x + C₁

Q4. Find the integrating factor of dy/dx + y cos x = sin x cos x.
Answer

P = cos x IF = e^(∫cos x dx) = e^(sin x)

Q5. Solve dy/dx = e^(x−y) given that y(0) = 0.
Answer

Variable separable: e^y dy = e^x dx ∫e^y dy = ∫e^x dx e^y = e^x + C At x = 0, y = 0: e⁰ = e⁰ + C → 1 = 1 + C → C = 0 e^y = e^x → y = x

Q6. Verify that y = 3 sin x + 4 cos x is a solution of d²y/dx² + y = 0.
Answer

dy/dx = 3 cos x − 4 sin x d²y/dx² = −3 sin x − 4 cos x = −(3 sin x + 4 cos x) = −y d²y/dx² + y = −y + y = 0 ✓ Verified.

Q7. Find the particular solution of dy/dx = 2y given y(0) = 3.
Answer

dy/y = 2 dx → ∫dy/y = 2∫dx → ln|y| = 2x + C → y = Ae^(2x) At x = 0, y = 3: 3 = Ae⁰ → A = 3 y = 3e^(2x)

VI. Answer in Detail

Q1. Solve the DE: dy/dx = (x² + y²)/(2xy).
Answer

Step 1: Identify as homogeneous dy/dx = (x² + y²)/(2xy) = (1 + (y/x)²)/(2(y/x)) This is of the form f(y/x), so it's a homogeneous DE. Step 2: Substitute y = vx, dy/dx = v + x(dv/dx) v + x(dv/dx) = (1 + v²)/(2v) x(dv/dx) = (1 + v²)/(2v) − v = (1 + v² − 2v²)/(2v) = (1 − v²)/(2v) Step 3: Separate variables (2v/(1 − v²)) dv = dx/x Step 4: Integrate both sides ∫(2v/(1 − v²)) dv = ∫dx/x Let u = 1 − v², du = −2v dv → −du = 2v dv −∫du/u = ∫dx/x → −ln|u| = ln|x| + C₁ −ln|1 − v²| = ln|x| + C₁ ln|x| + ln|1 − v²| = −C₁ ln|x(1 − v²)| = C₂ x(1 − v²) = C₃ Step 5: Substitute back v = y/x x(1 − y²/x²) = C₃ x · (x² − y²)/x² = C₃ (x² − y²)/x = C (or equivalently, x² − y² = Cx)

Q2. Solve the linear DE: dy/dx + 3y = e^(−3x).
Answer

Step 1: Identify P and Q P = 3, Q = e^(−3x) Step 2: Find integrating factor IF = e^(∫3 dx) = e^(3x) Step 3: Multiply DE by IF e^(3x) · dy/dx + 3e^(3x) · y = e^(3x) · e^(−3x) = 1 d/dx[y · e^(3x)] = 1 Step 4: Integrate y · e^(3x) = ∫1 dx = x + C y = (x + C)e^(−3x) Verification: dy/dx = e^(−3x) − 3(x + C)e^(−3x). Then dy/dx + 3y = e^(−3x) − 3(x+C)e^(−3x) + 3(x+C)e^(−3x) = e^(−3x) ✓

Q3. Solve: dy/dx + (2/x)y = x³, given y(1) = 1.
Answer

Step 1: Identify P and Q P = 2/x, Q = x³ Step 2: Find IF IF = e^(∫(2/x) dx) = e^(2 ln x) = e^(ln x²) = x² Step 3: Multiply by IF x² · dy/dx + 2x · y = x² · x³ = x⁵ d/dx[y · x²] = x⁵ Step 4: Integrate y · x² = ∫x⁵ dx = x⁶/6 + C y = x⁴/6 + C/x² Step 5: Apply initial condition y(1) = 1 1 = 1/6 + C/1 → C = 1 − 1/6 = 5/6 y = x⁴/6 + 5/(6x²)

Q4. Solve: (x² + 1) dy/dx + 2xy = 4x².
Answer

Step 1: Rewrite in standard form dy/dx + (2x/(x² + 1)) y = 4x²/(x² + 1) P = 2x/(x² + 1), Q = 4x²/(x² + 1) Step 2: Find IF IF = e^(∫(2x/(x²+1)) dx) = e^(ln(x²+1)) = x² + 1 Step 3: Multiply by IF (x² + 1) · dy/dx + 2xy = 4x² d/dx[y(x² + 1)] = 4x² Step 4: Integrate y(x² + 1) = ∫4x² dx = 4x³/3 + C y = (4x³/3 + C)/(x² + 1)

Q5. Form the differential equation from y = a sin x + b cos x + e^x, where a and b are arbitrary constants.
Answer

Step 1: dy/dx = a cos x − b sin x + e^x Step 2: d²y/dx² = −a sin x − b cos x + e^x Step 3: d²y/dx² − y = (−a sin x − b cos x + e^x) − (a sin x + b cos x + e^x) = −2a sin x − 2b cos x But we need to eliminate a and b. From the original: a sin x + b cos x = y − e^x From dy/dx: a cos x − b sin x = dy/dx − e^x From d²y/dx² = −(a sin x + b cos x) + e^x = −(y − e^x) + e^x = −y + 2e^x d²y/dx² + y = 2e^x

Q6. A bacterial culture grows at a rate proportional to its size. Initially there are 100 bacteria, and after 2 hours there are 300. Find (a) the population after 5 hours, and (b) when the population reaches 1000.
Answer

Step 1: Form the DE dN/dt = kN (growth is proportional to size) dN/N = k dt → ∫dN/N = ∫k dt → ln N = kt + C → N = N₀e^(kt) Step 2: Find constants N(0) = 100: 100 = N₀e⁰ → N₀ = 100 N(2) = 300: 300 = 100e^(2k) → e^(2k) = 3 → 2k = ln 3 → k = (ln 3)/2 N(t) = 100e^((ln 3)t/2) = 100 · 3^(t/2) (a) Population after 5 hours: N(5) = 100 · 3^(5/2) = 100 · 3^2 · 3^(1/2) = 100 · 9 · √3 ≈ 100 · 9 · 1.732 ≈ 1558 bacteria (b) When population reaches 1000: 1000 = 100 · 3^(t/2) 10 = 3^(t/2) ln 10 = (t/2) ln 3 t = 2 ln 10 / ln 3 ≈ 2(2.3026)/(1.0986) ≈ 4.19 hours

Common Mistakes to Avoid

1. Confusing order and degree: Order is the highest derivative; degree is its power. (d²y/dx²)³ has order 2, degree 3 — they are not the same thing.

2. Degree not defined for fractional powers: If the DE contains (dy/dx)^(1/2) or sqrt(dy/dx), the degree is not defined until you remove the fractional power by squaring — which changes the DE. Report "degree not defined" in such cases.

3. Forgetting the constant when integrating: In variable separable DEs, always add + C on one side after integration. A common error is adding C to both sides or forgetting it entirely.

4. Wrong substitution for homogeneous DEs: The correct substitution is y = vx (not x = vy or anything else in most cases). Don't forget that dy/dx = v + x(dv/dx).

5. Incorrect integrating factor calculation: When P is a fraction like 2/x, compute ∫P dx = ∫(2/x) dx = 2 lnx, so IF = e^(2 ln x) = x². Students often miscalculate the integral of P.

6. Not applying initial conditions: After finding the general solution, substitute the given condition (e.g., y(0) = 3) to find C. The particular solution is what examiners want, not just the general solution.

Exam Checklist

  • [ ] I can determine the order and degree of a DE
  • [ ] I can identify when a DE has undefined degree (fractional powers of derivatives)
  • [ ] I can form a DE from a given family of curves by differentiation
  • [ ] I can solve variable-separable DEs by separating and integrating
  • [ ] I can solve homogeneous DEs using the y = vx substitution
  • [ ] I can solve linear first-order DEs using the integrating factor method
  • [ ] I can apply initial/boundary conditions to find particular solutions
  • [ ] I can model real-world problems (growth, decay, etc.) as DEs and solve them

Frequently Asked Questions

The order of a differential equation is the highest order derivative present in the equation. For example, d²y/dx² + dy/dx + y = 0 is a second-order DE.
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