UG Applied Mathematics Booster Test 3 - Introduction to Differential Equations -

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UG Applied Mathematics Booster Test 3 - Introduction to Differential Equations

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

UG Applied Mathematics Booster Test 3 - Introduction to Differential Equations

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QUESTION 1 OF 20

Why does an equation modeling the decay of carbon-14 (dC/dt = -kC) qualify strictly as an ordinary differential equation?

Answer Review

1 Why does an equation modeling the decay of carbon-14 (dC/dt = -kC) qualify strictly as an ordinary differential equation?

A It involves partial derivatives with respect to space and time.

B It contains derivatives of the dependent variable with respect to a single independent variable (time).

C It contains no arbitrary constants in its fully differentiated form.

D It is an algebraic polynomial without transcendental functions.

💡 Short Explanation

One dependent variable (C) One independent variable (t) Ordinary differential equation definition

✅ Correct Answer Explanation

dC/dt = -kC involves derivative w.r.t. only time → Only one independent variable is present → Hence it is an ODE

❓ Why Other Options Are Incorrect

Option A → describes PDE, not ODE
Option C → irrelevant to classification
Option D → unrelated to differential equations

🎯 Strategy

Used: Definition Matching

Final Logic: Single independent variable → ODE

🧠 Memory Trick

"One variable → ODE."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Definition

2 Analytically match the resulting equation or factor to its respective mathematical model:

List I	List II
1. Compound Interest	a. T - A = λe^(-kt)
2. Newton's Cooling	b. C = C₀e^(-(ln 2 / 5700)t)
3. Carbon Dating	c. A = A₀e^(rt)
4. Drug Absorption	d. y = y₀e^(-kt)

A (1)-(c), (2)-(a), (3)-(b), (4)-(d)

B (1)-(b), (2)-(a), (3)-(c), (4)-(d)

C (1)-(c), (2)-(d), (3)-(a), (4)-(b)

D (1)-(c), (2)-(b), (3)-(a), (4)-(d)

💡 Short Explanation

Compound interest → exponential growth Cooling → temperature difference model Carbon dating → decay law Drug absorption → decay model

✅ Correct Answer Explanation

(1) Compound Interest → A = A₀e^(rt) → (c) → (2) Newton's Cooling → T − A = λe^(−kt) → (a) → (3) Carbon Dating → exponential decay → (b) → (4) Drug Absorption → y = y₀e^(−kt) → (d)

❓ Why Other Options Are Incorrect

Option B → swaps models incorrectly
Option C → mismatched decay forms
Option D → incorrect pairing

🎯 Strategy

Used: Concept Mapping

Final Logic: Standard physical laws → fixed exponential models

🧠 Memory Trick

"Growth, Cooling, Decay, Absorption = exponential quartet."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Applications

3 In xy(d²y/dx²) + x(dy/dx)² − y(dy/dx) = 0, analyzing the dependent variable y, which statements hold true?
1. y is differentiated with respect to x.
2. The degree of the highest derivative of y is 1.
3. The order of the differential equation concerning y is 2.
4. y can be ignored to form an algebraic constant.

A 1, 2, 4

B 1, 2

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

y is differentiated Highest derivative is second order Order is 2

✅ Correct Answer Explanation

1: y is differentiated → 2: highest derivative power is 1 → 3: order is 2 → 4 is false

❓ Why Other Options Are Incorrect

Option A → includes wrong 4
Option B → incomplete
Option D → includes false statement

🎯 Strategy

Used: Elimination

🧠 Memory Trick

"Check order, ignore irrelevant claims."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Order and Degree

4 Regarding the formation of differential equations, identify the INCORRECT analytical statement about independent variables:

A To eliminate n arbitrary constants, we typically differentiate n times with respect to the independent variable

B A valid differential equation must be entirely free of the independent variable (x)

C The independent variable often represents continuous physical constraints like time or space

D Both independent (x) and dependent (y) variables can be present alongside derivatives in the final differential equation

💡 Short Explanation

DE need not be free of x x often remains in equation Statement B is false

✅ Correct Answer Explanation

Differential equations generally include independent variable → It is not eliminated from final DE → So B is incorrect

❓ Why Other Options Are Incorrect

Option A → correct
Option C → correct
Option D → correct

🎯 Strategy

Used

Extreme Word Filter

🧠 Memory Trick

"x stays in DE."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation

5 Mixing algebraic powers and derivative orders can be tricky. For the equation (ds/dt)⁴ + 2s(d²s/dt²) = 0, what dictates the analytical order?

A The highest exponent, which is 4

B The presence of s acting as a scaling coefficient

C The highest order derivative present, which is d²s/dt².

D The arithmetic sum of the exponents of all derivatives.

💡 Short Explanation

Highest derivative is d²s/dt² Order depends only on derivative level Powers ignored

✅ Correct Answer Explanation

Highest derivative present is second derivative → Therefore order = 2

❓ Why Other Options Are Incorrect

Option A → power irrelevant
Option B → coefficient irrelevant
Option D → not a rule

🎯 Strategy

Used: Highest Derivative Rule

🧠 Memory Trick

"Ignore powers, take highest derivative."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Order

6 When evaluating the differential equation dy/dx = y²/(1-xy) in the constraint region xy ≠ 1, if we algebraically inverse the relation to dx/dy, what happens to its order?

A The order becomes 0.

B The order becomes negative.

C The order remains 1 as the highest derivative is still a first derivative.

D The degree becomes fundamentally undefined.

💡 Short Explanation

Order depends on highest derivative Reversal does not change order Still first order

✅ Correct Answer Explanation

Original equation has first derivative → Rewriting does not change order → Order remains 1

❓ Why Other Options Are Incorrect

Option A → incorrect
Option B → meaningless
Option D → unrelated

🎯 Strategy

Used: Structural Invariance

🧠 Memory Trick

"Rewriting doesn't change order."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Order Invariance

7 Structuring equations linearly is similar to fixed EMI logic. If a differential equation is not expressible as a polynomial in its derivatives, such as d²y/dx² + y² + e^(dy/dx) = 0, what is its degree analytically?

A It takes the value 1.

B It takes the value of the order.

C It is strictly undefined.

D It reverts to the exponent of the dependent variable.

💡 Short Explanation

Non-polynomial in derivatives Exponential term present Degree undefined

✅ Correct Answer Explanation

e^(dy/dx) breaks polynomial rule → Degree defined only for polynomial DE → Hence undefined

❓ Why Other Options Are Incorrect

Option A → irrelevant
Option B → order unrelated
Option D → irrelevant

🎯 Strategy

Used: Rule Violation

🧠 Memory Trick

"Exponential → no degree."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Degree

8 Assessing moving average growth changes relies on finding the highest impact point. In (y''')² + (y'')³ + (y')⁴ + y⁵ = 0, the degree is analytically defined by:

A The maximum power present, which is 5.

B The power of the highest order derivative, which is 2.

C The sum of powers of derivatives, which is 9.

D The product of order and degree.

💡 Short Explanation

Highest order derivative is y''' Power is 2 Degree = 2

✅ Correct Answer Explanation

Highest derivative is third order → Its power is 2 → Degree = 2

❓ Why Other Options Are Incorrect

Option A → wrong interpretation
Option C → wrong sum logic
Option D → irrelevant

🎯 Strategy

Used: Order-Power Rule

🧠 Memory Trick

"Power of highest derivative = degree."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Degree

9 If an equation is formulated by differentiating y = c e^x + d e^(-x) twice to form y'' - y = 0, what is the probability that the resulting differential equation is a polynomial in its derivatives?

A 0

B 0.25

C 0.5

D 1

💡 Short Explanation

Expression is polynomial Degree always defined Certainty case

✅ Correct Answer Explanation

All derivatives appear in polynomial form → Hence degree always defined → Probability = 1

❓ Why Other Options Are Incorrect

Option A → impossible
Option B → arbitrary
Option C → incorrect

🎯 Strategy

Used: Certainty Principle

🧠 Memory Trick

"Polynomial → guaranteed degree."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Degree

10 If y = log(x), forming the differential equation gives x y' = 1. Though logarithms act as vector-like non-linear transformations initially, is the degree of this formed differential equation defined?

A Yes, the degree is 1 because the derivative y' has a power of 1.

B No, because the original function log(x) is non-polynomial.

C No, because x is multiplied with y'.

D Yes, but the degree is 0

💡 Short Explanation

Polynomial in derivative y' power is 1 Degree defined

✅ Correct Answer Explanation

Equation is x y' = 1 → Derivative appears linearly → Degree = 1

❓ Why Other Options Are Incorrect

Option B → incorrect reasoning
Option C → incorrect restriction
Option D → meaningless

🎯 Strategy

Used: Direct Degree Check

🧠 Memory Trick

"Linear derivative → degree 1."

📚 Class XI – NCERT Mathematics – Chapter: Differential Equations – Topic: Degree Definition

11 To find an area bounding curve, we solve the first-order differential equation dy/dx = e^(x+y) + x²e^y. The solution requires integrating factors or variables separation. What is the integrated particular format?

A e^(-y) = e^x + x³/3 + c

B -e^(-y) = e^x + x³/3 + c

C e^y = e^x + x² + c

D -e^y = e^x + x³/3 + c

💡 Short Explanation

Requires substitution separation Integration of mixed exponential terms Correct sign emerges after rearrangement

✅ Correct Answer Explanation

Rearranging dy/dx = e^(x+y) + x²e^y → Multiply by e^(-y) gives separable form → Integrate both sides → Final integrated form becomes −e^(−y) = e^x + x³/3 + c

❓ Why Other Options Are Incorrect

Option A → wrong sign
Option C → incorrect transformation
Option D → wrong exponent handling

🎯 Strategy

Used: Variable Separation + Transformation

🧠 Memory Trick

"Multiply by e^(-y) to simplify."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: First Order Linear Equations

12 Forming the differential equation of the family of ellipses having foci on the y-axis and center at origin x²/b² + y²/a² = 1 leads to an integral-free relation xy y'' + x(y')² - y y' = 0. What is the order of this differential equation?

A 1

B 2

C 3

D 4

💡 Short Explanation

Family of parabola One parameter elimination Leads to second order equation

✅ Correct Answer Explanation

x² = 4ay has one parameter (a) → Differentiate twice to eliminate a → Highest derivative becomes second order

❓ Why Other Options Are Incorrect

Option A → insufficient differentiation
Option C → overestimation
Option D → irrelevant

🎯 Strategy

Used: Parameter Elimination

🧠 Memory Trick

"One parameter → two differentiations → order 2."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of DE

13 A family of curves given by y = A e^(2x) + B e^(-x) + C sin(x) with three arbitrary constants will theoretically lead to a differential equation of what order?

A 1

B 2

C 3

D 4

💡 Short Explanation

Order equals constants Third order → 3 constants Must eliminate 3 constants

✅ Correct Answer Explanation

General solution of order n has n constants → For third order, 3 constants exist → Hence eliminate 3 constants

❓ Why Other Options Are Incorrect

Option A → undercount
Option B → incomplete
Option D → overcount

🎯 Strategy

Used: Order-Constant Correspondence

🧠 Memory Trick

"Order = number of constants."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Order and Constants

14 In forming the equation x(y y₂ + y₁²) = y y₁, what technique is analytically utilized to eliminate the arbitrary constants from the original relation?

A Partial differentiation

B Successive differentiation with respect to x twice

C Integration by parts

D Separation of variables

💡 Short Explanation

Arbitrary constants are eliminated by differentiating the given relation repeatedly until all constants disappear, resulting in the required differential equation.

✅ Correct Answer Explanation

To form a differential equation from a relation containing arbitrary constants: Differentiate the original equation with respect to \(x\). Continue differentiating until the number of arbitrary constants is matched by the number of differentiations required. Since the given equation involves second derivatives (\(y^{''}\)), the original relation must have been differentiated twice. This process successfully eliminates the arbitrary constants and produces the required differential equation. Therefore, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) Partial differentiation – Incorrect because partial differentiation is used for functions of two or more independent variables, whereas this problem involves an ordinary differential equation with a single independent variable.
Option C) Integration by parts – Incorrect because integration by parts is a technique for evaluating integrals, not for eliminating arbitrary constants.
Option D) Separation of variables – Incorrect because separation of variables is a method used to solve certain differential equations, not to form a differential equation from a given relation.

🎯 Strategy

Used

Concept Identification

Application: Identify the method used in forming a differential equation by eliminating arbitrary constants.

Final Logic: The presence of a second derivative indicates that the original relation has been differentiated twice, making Option B the correct answer.

🧠 Memory Trick

"One Constant → Differentiate Once; Two Constants → Differentiate Twice."

📚 Chapter: Differential Equations

15 Analytically, why does the general solution of an nth order differential equation necessarily contain exactly n independent arbitrary constants?

A Because constants cancel out precisely n times during partial differentiation.

B Because each order of derivative corresponds to one step of indefinite integration required to reverse the differentiation.

C Because the degree of the final equation must match n.

D It is an empirical observation without direct mathematical proof.

💡 Short Explanation

An nth-order differential equation requires n successive integrations to obtain its general solution. Each indefinite integration introduces one independent arbitrary constant, giving a total of n arbitrary constants.

✅ Correct Answer Explanation

A differential equation of order n contains the nth derivative of the dependent variable. To recover the original function: The equation must be integrated n times. Every indefinite integration introduces one arbitrary constant of integration. Therefore, after n integrations, the general solution contains exactly n independent arbitrary constants. This is the fundamental reason why the general solution of an nth-order differential equation always has n arbitrary constants. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) Incorrect because arbitrary constants are introduced through integration, not eliminated by partial differentiation.
Option C) Incorrect because the degree of a differential equation is unrelated to the number of arbitrary constants in its general solution.
Option D) Incorrect because this result follows directly from the mathematical process of successive integration and is not merely an empirical observation.

🎯 Strategy

Used

Concept Identification

Application: Recall the relationship between the order of a differential equation and the number of integrations needed to obtain its general solution.

Final Logic: Since each integration introduces one arbitrary constant, an nth-order differential equation has n independent arbitrary constants. Therefore, Option B is correct.

🧠 Memory Trick

"n Order → n Integrations → n Constants."

📚 Chapter: Differential Equations

16 The differential equation x + y y' = 0 resulting from x² + y² = a² represents which family of curves analytically?

A Parabolas with vertices at origin

B Circles having centers at origin

C Straight lines passing through origin

D Hyperbolas on the x-axis

💡 Short Explanation

Derived from x² + y² = a² Circle family Center at origin

✅ Correct Answer Explanation

Implicit differentiation of x² + y² = a² → Gives x + y y' = 0 → Represents circle family centered at origin

❓ Why Other Options Are Incorrect

Option A → incorrect shape
Option C → wrong geometry
Option D → unrelated

🎯 Strategy

Used: Geometric Identification

🧠 Memory Trick

"x² + y² → circle."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Family of Curves

17 In Newton's cooling model T - 75 = 110e^(-kt), substituting the boundary condition t = 30 and T = 150 fixes the value of k as a constant. What is the exact resulting evaluation of e^(-30k)?

A 150/110

B 75/110

C 0.5

D 0.75

💡 Short Explanation

Substitute values Solve exponential ratio Direct algebra

✅ Correct Answer Explanation

T − 75 = 110e^(−kt) → Substitute t = 30, T = 150 → 150 − 75 = 110e^(−30k) → 75 = 110e^(−30k) → e^(−30k) = 75/110

❓ Why Other Options Are Incorrect

Option A → reversed ratio
Option C → incorrect value
Option D → wrong simplification

🎯 Strategy

Used: Substitution Method

🧠 Memory Trick

"Plug values, isolate exponential."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Newton's Law of Cooling

18 Determine the specific solution curve for dy/dx = (2x log x + x) if integration yields y = x² log x - x²/2 + x²/2 + c. Which correctly simplifies to:

A y = x² log x + c

B y = log x + c

C y = x log x + c

D y = x² + c

💡 Short Explanation

Cancel redundant terms Simplify expression Final compact form

✅ Correct Answer Explanation

y = x² log x − x²/2 + x²/2 + c → Middle terms cancel → Final form: y = x² log x + c

❓ Why Other Options Are Incorrect

Option B → wrong structure
Option C → incorrect power
Option D → incomplete

🎯 Strategy

Used: Algebraic Simplification

🧠 Memory Trick

"Cancel + combine constants."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Particular Solution

📄 Passage

PASSAGE To obtain the differential equation that governs the variation in the amount of money A in the bank with time t, we follow these steps. During a short time interval dt the amount of interest added to the account is approximately given by dA = (r)A dt. The differential equation is dA/dt = rA. Integrating both sides gives ∫ (1/A) dA = ∫ r dt. Therefore, log A = rt + c, which yields A = A₀e^(rt).

How does substituting t=0 structurally transition the general solution to a particular solution in this mathematical banking model?

A It eliminates the interest rate r from the equation entirely

B It sets the exponential term e^(rt) to 1, mapping the arbitrary constant e^c directly to the initial deposit A₀.

C It makes the differential equation strictly linear

D It proves the continuous compounding fails at boundary t=0.

💡 Short Explanation

Initial condition applied Exponential becomes 1 Constant fixed

✅ Correct Answer Explanation

f(t) = A₀e^(rt) → At t = 0 → e^0 = 1 → f(0) = A₀ = e^c → Converts general to particular solution

❓ Why Other Options Are Incorrect

Option A → incorrect
Option C → irrelevant
Option D → false

🎯 Strategy

Used: Initial Condition Substitution

🧠 Memory Trick

"t=0 → exponential becomes 1."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Exponential Growth

📄 Passage

If the continuous rate of decrease of an oil well is 10% per year, using the derived exponential integration technique for dP/dt = -0.1P, what is the analytical function representing output P(t)?

A P(t) = P₀e^(0.1t)

B P(t) = P₀e^(-0.1t)

C P(t) = 0.1 e^(P₀t)

D P(t) = -0.1 e^(P₀t)

💡 Short Explanation

Standard decay form Negative rate constant Exponential decay solution

✅ Correct Answer Explanation

dP/dt = -0.1P → Solution P = P₀e^(kt) → k = -0.1 → P = P₀e^(−0.1t)

❓ Why Other Options Are Incorrect

Option A → wrong sign
Option C → invalid form
Option D → incorrect structure

🎯 Strategy

Used: Standard Formula Application

🧠 Memory Trick

"Negative k → decay exponential."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Exponential Decay

UG Applied Mathematics Booster Test 3 - Introduction to Differential Equations

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