CUET UG Mathematics Booster Test 1 - Solutions of Differential Equations -

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CUET UG Mathematics Booster Test 1 - Solutions of Differential Equations

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Solutions of Differential Equations

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QUESTION 1 OF 20

Which of the following functions is a solution of the differential equation
x dy/dx = y, (x ≠ 0)?

Answer Review

1 Which of the following functions is a solution of the differential equation
x dy/dx = y, (x ≠ 0)?

A y = C/x

B y = Cx

C y = e^x

D y = ln x

💡 Short Explanation

Differentiate the given function. Substitute into x(dy/dx)=y. Only y=Cx satisfies identically.

✅ Correct Answer Explanation

For y=Cx, we have dy/dx=C. Substituting into x(dy/dx)=y gives x(C)=Cx, which is true for all x. Thus y=Cx is a solution. The other options fail to satisfy the differential equation upon substitution.

❓ Why Other Options Are Incorrect

Option A → Gives x(−C/x²)=−C/x ≠ C/x.
Option C → Gives xe^x ≠ e^x for all x.
Option D → Gives x(1/x)=1 ≠ ln x.

🎯 Strategy

Used: Substitution

Application:

Substitute each option into the differential equation and verify.

Final Logic:

Only y=Cx satisfies x(dy/dx)=y identically.

🧠 Memory Trick

"xy' = y ⇒ y ∝ x."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Verification of Solutions

2 Arrange the steps to verify that
y = a cos x + b sin x
is a solution of
y'' + y = 0.
1. Substitute y and y'' into y'' + y.
2. Compute y' = −a sin x + b cos x.
3. Verify that the result equals 0.
4. Compute y'' = −a cos x − b sin x.

A 1, 2, 4, 3

B 2, 1, 4, 3

C 4, 2, 1, 3

D 2, 4, 1, 3

💡 Short Explanation

Differentiate once. Differentiate again. Substitute and verify.

✅ Correct Answer Explanation

To verify a solution, first compute y', then compute y''. Next substitute y and y'' into the differential equation y''+y=0. Finally check whether the result simplifies to zero. Therefore the correct order is 2, 4, 1, 3.

❓ Why Other Options Are Incorrect

Option A → Substitution cannot occur before differentiation.
Option B → Second derivative must be computed before substitution.
Option C → Cannot find y'' before finding y'.

🎯 Strategy

Used: Option Grouping

Application:

Arrange verification steps in their natural mathematical sequence.

Final Logic:

Differentiate → Differentiate → Substitute → Verify.

🧠 Memory Trick

"D-D-S-V" (Differentiate, Differentiate, Substitute, Verify).

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Verification of Solutions

3 The integral curves of
dy/dx = −x/y
are given by x²+y²=C.
If the curve passes through (3,4), find the diameter of the corresponding circle.

A 10 units

B 5 units

C 25 units

D 20 units

💡 Short Explanation

Substitute the given point. Find C. Determine radius and diameter.

✅ Correct Answer Explanation

Substituting (3,4) into x²+y²=C gives: 3²+4²=9+16=25 Thus C=25 and radius = √25 = 5. Therefore diameter = 2×5 = 10 units. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Radius, not diameter.
Option C → Value of C, not diameter.
Option D → Double the actual diameter.

🎯 Strategy

Used: Substitution

Application:

Use the point to determine the circle.

Final Logic:

C=25 ⇒ r=5 ⇒ diameter=10.

🧠 Memory Trick

"3-4-5 triangle ⇒ radius 5."

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

4 For the differential equation
dy/dx = 2x,
identify the incorrect statement.

A The solutions form a family of parabolas.

B The curves are symmetric about the y-axis.

C The curves intersect each other.

D The curves open upward.

💡 Short Explanation

Integrate the equation. Obtain y=x²+C. Different solution curves never intersect.

✅ Correct Answer Explanation

Integrating gives y=x²+C, a family of upward-opening parabolas symmetric about the y-axis. Different values of C produce vertically shifted curves. By uniqueness of solutions, two distinct solution curves cannot intersect. Therefore Option C is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct; y=x²+C is a parabola family.
Option B → Correct symmetry property.
Option D → Coefficient of x² is positive.

🎯 Strategy

Used: Elimination

Application:

Find the general solution and analyze its graph.

Final Logic:

Solution curves of a differential equation do not intersect.

🧠 Memory Trick

"Different solution curves never cross."

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

5 The general solution of y''=0 is y=mx+c.
For m=1, c=0, find the area bounded by the line, the x-axis, and x=0 to x=4.

A 8 sq units

B 4 sq units

C 16 sq units

D 2 sq units

💡 Short Explanation

Equation becomes y=x. Find area under line. Evaluate definite integral.

✅ Correct Answer Explanation

For m=1 and c=0, the line is y=x. Required area: ∫₀⁴ x dx = [x²/2]₀⁴ = 16/2 = 8 sq units. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Half of the correct area.
Option C → Equals x² at x=4, not area.
Option D → Far smaller than actual value.

🎯 Strategy

Used: Substitution

Application:

Substitute constants and integrate.

Final Logic:

Area under y=x from 0 to 4 equals 8.

🧠 Memory Trick

"Area under y=x = x²/2."

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

6 For the family y=Ce^x, find the unit vector parallel to the tangent at the point (0,1).

A 1/√2 î − 1/√2 ĵ

B 1/√2 î + 1/√2 ĵ

C î + ĵ

D ĵ

💡 Short Explanation

Determine slope. Form direction vector. Normalize it.

✅ Correct Answer Explanation

At (0,1), C=1. Thus y=e^x and dy/dx=e^x. At x=0, slope=1. A direction vector is (1,1). Its magnitude is √2. Hence the unit vector is: (1/√2)î + (1/√2)ĵ.

❓ Why Other Options Are Incorrect

Option A → Corresponds to slope −1.
Option C → Not a unit vector.
Option D → Vertical direction, slope undefined.

🎯 Strategy

Used: Substitution

Application:

Find slope and normalize the tangent vector.

Final Logic:

Slope 1 ⇒ unit vector (1/√2,1/√2).

🧠 Memory Trick

"Slope 1 ⇒ (1,1) ⇒ divide by √2."

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

7 Let y=x²+C and y(0)=k, where k is chosen randomly from {1,2,3,4,5}.
What is the probability that C is a prime number?

A 1/2

B 1/5

C 3/10

D 2/5

💡 Short Explanation

Use y(0)=k. Hence C=k. Count prime values.

✅ Correct Answer Explanation

Since y=x²+C and y(0)=k, k=C. Possible values of C are {1,2,3,4,5}. Prime values are 2,3,5. Thus favorable outcomes = 3 and total outcomes = 5. Probability = 3/5. The provided answer is incorrect.

❓ Why Other Options Are Incorrect

Option A → Equals 1/2, not 3/5.
Option B → Counts only one prime.
Option D → Gives 2/5 instead of 3/5.

🎯 Strategy

Used: Option Grouping

Application:

List possible values and count primes.

Final Logic:

Primes = {2,3,5} ⇒ 3/5.

🧠 Memory Trick

"2,3,5 are primes."

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

8 Given
x dy=(2x²+1)dx, y(1)=1,
the particular solution is
y=x²+ln|x|.
Evaluate y(e).
I. e²
II. e²+1
III. e²+ln e

A I only

B I and II

C II and III

D III only

💡 Short Explanation

Substitute x=e. Use ln e=1. Compare statements.

✅ Correct Answer Explanation

Given y=x²+ln|x|, y(e)=e²+ln e =e²+1. Thus Statements II and III are both correct. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores logarithmic term.
Option B → Omits Statement III.
Option D → Omits Statement II.

🎯 Strategy

Used: Substitution

Application:

Evaluate the expression directly.

Final Logic:

ln e = 1.

🧠 Memory Trick

"ln e = 1."

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

9 Given y_g=2x+C and y_p=2x+1, evaluate:
∫₀¹ (y_g−y_p) dx

A C+1

B C/2

C 1−C

D C−1

💡 Short Explanation

Simplify integrand. Integrate constant. Evaluate limits.

✅ Correct Answer Explanation

y_g−y_p=(2x+C)−(2x+1)=C−1. Therefore, ∫₀¹(C−1)dx=(C−1)[x]₀¹=C−1. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Wrong sign.
Option B → Constant integration done incorrectly.
Option C → Reverse of correct answer.

🎯 Strategy

Used: Substitution

Application:

Reduce the expression before integrating.

Final Logic:

Integral of constant C−1 over length 1 equals C−1.

🧠 Memory Trick

"Constant × interval length."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: General and Particular Solutions

10 Match each differential equation with the corresponding constant value.

List I	List II
1. dy=dx, y(0)=1	a. C=2
2. dy=2x dx, y(0)=2	b. C=1
3. dy/y=dx, y(0)=e	c. C=0
4. dy=−dx, y(0)=0	d. C=e

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

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

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

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

💡 Short Explanation

Integrate each equation. Apply initial conditions. Match constants.

✅ Correct Answer Explanation

1. y=x+C, y(0)=1 ⇒ C=1 (b) 2. y=x²+C, y(0)=2 ⇒ C=2 (a) 3. ln y=x+C, y(0)=e ⇒ C=1, but the intended matching uses y=Ce^x giving C=e (d) 4. y=−x+C, y(0)=0 ⇒ C=0 (c) Hence Option A is the intended answer.

❓ Why Other Options Are Incorrect

Option B → Several constants mismatched.
Option C → First equation incorrectly matched.
Option D → Second and third equations mismatched.

🎯 Strategy

Used: Option Grouping

Application:

Solve each differential equation and identify the constant.

Final Logic:

Matching gives Option A.

🧠 Memory Trick

"Integrate first, apply condition later."

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

11 Assertion (A):
y = A sin x satisfies the differential equation
y'' + y = 0.
Reason (R):
Differentiating twice gives y'' = −A sin x = −y.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R explains A

D A is false, R is true

💡 Short Explanation

Differentiate twice. Obtain y'' = −y. Substitution gives zero.

✅ Correct Answer Explanation

For y = A sin x, we get y' = A cos x and y'' = −A sin x = −y. Therefore y'' + y = −y + y = 0. The assertion is true. The reason correctly explains why the function satisfies the differential equation.

❓ Why Other Options Are Incorrect

Option A → Both statements are actually true.
Option B → Reason is not false.
Option D → Assertion is not false.

🎯 Strategy

Used: Substitution

Application:

Differentiate and substitute into the given equation.

Final Logic:

y'' = −y directly implies y'' + y = 0.

🧠 Memory Trick

"Sin becomes minus sin after two derivatives."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Verification of Solutions

12 For y = e^(kx) to satisfy
y'' − 5y' + 6y = 0,
find the sum of all possible values of k.

A -5

B 5

C 6

D -6

💡 Short Explanation

Assume y = e^(kx). Form auxiliary equation. Add roots.

✅ Correct Answer Explanation

Substituting y = e^(kx) gives: k²e^(kx) − 5ke^(kx) + 6e^(kx) = 0 ⇒ k² − 5k + 6 = 0 ⇒ (k−2)(k−3)=0 Thus k = 2, 3. Their sum is 5. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect sign.
Option C → Product of roots, not sum.
Option D → Does not satisfy the quadratic relation.

🎯 Strategy

Used: Substitution

Application:

Replace y by e^(kx) and solve for k.

Final Logic:

Roots 2 and 3 give sum 5.

🧠 Memory Trick

"Auxiliary roots: add for sum."

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

13 Which differential equation has general solution
y = C₁ cos 2x + C₂ sin 2x?

A y'' + 2y = 0

B y'' + 4y = 0

C y'' − 4y = 0

D y' + 4y = 0

💡 Short Explanation

Frequency equals 2. Characteristic roots are ±2i. Equation becomes y''+4y=0.

✅ Correct Answer Explanation

The solution y = C₁ cos 2x + C₂ sin 2x corresponds to characteristic roots ±2i. Therefore the auxiliary equation is m²+4=0, giving the differential equation y''+4y=0. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Produces cos x and sin x.
Option C → Gives exponential solutions.
Option D → First-order equation only.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Match the standard trigonometric solution form.

Final Logic:

Frequency 2 implies coefficient 4.

🧠 Memory Trick

"cos(ax), sin(ax) ⇒ y''+a²y=0."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Trigonometric Solutions

14 For y(t)=a cos t+b sin t with a=3, b=4, find the amplitude.

A 5

B 7

C 1

D 12

💡 Short Explanation

Use amplitude formula. √(a²+b²). Evaluate numerically.

✅ Correct Answer Explanation

For y(t)=a cos t+b sin t, amplitude is: √(a²+b²) = √(3²+4²) = √25 = 5. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Sum of coefficients.
Option C → Not from amplitude formula.
Option D → Product of coefficients.

🎯 Strategy

Used: Substitution

Application:

Insert values into the standard amplitude formula.

Final Logic:

√(9+16)=5.

🧠 Memory Trick

"Amplitude = Pythagoras."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Trigonometric Solutions

15 Which differential equation is satisfied by y=e^(mx), where m is a constant?

A y''=y

B x dy/dx=y

C y y''=(y')³

D y y''=(y')²

💡 Short Explanation

Differentiate y=e^(mx). Compute y' and y''. Compare both sides.

✅ Correct Answer Explanation

For y=e^(mx), y'=me^(mx)=my y''=m²e^(mx)=m²y Then: yy'' = y(m²y)=m²y² and (y')²=(my)²=m²y² Hence yy''=(y')². Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → True only when m²=1.
Option B → Not generally satisfied.
Option C → Gives mismatched powers.

🎯 Strategy

Used: Substitution

Application:

Evaluate y', y'' and test each option.

Final Logic:

yy'' and (y')² are identical.

🧠 Memory Trick

"Exponential: yy''=(y')²."

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

16 Given
dP/dt=0.05P,
the doubling time is t=20 ln 2.
Using ln 2≈0.6931, find t.

A 10.5 years

B 12.3 years

C 13.86 years

D 14.5 years

💡 Short Explanation

Use given formula. Substitute ln 2. Calculate product.

✅ Correct Answer Explanation

Given: t = 20 ln 2 Using ln 2 ≈ 0.6931, t = 20 × 0.6931 = 13.862 ≈ 13.86 years. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Product evaluated incorrectly.
Option B → Smaller than actual value.
Option D → Overestimates the product.

🎯 Strategy

Used: Substitution

Application:

Insert the numerical value into the formula.

Final Logic:

20×0.6931=13.86.

🧠 Memory Trick

"20 × 0.69 ≈ 14."

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

17 Identify the incorrect statement.

A A first-order equation has one arbitrary constant in its general solution.

B A second-order equation has two arbitrary constants.

C A particular solution contains no arbitrary constants.

D A third-order equation's particular solution contains three arbitrary constants.

💡 Short Explanation

General solutions contain constants. Particular solutions do not. Option D contradicts definition.

✅ Correct Answer Explanation

A first-order, second-order, and third-order differential equation generally contain one, two, and three arbitrary constants respectively in their general solutions. However, a particular solution is obtained after assigning values to those constants. Hence Option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct statement.
Option B → Correct statement.
Option C → Correct definition of a particular solution.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check whether the statement contradicts a standard definition.

Final Logic:

Particular solutions never contain arbitrary constants.

🧠 Memory Trick

"Particular ⇒ No arbitrary C."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: General and Particular Solutions

18 Which equations have exactly two arbitrary constants in their general solution?
I. (d²y)/(dx²)=0
II. dy/dx+y=0
III. y''+2y'+y=sin x

A I and III

B II only

C I only

D I, II, and III

💡 Short Explanation

Order determines constants. Second-order gives two constants. First-order gives one.

✅ Correct Answer Explanation

Equation I is second-order, so its general solution contains two arbitrary constants. Equation II is first-order and contains one arbitrary constant. Equation III is also second-order and therefore contains two arbitrary constants. Hence I and III are correct.

❓ Why Other Options Are Incorrect

Option B → First-order equation has only one constant.
Option C → Ignores Equation III.
Option D → Incorrectly includes Equation II.

🎯 Strategy

Used: Option Grouping

Application:

Determine the order of each equation.

Final Logic:

Second-order equations contain two arbitrary constants.

🧠 Memory Trick

"Order = Number of constants."

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

19 If a differential equation is separable, it can be written as
1/h(y) dy = g(x) dx
and integrated to obtain H(y)=G(x)+C.
For
dy/dx=(1+y²)/(1+x²),
after separation and integration, the result is:

A sin⁻¹y = sin⁻¹x

B ln(1+y²)=ln(1+x²)

C tan⁻¹y = tan⁻¹x + C

D e^y=e^x

💡 Short Explanation

Separate variables. Integrate both sides. Obtain inverse tangent functions.

✅ Correct Answer Explanation

Separating variables: dy/(1+y²)=dx/(1+x²) Integrating: ∫dy/(1+y²)=∫dx/(1+x²) gives tan⁻¹y = tan⁻¹x + C. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Integration gives tan⁻¹, not sin⁻¹.
Option B → Logarithmic integration is inappropriate here.
Option D → Exponential form does not result.

🎯 Strategy

Used: Substitution

Application:

Recognize the standard integral ∫dx/(1+x²).

Final Logic:

Integral yields inverse tangent.

🧠 Memory Trick

"1+x² ⇒ tan⁻¹x."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Variable Separable Equations

20 If the solution passes through (0,1), find the constant C.

A 0

B π/4

C π/2

D 1

💡 Short Explanation

Use result from Question 19. Substitute point (0,1). Solve for C.

✅ Correct Answer Explanation

From Question 19, tan⁻¹y = tan⁻¹x + C. Using (0,1): tan⁻¹(1)=tan⁻¹(0)+C π/4 = 0 + C Therefore C = π/4. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Does not satisfy the point condition.
Option C → Larger than required value.
Option D → Numerical value differs from π/4.

🎯 Strategy

Used: Substitution

Application:

Insert the given point into the general solution.

Final Logic:

π/4 = C.

🧠 Memory Trick

"tan⁻¹(1)=π/4."

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

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