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Differential Equations

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Chapter Notes

Overview
This chapter introduces differential equations, which are equations involving derivatives of a dependent variable with respect to an independent variable. It is a crucial topic in mathematics because differential equations model real-world phenomena in physics, chemistry, biology, and economics. The one big idea a student must walk away with is that a differential equation describes a relationship between a function and its derivatives, and solving it means finding the function (or family of functions) that satisfies that relationship.
Key Concepts (12)
Differential Equation
An equation involving derivatives of the dependent variable with respect to the independent variable is called a differential equation.
e.g. dy/dx + y = 0 is a differential equation because it involves the derivative dy/dx.
Ordinary Differential Equation
A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation.
e.g. d²y/dx² + (dy/dx)³ = 0 is an ordinary differential equation because it involves derivatives with respect to x only.
Order of a Differential Equation
The order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation.
e.g. In d³y/dx³ + (d²y/dx²)² = 0, the highest order derivative is d³y/dx³, so the order is 3.
Degree of a Differential Equation
The degree of a differential equation, when it is a polynomial equation in derivatives, is the highest power (positive integral index) of the highest order derivative involved in the given differential equation.
e.g. In (d²y/dx²)³ + dy/dx = 0, the highest order derivative is d²y/dx², and its highest power is 3, so the degree is 3.
General Solution
The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation.
e.g. y = a cos x + b sin x is the general solution of d²y/dx² + y = 0, as it contains two arbitrary constants a and b.
Particular Solution
The solution free from arbitrary constants, obtained from the general solution by giving particular values to the arbitrary constants, is called a particular solution of the differential equation.
e.g. y = 2 sin(x + π/4) is a particular solution of d²y/dx² + y = 0, obtained by setting a = 2 and b = π/4 in the general solution.
Variable Separable Method
A first order-first degree differential equation of the form dy/dx = g(x)h(y) can be solved by separating the variables as (1/h(y)) dy = g(x) dx and then integrating both sides.
e.g. For dy/dx = (x+1)/(2-y), we write (2-y) dy = (x+1) dx and integrate.
Homogeneous Differential Equation
A differential equation of the form dy/dx = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, i.e., F(λx, λy) = λ⁰ F(x, y) = F(x, y).
e.g. dy/dx = (x+2y)/(x-y) is homogeneous because F(λx, λy) = (λx+2λy)/(λx-λy) = (x+2y)/(x-y) = F(x, y).
Homogeneous Function
A function F(x, y) is said to be homogeneous of degree n if F(λx, λy) = λⁿ F(x, y) for any nonzero constant λ.
e.g. F(x, y) = y² + 2xy is homogeneous of degree 2 because F(λx, λy) = λ²(y² + 2xy) = λ² F(x, y).
Linear Differential Equation
A differential equation of the form dy/dx + Py = Q, where P and Q are constants or functions of x only, is known as a first order linear differential equation.
e.g. dy/dx + y/x = x² is a linear differential equation with P = 1/x and Q = x².
Integrating Factor (I.F.)
For a linear differential equation dy/dx + Py = Q, the integrating factor is given by I.F. = e^(∫P dx). Multiplying the equation by I.F. makes the left-hand side a derivative of y * I.F.
e.g. For dy/dx + (2/x)y = x, P = 2/x, so I.F. = e^(∫(2/x) dx) = e^(2 log x) = x².
Solution Curve (Integral Curve)
The curve y = φ(x) that satisfies a given differential equation is called the solution curve or integral curve of the differential equation.
e.g. For dy/dx = 2x, y = x² + C is a family of solution curves; each specific C gives one curve.
Formulas & Equations (10)
dy/dx = g(x)h(y)
Reads as: dy by dx equals g of x times h of y
Variables: y is the dependent variable, x is the independent variable, g(x) is a function of x only, h(y) is a function of y only.
Use when: Use this form to identify a variable separable differential equation, then separate variables as (1/h(y)) dy = g(x) dx.
⚠ Students forget to check if h(y) ≠ 0 before separating, leading to division by zero errors.
∫ (1/h(y)) dy = ∫ g(x) dx + C
Reads as: Integral of 1 over h of y dy equals integral of g of x dx plus constant C
Variables: h(y) is a function of y, g(x) is a function of x, C is the constant of integration.
Use when: Apply this after separating variables in a variable separable differential equation to find the general solution.
⚠ Students often forget to add the constant of integration C, or add it to only one side.
F(λx, λy) = λⁿ F(x, y)
Reads as: F of lambda x comma lambda y equals lambda to the power n times F of x comma y
Variables: F is a function of x and y, λ is a nonzero constant, n is the degree of homogeneity.
Use when: Use this to test if a function F(x, y) is homogeneous of degree n.
⚠ Students incorrectly apply the test to non-polynomial functions like sin(x+y) and claim they are homogeneous.
dy/dx = g(y/x)
Reads as: dy by dx equals g of y over x
Variables: y is the dependent variable, x is the independent variable, g is a function of y/x only.
Use when: Use this form to identify a homogeneous differential equation, then substitute y = vx to solve.
⚠ Students forget to also differentiate y = vx correctly: dy/dx = v + x(dv/dx).
y = vx, dy/dx = v + x(dv/dx)
Reads as: y equals v times x, dy by dx equals v plus x times dv by dx
Variables: v = y/x is a new dependent variable, x is the independent variable.
Use when: Use this substitution to solve a homogeneous differential equation of the form dy/dx = g(y/x).
⚠ Students substitute y = vx but forget to replace dy/dx with v + x(dv/dx), leading to wrong equations.
I.F. = e^(∫P dx)
Reads as: Integrating factor equals e to the power of integral of P with respect to x
Variables: P is a function of x only from the linear differential equation dy/dx + Py = Q.
Use when: Use this to find the integrating factor for a first order linear differential equation in the form dy/dx + Py = Q.
⚠ Students forget to ensure the equation is in the standard form dy/dx + Py = Q before finding I.F., leading to incorrect P.
y * (I.F.) = ∫ Q * (I.F.) dx + C
Reads as: y times integrating factor equals integral of Q times integrating factor dx plus constant C
Variables: y is the dependent variable, I.F. is the integrating factor, Q is a function of x from the linear differential equation, C is the constant of integration.
Use when: Use this formula to write the general solution of a first order linear differential equation after finding the I.F.
⚠ Students forget to multiply Q by the I.F. before integrating, or they integrate y * I.F. incorrectly.
dx/dy + P₁x = Q₁
Reads as: dx by dy plus P one times x equals Q one
Variables: x is the dependent variable, y is the independent variable, P₁ and Q₁ are functions of y only.
Use when: Use this form when the differential equation is linear in x as a function of y, e.g., dx/dy - (1/y)x = 2y.
⚠ Students try to solve such equations using the form dy/dx + Py = Q, leading to confusion with variables.
I.F. = e^(∫P₁ dy)
Reads as: Integrating factor equals e to the power of integral of P one with respect to y
Variables: P₁ is a function of y only from the linear differential equation dx/dy + P₁x = Q₁.
Use when: Use this to find the integrating factor for a linear differential equation in the form dx/dy + P₁x = Q₁.
⚠ Students mistakenly use ∫P dx instead of ∫P₁ dy when the equation is in terms of x and y swapped.
x * (I.F.) = ∫ Q₁ * (I.F.) dy + C
Reads as: x times integrating factor equals integral of Q one times integrating factor dy plus constant C
Variables: x is the dependent variable, I.F. is the integrating factor, Q₁ is a function of y, C is the constant of integration.
Use when: Use this formula to write the general solution of a linear differential equation in the form dx/dy + P₁x = Q₁.
⚠ Students forget to integrate with respect to y, not x, when using this formula.
How It Works
Determining Order and Degree of a Differential Equation
  1. Step 1: Identify the highest order derivative present in the differential equation.
  2. Step 2: The order of that highest derivative is the order of the differential equation.
  3. Step 3: Check if the differential equation is a polynomial equation in derivatives (y', y'', y''', etc.).
  4. Step 4: If it is a polynomial, find the highest power (positive integral index) of the highest order derivative. This is the degree.
  5. Step 5: If the equation is not a polynomial in derivatives, the degree is not defined.
💡 Order and degree (if defined) are always positive integers. Do not confuse order (which derivative is highest) with degree (the power of that highest derivative).
Verifying a Solution of a Differential Equation
  1. Step 1: Given a function y = φ(x), find its derivatives (dy/dx, d²y/dx², etc.) as required by the differential equation.
  2. Step 2: Substitute the function and its derivatives into the given differential equation.
  3. Step 3: Simplify the left-hand side (L.H.S.) of the equation.
  4. Step 4: Check if L.H.S. equals the right-hand side (R.H.S.).
  5. Step 5: If L.H.S. = R.H.S., the function is a solution of the differential equation.
💡 A solution must satisfy the equation for all x in its domain. A common mistake is to only check at one point instead of verifying the identity.
Solving a Variable Separable Differential Equation
  1. Step 1: Write the differential equation in the form dy/dx = g(x)h(y).
  2. Step 2: If h(y) ≠ 0, separate the variables: (1/h(y)) dy = g(x) dx.
  3. Step 3: Integrate both sides: ∫ (1/h(y)) dy = ∫ g(x) dx + C.
  4. Step 4: Perform the integrations to get H(y) = G(x) + C, where H and G are antiderivatives.
  5. Step 5: If initial conditions are given, substitute to find the particular value of C.
💡 Always include the constant of integration C. Do not forget to check for the case h(y) = 0, which might be a singular solution.
Solving a Homogeneous Differential Equation
  1. Step 1: Verify that the differential equation is homogeneous, i.e., dy/dx = F(x, y) where F(λx, λy) = F(x, y).
  2. Step 2: Write the equation in the form dy/dx = g(y/x).
  3. Step 3: Substitute y = vx, so dy/dx = v + x(dv/dx).
  4. Step 4: Replace y and dy/dx in the equation to get v + x(dv/dx) = g(v).
  5. Step 5: Rearrange to separate variables: dv/(g(v) - v) = dx/x.
  6. Step 6: Integrate both sides: ∫ dv/(g(v) - v) = ∫ dx/x + C.
  7. Step 7: After integration, replace v by y/x to get the general solution in terms of x and y.
💡 The substitution y = vx is key. A common mistake is to forget to replace dy/dx with v + x(dv/dx). Also, if the equation is of the form dx/dy = h(x/y), substitute x = vy instead.
Solving a First Order Linear Differential Equation (dy/dx + Py = Q)
  1. Step 1: Write the differential equation in the standard form dy/dx + Py = Q, where P and Q are functions of x only.
  2. Step 2: Find the Integrating Factor (I.F.) = e^(∫P dx).
  3. Step 3: Multiply both sides of the equation by the I.F.: I.F. * dy/dx + P * I.F. * y = Q * I.F.
  4. Step 4: Notice that the left-hand side is d/dx (y * I.F.).
  5. Step 5: Write the equation as d/dx (y * I.F.) = Q * I.F.
  6. Step 6: Integrate both sides with respect to x: y * I.F. = ∫ (Q * I.F.) dx + C.
  7. Step 7: Solve for y: y = (1/I.F.) * [∫ (Q * I.F.) dx + C].
💡 Always ensure the equation is in the exact form dy/dx + Py = Q. If the coefficient of dy/dx is not 1, divide the entire equation by that coefficient first.
Solving a First Order Linear Differential Equation (dx/dy + P₁x = Q₁)
  1. Step 1: Write the differential equation in the standard form dx/dy + P₁x = Q₁, where P₁ and Q₁ are functions of y only.
  2. Step 2: Find the Integrating Factor (I.F.) = e^(∫P₁ dy).
  3. Step 3: Multiply both sides of the equation by the I.F.: I.F. * dx/dy + P₁ * I.F. * x = Q₁ * I.F.
  4. Step 4: Notice that the left-hand side is d/dy (x * I.F.).
  5. Step 5: Write the equation as d/dy (x * I.F.) = Q₁ * I.F.
  6. Step 6: Integrate both sides with respect to y: x * I.F. = ∫ (Q₁ * I.F.) dy + C.
  7. Step 7: Solve for x: x = (1/I.F.) * [∫ (Q₁ * I.F.) dy + C].
💡 This form is used when the equation is linear in x as a function of y. Do not confuse with the dy/dx form. The integrating factor uses dy, not dx.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Order and Degree
✗ For the equation (d²y/dx²)³ + (dy/dx)² = 0, students write 'Order = 3, Degree = 3'.
→ The order is the highest derivative, which is d²y/dx² (order 2), not the power 3. The degree is the power of that highest derivative, which is 3.
✓ Order = 2, Degree = 3.
⚠ Forgetting to Check Polynomial Condition for Degree
✗ For the equation sin(dy/dx) + y = 0, students write 'Degree = 1'.
→ The degree is defined only if the differential equation is a polynomial equation in derivatives. sin(dy/dx) is not a polynomial in dy/dx, so the degree is not defined.
✓ Degree is not defined.
⚠ Incorrect Substitution in Homogeneous Equations
✗ For dy/dx = (x+y)/(x-y), students substitute y = vx but write dy/dx = v + x(dv/dx) incorrectly as dy/dx = v.
→ When y = vx, differentiating gives dy/dx = v + x(dv/dx) by the product rule. Forgetting the term x(dv/dx) leads to an incorrect equation.
✓ Write dy/dx = v + x(dv/dx) correctly.
⚠ Forgetting to Divide by Coefficient in Linear Equations
✗ For 2(dy/dx) + y = x, students directly find I.F. = e^(∫1 dx) = e^x.
→ The standard form is dy/dx + Py = Q. The coefficient of dy/dx must be 1. Here, we must first divide by 2: dy/dx + (1/2)y = x/2, so P = 1/2, not 1.
✓ First divide by 2: dy/dx + (1/2)y = x/2, then I.F. = e^(∫(1/2) dx) = e^(x/2).
⚠ Mixing Up Variables in Linear Equations of the Form dx/dy
✗ For dx/dy - (1/y)x = 2y, students find I.F. = e^(∫(-1/y) dx) instead of e^(∫(-1/y) dy).
→ In the form dx/dy + P₁x = Q₁, P₁ is a function of y, so the integrating factor is e^(∫P₁ dy), not e^(∫P₁ dx).
✓ I.F. = e^(∫(-1/y) dy) = e^(-log|y|) = 1/y.
⚠ Omitting the Constant of Integration
✗ For dy/dx = 2x, students write the solution as y = x².
→ The general solution of a first order differential equation must include one arbitrary constant. The correct general solution is y = x² + C.
✓ y = x² + C, where C is an arbitrary constant.
Compare & Contrast
Number of arbitrary constants
General Solution
Particular Solution
The general solution contains arbitrary constants (e.g., y = a cos x + b sin x), while the particular solution is obtained by assigning specific values to those constants (e.g., y = 2 sin(x + π/4)).
What they measure
Order of a Differential Equation
Degree of a Differential Equation
Order is the order of the highest derivative present (e.g., in d²y/dx² + y = 0, order is 2). Degree is the highest power of that highest derivative, provided the equation is a polynomial in derivatives (e.g., in (d²y/dx²)³ + y = 0, degree is 3).
Form and solution method
Homogeneous Differential Equation
Linear Differential Equation
A homogeneous equation has the form dy/dx = g(y/x) and is solved by substituting y = vx. A linear equation has the form dy/dx + Py = Q and is solved using an integrating factor I.F. = e^(∫P dx).
Exam Focus
Marks pattern: Typically 1-2 MCQs (1 mark each) on order/degree, 1 short answer (3 marks) on solving a variable separable or homogeneous equation, and 1 long answer (5 marks) on solving a linear differential equation or an application problem.
Most asked: Finding order and degree of a differential equation · Verifying whether a given function is a solution of a differential equation · Solving variable separable differential equations · Solving homogeneous differential equations · Solving linear differential equations using integrating factor · Forming differential equations from given family of curves (though not explicitly in this text, it is a common theme)
Don't skip: Solving a first order linear differential equation (dy/dx + Py = Q) with initial conditions to find a particular solution; Solving a homogeneous differential equation using substitution y = vx; Application problems involving growth/decay (e.g., bank principal, bacteria count)
One-Line Revision (read aloud before the exam)
  1. A differential equation involves derivatives of a dependent variable with respect to an independent variable.
  2. Order of a differential equation is the order of the highest derivative present.
  3. Degree of a differential equation is the highest power of the highest order derivative, provided it is a polynomial in derivatives.
  4. Order and degree (if defined) are always positive integers.
  5. General solution of a differential equation contains arbitrary constants.
  6. Particular solution is obtained by giving specific values to the arbitrary constants in the general solution.
  7. Variable separable method: write dy/dx = g(x)h(y), separate to (1/h(y)) dy = g(x) dx, then integrate.
  8. A homogeneous differential equation has the form dy/dx = g(y/x) and is solved by substituting y = vx.
  9. For a homogeneous equation, after substituting y = vx, remember dy/dx = v + x(dv/dx).
  10. A linear differential equation has the form dy/dx + Py = Q, where P and Q are functions of x only.
  11. The integrating factor for a linear differential equation is I.F. = e^(∫P dx).
  12. The solution of a linear differential equation is y * I.F. = ∫ Q * I.F. dx + C.
  13. If the equation is linear in x as a function of y, use the form dx/dy + P₁x = Q₁ with I.F. = e^(∫P₁ dy).
  14. Always check if the differential equation is a polynomial in derivatives before finding its degree.
  15. To verify a solution, substitute the function and its derivatives into the differential equation and check if L.H.S. = R.H.S.

MCQs — 10 Questions with Answers

Q1. The order of the differential equation d⁴y/dx⁴ + d²y/dx² − 5dy/dx + y = 0 is:

  • A. 4 ✓
  • B. 2
  • C. 1
  • D. 0

Answer: A — The highest derivative present is d⁴y/dx⁴, so the order is 4.

Q2. The degree of (d²y/dx²)³ + (dy/dx)² + sin(y) = 0 is:

  • A. 2
  • B. 3
  • C. 1
  • D. Not defined ✓

Answer: D — The equation is not polynomial in derivatives due to sin(y) term, so degree is not defined.

Q3. Which of the following is an ordinary differential equation?

  • A. ∂u/∂x + ∂u/∂y = 0
  • B. dy/dx + 2y = e^x ✓
  • C. ∂²u/∂x² + ∂²u/∂y² = 0
  • D. ∂z/∂t = ∂²z/∂x²

Answer: B — An ordinary differential equation has derivatives with respect to only one independent variable; option B has only dy/dx (derivative with respect to x only).

Q4. The order and degree of d²y/dx² + (dy/dx)² − 3y = 0 are respectively:

  • A. 1, 2
  • B. 2, 1
  • C. 2, 2 ✓
  • D. 1, 1

Answer: C — Highest derivative is d²y/dx² (order 2); highest power of d²y/dx² is 1, but highest power of dy/dx is 2, so degree is 2.

Q5. For the differential equation sin(d²y/dx²) + d²y/dx² − y = 0, the degree is:

  • A. 1
  • B. 2
  • C. 3
  • D. Not defined ✓

Answer: D — The presence of sin(d²y/dx²) makes the equation non-polynomial in derivatives, so degree is not defined.

Q6. Which differential equation has order 3 and degree 2?

  • A. (d³y/dx³)² + dy/dx = 0 ✓
  • B. d³y/dx³ + (d²y/dx²)² = 0
  • C. (d²y/dx²)³ + dy/dx = 0
  • D. d³y/dx³ + d²y/dx² + dy/dx = 0

Answer: A — Option A has highest derivative d³y/dx³ raised to power 2, giving order 3 and degree 2.

Q7. The equation dy/dx + xy = cos(x) is a differential equation of order and degree:

  • A. Order 2, Degree 1
  • B. Order 1, Degree 1 ✓
  • C. Order 1, Degree 2
  • D. Order 2, Degree 2

Answer: B — Highest derivative is dy/dx (order 1); highest power of dy/dx is 1 (degree 1).

Q8. Which statement is NOT correct about the order and degree of differential equations?

  • A. Order is always a positive integer
  • B. Degree may not exist for all differential equations
  • C. Degree is always greater than or equal to order ✓
  • D. Degree is the highest power of the highest derivative only when polynomial in derivatives

Answer: C — Statement C is incorrect; degree need not be greater than or equal to order—for example, d³y/dx³ + y = 0 has order 3 but degree 1.

Q9. Given the differential equation [d²y/dx² − (dy/dx)³]² + y = 0, identify its order and degree: (i) Order is 2 (ii) Degree is 2

  • A. Both (i) and (ii) are correct
  • B. Only (i) is correct ✓
  • C. Only (ii) is correct
  • D. Neither is correct

Answer: B — Order is 2 (highest derivative d²y/dx²); degree is 6 because when expanded, the highest power of d²y/dx² is 2 × 3 = 6 from (dy/dx)³ inside the square bracket.

Q10. A general solution of a differential equation d²y/dx² + y = 0 is y = A cos(x) + B sin(x). The number of arbitrary constants in this solution and its significance is:

  • A. 2 constants; equals the degree of the equation
  • B. 2 constants; equals the order of the equation ✓
  • C. 1 constant; equals the order minus 1
  • D. Cannot be determined without solving the equation

Answer: B — A general solution of an nth order differential equation contains n arbitrary constants; here order 2 means 2 arbitrary constants (A and B).

Flashcards

What is a differential equation?

An equation involving a dependent variable, an independent variable, and derivatives of the dependent variable with respect to the independent variable.

Define order of a differential equation.

The order is the highest order derivative of the dependent variable present in the differential equation.

Define degree of a differential equation.

When a differential equation is a polynomial in derivatives, the degree is the highest power of the highest order derivative involved.

When is the degree of a differential equation not defined?

The degree is not defined when the differential equation is not a polynomial equation in its derivatives.

What is the difference between general and particular solutions?

A general solution contains arbitrary constants and represents a family of solution curves; a particular solution has specific values for these constants and represents a single curve.

Example of finding order: What is the order of d²y/dx² + dy/dx = 0?

The order is 2 because the highest derivative present is d²y/dx² (second order).

Example of finding degree: What is the degree of (d²y/dx²)³ + (dy/dx)² + y = 0?

The degree is 3 because the highest power of the highest order derivative d²y/dx² is 3.

Why are arbitrary constants important in general solutions?

Arbitrary constants represent the infinitely many solution curves of a differential equation; each value combination produces a different particular solution.

What must be true for a differential equation to have a defined degree?

The equation must be a polynomial equation in the derivatives (y′, y″, y‴, etc.) when written in simplified form.

Example: Find order and degree of y′′′ + 2y″ + y′ = 0.

Order is 3 (highest derivative is y′′′) and degree is 1 (highest power of y′′′ is 1, and it is polynomial in all derivatives).

Important Board Questions

Define a differential equation and classify the equation dy/dx + 2xy = e^(−x²) as ordinary or partial. Also find its order and degree. [2 marks]

Define DE as an equation involving derivatives of dependent variable. This is ordinary (one independent variable x only). Identify highest derivative for order and its power for degree.

Find the order and degree (if defined) of each of the following differential equations and justify your answer: (i) (d²y/dx²)³ + 2(dy/dx) + y = 0 (ii) sin(dy/dx) + cos(dy/dx) = x (iii) [d²y/dx² − (dy/dx)²]² + y = 0 [5 marks]

For each, identify the highest order derivative (order), check if polynomial in derivatives, then find highest power of highest derivative (degree). For (ii), recognize it is non-polynomial and degree is undefined. Show all steps.

Explain what is meant by the general solution and particular solution of a differential equation. If y = C₁e^(2x) + C₂e^(−x) is the general solution of a differential equation of order 2, find a particular solution satisfying the initial conditions y(0) = 5 and y'(0) = 4. Show all working. [6 marks]

General solution has arbitrary constants (C₁, C₂ here); particular solution assigns specific values to these constants. Use initial conditions y(0) = 5 and y'(0) = 4 to set up two equations in C₁ and C₂, solve simultaneously to find specific values, then substitute back into general form.

Assertion-Reason Questions for Differential Equations

Q1. Assertion (A):

The differential equation $\frac{d^3y}{dx^3} + \left(\frac{dy}{dx}\right)^2 = x$ has order 3 and degree 1.

Reason (R):

The order of a differential equation is the highest order of derivatives, and degree is the highest power of the highest order derivative when the equation is polynomial in derivatives.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true. The highest derivative is $\frac{d^3y}{dx^3}$ (order 3), and it appears to power 1, making degree 1. The reason correctly explains why the assertion is true.

Q2. Assertion (A):

The differential equation $\sin\left(\frac{d^2y}{dx^2}\right) + y = 0$ has order 2 and degree 1.

Reason (R):

Order is determined by the highest order derivative, while degree requires the differential equation to be polynomial in derivatives.

Answer: D — A is false but R is true — The assertion is false. While order = 2 is correct, degree is NOT defined because the equation is not polynomial in derivatives (the highest order derivative is inside a sine function). The reason is correct but doesn't explain the false assertion.

Q3. Assertion (A):

If a differential equation's general solution is $y = Ae^{3x} + Be^{-3x} + Ce^x$, then the differential equation is of order 3.

Reason (R):

The order of a differential equation equals the number of arbitrary constants in its general solution.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true and directly related. The general solution has 3 arbitrary constants (A, B, C), so the order is 3. This is the fundamental principle connecting general solutions to DE order.

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Case-Based Questions (CBQ) for Differential Equations

Case Study:

A pharmaceutical company is modelling the concentration of a drug in the bloodstream over time. Initially (at t = 0), a patient is given an injection with concentration $C_0 = 100$ mg/mL. The rate of change of drug concentration follows the relationship: $\frac{dC}{dt} = -kC$, where k is a positive constant representing the elimination rate. The general solution to this differential equation is $C(t) = Ae^{-kt}$, where A is an arbitrary constant determined by initial conditions.

Q1. What is the order of the differential equation $\frac{dC}{dt} = -kC$?

  • A. 0
  • B. 1 ✓
  • C. 2
  • D. 3

Answer: B

Q2. What is the degree of the differential equation $\frac{dC}{dt} = -kC$?

  • A. Not defined
  • B. 0
  • C. 1 ✓
  • D. 2

Answer: C

Q3. How many arbitrary constants does the general solution $C(t) = Ae^{-kt}$ contain?

  • A. 0
  • B. 1 ✓
  • C. 2
  • D. 3

Answer: B

Q4. If the initial condition is $C(0) = 100$ mg/mL, what is the particular solution?

  • A. $C(t) = 100e^{-kt}$ ✓
  • B. $C(t) = Ae^{kt}$
  • C. $C(t) = 100 - kt$
  • D. $C(t) = e^{-kt}$

Answer: A

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Previous Year Questions (PYQ) for Differential Equations

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 65/1/1 · 1 mark

The integrating factor of differential equation R(dy/dx) + Px = Q where P, Q, R are functions of y is

  • e^(∫Q/P dy)
  • e^(∫P dy)
  • e^(∫R/P dy)
  • e^(∫R/P dx)

CBSE 2026 · Set 65/1/1 · 1 mark

The order and degree of the differential equation d/dx(e^y) = 0 respectively are

  • 0, 1
  • 1, 1
  • 2, 1
  • 1, not defined

CBSE 2026 · Set 65/1/1 · 3 marks

Find the general solution of the following differential equation: x² dy/dx = x² + xy + y².

CBSE 2026 · Set 65/1/2 · 1 mark

The order and degree of the differential equation: d/dx(sin y) = y² respectively are

  • 1, 1
  • 2, 1
  • 2, 2
  • 1, 2

CBSE 2026 · Set 65/1/2 · 3 marks

Find the general solution of the differential equation: y²dx + (x² – xy + y²)dy = 0

CBSE 2026 · Set 65/1/3 · 1 mark

The order and degree of the differential equation d/dx(y')³ + (y')³ = 1 respectively are, where y' = dy/dx

  • 1, 3
  • 2, 1
  • 3, 1
  • 3, 2

CBSE 2026 · Set 65/1/3 · 3 marks

Find the general solution of the differential equation (y² – x²)dx = 2xy dy

CBSE 2026 · Set 65/2/1 · 1 mark

The general solution of the differential equation dy/dx = y/x is

  • log y = log x + C
  • y + x = C
  • y - x = C
  • log y + log x = C

CBSE 2026 · Set 65/2/1 · 1 mark

The integrating factor of the differential equation 2x(dy/dx) - y = 3 is

  • x
  • 1/x
  • eˣ
  • e⁻ˣ

CBSE 2026 · Set 65/2/1 · 3 marks

Solve the following differential equation: x(dy/dx) = y - x sin²(y/x), given that y(1) = π/6.

CBSE 2026 · Set 65/2/2 · 1 mark

The general solution of the differential equation xdy – ydx = 0 is

  • x² – y² = k
  • xy = k
  • x = ky
  • log y + log x = k

CBSE 2026 · Set 65/2/2 · 1 mark

The integrating factor of the differential equation 2x(dy/dx) – y = 3 is

  • x
  • 1/x
  • e^x
  • e^(–x)

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Class 12 Maths Formula Sheet

Every formula from all 13 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Mathematics Previous Year Questions, Chapter-wise

All 1048 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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