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Differential Equations Exercise Solutions | KalviGavani

Samacheer Kalvi 11th Business Maths Differential Equations exercise solutions — formation, order, degree, first order first degree, solving methods, worked examples, MCQs, full answers.

Updated 15 June 202617 min read

How to Use This Page

This page covers Samacheer Kalvi Class 11 Business Maths — Chapter: Differential Equations with complete exercise solutions.

  • Quick Recap tables refresh every formula and definition you need.
  • MCQs, Fill-ups, True/False, Match the Following give fast practice for objective sections.
  • Brief and Detailed Answers provide step-by-step worked solutions.
  • Common Mistakes and Exam Checklist help you avoid losing marks.

Scroll to any section or use Ctrl+F to jump to a specific question number.


Key Concepts (Quick Recap)

What Is a Differential Equation?

A differential equation (DE) is an equation that involves one or more derivatives of an unknown function with respect to one or more independent variables.

TermDefinition
Ordinary DE (ODE)A DE involving derivatives with respect to a single independent variable
Partial DE (PDE)A DE involving partial derivatives with respect to two or more independent variables
OrderThe highest order of the derivative present in the DE
DegreeThe power of the highest-order derivative after the DE is made polynomial in derivatives (no fractional or negative powers on derivatives)

Order and Degree — Rules

RuleExplanation
Order is always a positive integerIt equals the order of the highest derivative: d³y/dx³ means order 3
Degree is defined only when the DE is polynomial in derivativesIf dy/dx = sin(x), degree is 1; if (dy/dx)² + y = 0, degree is 2
No fractional or negative powers on derivatives for degreeIf (d²y/dx²)^(3/2) = x, first square both sides then find degree
Order must be determined before degreeAlways find order first, then determine degree

Formation of a Differential Equation

To form a DE from a family of curves:

StepAction
1Write the equation of the family with arbitrary constants
2Differentiate as many times as the number of arbitrary constants
3Eliminate the arbitrary constants using the original and differentiated equations
4The resulting DE has order = number of arbitrary constants eliminated

Key point: A DE with order n is formed by eliminating n arbitrary constants.

Types of First-Order First-Degree DEs and Solving Methods

TypeFormMethod
Variable Separabledy/dx = f(x) · g(y)Separate: (1/g(y))dy = f(x)dx, then integrate
Homogeneousdy/dx = f(y/x)Substitute y = vx, so dy/dx = v + x(dv/dx)
Linear (First Order)dy/dx + Py = QFind IF = e^(∫P dx); Solution: y · IF = ∫(Q · IF) dx + C

Integrating Factor (IF)

For dy/dx + P(x)·y = Q(x): IF = e^(∫P dx) General solution: y · IF = ∫(Q · IF) dx + C

General Solution vs Particular Solution

TypeDefinition
General SolutionContains arbitrary constant(s) — represents a family of curves
Particular SolutionObtained by substituting given initial/boundary conditions to find the constant

I. Choose the Best Answer (MCQs)

1. The order of the differential equation (d²y/dx²) + 3(dy/dx) + 2y = 0 is:

(A) 1 (B) 2 (C) 3 (D) 0

Answer(B) 2

The highest derivative is d²y/dx², which is second order. Hence the order is 2.


2. The degree of the differential equation (dy/dx)² + y = x is:

(A) 1 (B) 2 (C) 3 (D) Not defined

Answer(B) 2

The equation is already polynomial in derivatives. The highest-order derivative dy/dx is raised to the power 2, so degree = 2.


3. The integrating factor of dy/dx + (2/x)y = x is:

(A) x² (B) e^(x²) (C) x⁻² (D) e^(2x)

Answer(A) x²
P = 2/x ∫P dx = ∫(2/x) dx = 2 ln|x| = ln(x²) IF = e^(ln x²) = x²

4. The differential equation representing the family y = ae^x + be^(−x) is:

(A) d²y/dx² + y = 0 (B) d²y/dx² − y = 0 (C) d²y/dx² + 2y = 0 (D) dy/dx = y

Answer(B) d²y/dx² − y = 0
y = ae^x + be^(−x) ... (1) dy/dx = ae^x − be^(−x) ... (2) d²y/dx² = ae^x + be^(−x) = y (from equation 1) Therefore: d²y/dx² − y = 0

5. If dy/dx = x/y, then the general solution is:

(A) x² + y² = C (B) x² − y² = C (C) y² − x² = C (D) xy = C

Answer(C) y² − x² = C
dy/dx = x/y y dy = x dx Integrating both sides: ∫y dy = ∫x dx y²/2 = x²/2 + C₁ y² − x² = 2C₁ = C Therefore: y² − x² = C

6. A homogeneous differential equation is one in which:

(A) All terms are constants (B) f(x, y) can be written as f(y/x) (C) The degree of each term is different (D) The equation has no solution

Answer(B) f(x, y) can be written as f(y/x)

A first-order DE dy/dx = f(x,y) is homogeneous if f(tx, ty) = f(x,y) for all t ≠ 0, which means f can be expressed as a function of y/x.


7. The order and degree of [1 + (dy/dx)²]^(3/2) = d²y/dx² are respectively:

(A) 2, 3 (B) 2, 3/2 (C) 2, 2 (D) 2, 1

Answer(C) 2, 2
The highest-order derivative is d²y/dx² → order = 2. To find degree, make it polynomial in derivatives: [1 + (dy/dx)²]^(3/2) = d²y/dx² Square both sides: [1 + (dy/dx)²]³ = (d²y/dx²)² The highest-order derivative is d²y/dx², raised to power 2. Degree = 2.

II. Fill in the Blanks

1. The order of a differential equation is the highest ________ of the derivative present in it.

Answerorder

The order of a differential equation is the highest order of the derivative present in it.


2. The degree of a differential equation is defined only when it is ________ in its derivatives.

Answerpolynomial

The degree is defined only when the DE is expressible as a polynomial in derivatives — no fractional or negative powers on derivatives.


3. To eliminate n arbitrary constants from a family of curves, we need to differentiate ________ times.

Answern

The number of arbitrary constants equals the order of the DE formed, which equals the number of times we differentiate.


4. The integrating factor for the linear DE dy/dx + P(x)y = Q(x) is ________.

Answere^(∫P dx)

IF = e^(∫P dx), where P is the coefficient of y in the standard linear form.


5. If dy/dx = e^(x−y), then separating variables gives e^y dy = ________ dx.

Answere^x
dy/dx = e^(x−y) = e^x · e^(−y) e^y dy = e^x dx

6. In a homogeneous DE, the substitution y = vx transforms dy/dx into ________.

Answerv + x(dv/dx)

With y = vx, differentiating by the product rule: dy/dx = v + x(dv/dx).


7. The DE dy/dx = (x + y)/(x − y) is ________ (homogeneous / non-homogeneous).

Answerhomogeneous

f(x,y) = (x+y)/(x−y). Check: f(tx,ty) = (tx+ty)/(tx−ty) = (x+y)/(x−y) = f(x,y). Since f is homogeneous of degree 0, the DE is homogeneous.


8. The general solution of dy/dx = 1 contains ________ arbitrary constant(s).

Answerone

Integrating: y = x + C. One arbitrary constant because the order is 1.


9. For the DE dy/dx + y/x = x², the integrating factor IF = ________.

Answerx
P = 1/x ∫P dx = ∫(1/x) dx = ln|x| IF = e^(ln x) = x

10. A particular solution is obtained from the general solution by substituting ________ conditions.

Answerinitial (or boundary)

A particular solution is found by applying given initial/boundary conditions to determine the value of the arbitrary constant.


11. The differential equation of the family y = mx + c (m and c arbitrary) has order ________.

Answer2

Two arbitrary constants (m and c) means we differentiate twice and eliminate both constants, so the resulting DE has order 2.


12. If IF is the integrating factor, the solution of dy/dx + Py = Q is y · IF = ________ + C.

Answer∫(Q · IF) dx

The standard formula: y · IF = ∫(Q · IF) dx + C.


III. True or False

1. The degree of a differential equation is always defined.

AnswerFalse

The degree is defined only when the DE is polynomial in its derivatives. For example, sin(dy/dx) + y = 0 has no defined degree because the derivative appears inside a trigonometric function.


2. The order of a differential equation can be zero.

AnswerFalse

A differential equation must contain at least one derivative; otherwise it is not a differential equation. Hence the order is always a positive integer (1, 2, 3, ...).


3. The differential equation formed from y = ae^(2x) + be^(3x) has order 2.

AnswerTrue
Two arbitrary constants a and b → differentiate twice → order = 2. y = ae^(2x) + be^(3x) y' = 2ae^(2x) + 3be^(3x) y'' = 4ae^(2x) + 9be^(3x) Eliminating a and b gives y'' − 5y' + 6y = 0, which is order 2.

4. The integrating factor for dy/dx + y = e^x is e^x.

AnswerTrue
P = 1 ∫P dx = ∫1 dx = x IF = e^(x) = e^x The statement is correct.

5. A first-order linear DE must be of the form dy/dx + Py = Q where P and Q are constants.

AnswerFalse

P and Q can be functions of x — they need not be constants. For example, dy/dx + (1/x)y = x² has P = 1/x and Q = x², which are functions of x.


IV. Match the Following

Match the items in Column A with the correct items in Column B.

Column AColumn B
(A) Order of (d³y/dx³) + y = 0(p) e^(∫P dx)
(B) Degree of (d²y/dx²)³ + y = 0(q) 3
(C) Integrating factor(r) y = vx
(D) Homogeneous DE substitution(s) ∫f(x)dx = ∫g(y)dy
(E) Variable separable method(t) 3
(F) General solution constant(u) arbitrary
Answer
Column AMatchReason
(A) Order of (d³y/dx³) + y = 0(q) 3The highest derivative is third order
(B) Degree of (d²y/dx²)³ + y = 0(t) 3The highest-order derivative (d²y/dx²) is raised to power 3
(C) Integrating factor(p) e^(∫P dx)Standard formula for linear first-order DE
(D) Homogeneous DE substitution(r) y = vxStandard substitution for homogeneous DEs
(E) Variable separable method(s) ∫f(x)dx = ∫g(y)dySeparate and integrate both sides
(F) General solution constant(u) arbitraryGeneral solution always contains an arbitrary constant

V. Answer Briefly

1. Define the order and degree of a differential equation with an example.

Answer
  • Order: The order of a DE is the order of the highest derivative present in the equation.
  • Degree: The degree of a DE is the exponent of the highest-order derivative, provided the equation is polynomial in its derivatives (no fractional or negative powers on derivatives).

Example: Consider the DE:

[1 + (dy/dx)²]³/² = d²y/dx²
  • Order = 2 (highest derivative is d²y/dx²)
  • To find degree, square both sides: [1 + (dy/dx)²]³ = (d²y/dx²)²
  • Degree = 2 (power of the highest-order derivative d²y/dx²)

2. Form the differential equation from y = mx + c, where m and c are arbitrary constants.

Answer
y = mx + c ... (1) Differentiating w.r.t. x: dy/dx = m ... (2) Differentiating again: d²y/dx² = 0 The required DE is: d²y/dx² = 0

Since there are two arbitrary constants (m and c), we differentiate twice. The resulting DE is d²y/dx² = 0, which is of order 2.


3. Write the standard form of a first-order linear differential equation and give the formula for its solution.

Answer

The standard form of a first-order linear DE is:

dy/dx + P(x)·y = Q(x) where P and Q are functions of x (or constants). Integrating Factor: IF = e^(∫P dx) General Solution: y · IF = ∫(Q · IF) dx + C Or equivalently: y = [∫(Q · IF) dx + C] / IF

4. Verify that y = 2e^(3x) is a solution of d²y/dx² − 5(dy/dx) + 6y = 0.

Answer
Given: y = 2e^(3x) dy/dx = 6e^(3x) d²y/dx² = 18e^(3x) Substitute into LHS: d²y/dx² − 5(dy/dx) + 6y = 18e^(3x) − 5(6e^(3x)) + 6(2e^(3x)) = 18e^(3x) − 30e^(3x) + 12e^(3x) = (18 − 30 + 12)e^(3x) = 0 · e^(3x) = 0 = RHS ✓ Hence y = 2e^(3x) is a solution.

VI. Answer in Detail

1. Finding Order and Degree, and Forming a DE from a Given Equation

Problem: Find the order and degree of the DE formed by eliminating the arbitrary constants from y = ae^(2x) + bxe^(2x), where a and b are arbitrary constants.

Worked Solution:

Given: y = ae^(2x) + bxe^(2x) ... (1) Step 1: Differentiate once w.r.t. x dy/dx = 2ae^(2x) + b[e^(2x) + 2xe^(2x)] = 2ae^(2x) + be^(2x) + 2bxe^(2x) = (2a + b)e^(2x) + 2bxe^(2x) ... (2) Step 2: Differentiate again d²y/dx² = 2(2a + b)e^(2x) + 2b[e^(2x) + 2xe^(2x)] = (4a + 2b)e^(2x) + 2be^(2x) + 4bxe^(2x) = (4a + 4b)e^(2x) + 4bxe^(2x) ... (3) Step 3: Eliminate a and b From (1): y = ae^(2x) + bxe^(2x) From (2): y' = (2a + b)e^(2x) + 2bxe^(2x) We check whether y'' − 4y' + 4y = 0: y'' = (4a + 4b)e^(2x) + 4bxe^(2x) 4y' = 4(2a + b)e^(2x) + 8bxe^(2x) = (8a + 4b)e^(2x) + 8bxe^(2x) 4y = 4ae^(2x) + 4bxe^(2x) y'' − 4y' + 4y = [(4a+4b) − (8a+4b) + 4a]e^(2x) + [4b − 8b + 4b]xe^(2x) = 0 · e^(2x) + 0 · xe^(2x) = 0 ✓ Result: d²y/dx² − 4(dy/dx) + 4y = 0 Order = 2 (highest derivative is d²y/dx²) Degree = 1 (power of d²y/dx² is 1)

2. Solving a Variable-Separable Differential Equation Step-by-Step

Problem: Solve dy/dx = e^(x−y) + x²e^(−y) given that y(0) = 0.

Worked Solution:

Given: dy/dx = e^(x−y) + x²e^(−y) Step 1: Factor out e^(−y) on the RHS dy/dx = e^(−y)[e^x + x²] Step 2: Separate variables — multiply both sides by e^y e^y dy = (e^x + x²) dx Step 3: Integrate both sides ∫ e^y dy = ∫ (e^x + x²) dx e^y = e^x + x³/3 + C ... (*) Step 4: Apply initial condition y(0) = 0 e^0 = e^0 + 0 + C 1 = 1 + C C = 0 Step 5: Write the particular solution e^y = e^x + x³/3 Step 6: Express y explicitly (if required) y = ln(e^x + x³/3) Answer: e^y = e^x + x³/3 or equivalently: y = ln(e^x + x³/3)

3. Solving a Linear First-Order DE Using Integrating Factor

Problem: Solve dy/dx + (2x/(1 + x²))y = 1/(1 + x²) given y(0) = 1.

Worked Solution:

Given: dy/dx + (2x/(1 + x²))y = 1/(1 + x²) This is in the standard form dy/dx + Py = Q where: P = 2x/(1 + x²) Q = 1/(1 + x²) Step 1: Find the Integrating Factor IF = e^(∫P dx) = e^(∫ 2x/(1+x²) dx) Let u = 1 + x², then du = 2x dx ∫ 2x/(1+x²) dx = ∫ du/u = ln|u| = ln(1 + x²) IF = e^(ln(1 + x²)) = 1 + x² Step 2: Multiply both sides by IF (1 + x²)(dy/dx) + 2xy = 1 The left side is d/dx[(1 + x²)y]: d/dx[(1 + x²)y] = 1 Step 3: Integrate both sides (1 + x²)y = ∫ 1 dx = x + C Step 4: Apply initial condition y(0) = 1 (1 + 0) · 1 = 0 + C C = 1 Step 5: Write the particular solution (1 + x²)y = x + 1 y = (x + 1)/(1 + x²) Answer: y = (x + 1)/(1 + x²)

Common Mistakes to Avoid

1. Confusing order and degree — Order is the highest derivative present (1st, 2nd, 3rd...). Degree is the power of that highest-order derivative after making the equation polynomial in derivatives. They are not the same thing.

2. Forgetting to check if degree is defined — Not every DE has a defined degree. If derivatives appear inside trigonometric, logarithmic, or exponential functions, or have fractional/negative powers, the degree is not defined until you algebraically remove these.

3. Wrong integrating factor — Always ensure the DE is in standard form dy/dx + Py = Q before computing IF. If the coefficient of dy/dx is not 1, divide the entire equation by that coefficient first.

4. Forgetting the constant of integration — Every indefinite integration introduces a constant C. Omitting it means your solution is incomplete and will be marked wrong.

5. Incorrect homogeneous substitution — After substituting y = vx, remember that dy/dx = v + x(dv/dx), not just dv/dx. The product rule gives the extra v term.

6. Sign errors in linear DE — When rearranging to standard form dy/dx + Py = Q, pay attention to signs. A term like dy/dx − 3y = x must be written as dy/dx + (−3)y = x, so P = −3, not 3.

7. Not applying initial conditions — Finding the general solution is only half the work. If an initial condition is given, you must substitute it to find C and write the particular solution.

8. Incomplete separation of variables — When solving variable-separable DEs, ensure ALL y-terms are on one side with dy and ALL x-terms on the other with dx. Do not leave terms mixed on one side.


Exam Checklist

  • [ ] I can define a differential equation and distinguish ODE from PDE
  • [ ] I can identify the order of any given DE (highest derivative)
  • [ ] I can determine the degree of a DE after making it polynomial in derivatives
  • [ ] I know when degree is not defined (fractional/negative powers on derivatives, or derivatives inside non-polynomial functions)
  • [ ] I can form a DE from a family of curves by eliminating arbitrary constants (differentiate n times for n constants)
  • [ ] I can solve variable-separable DEs by separating x and y terms and integrating
  • [ ] I can identify and solve homogeneous DEs using the substitution y = vx
  • [ ] I can identify and solve linear first-order DEs using the integrating factor method (IF = e^(∫P dx))
  • [ ] I remember the solution formula: y · IF = ∫(Q · IF) dx + C
  • [ ] I can find a particular solution by substituting initial/boundary conditions into the general solution to find C

Frequently Asked Questions

Order is the highest derivative present. Degree is the power of the highest-order derivative after making the equation polynomial in derivatives (no fractional powers).
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