UG Applied Mathematics Booster Test 2 - Formation of Differential Equations -

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UG Applied Mathematics Booster Test 2 - Formation of Differential Equations

⏱️ 30 Minutes

❓ 20 Questions

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

UG Applied Mathematics Booster Test 2 - Formation of Differential Equations

1 / 20

QUESTION 1 OF 20

If
\(y=Ae^{3x}+Be^{-2x},\)
eliminating the constants \(A\) and \(B\) requires forming a structural equation that strictly depends on:

Answer Review

1 If
\(y=Ae^{3x}+Be^{-2x},\)
eliminating the constants \(A\) and \(B\) requires forming a structural equation that strictly depends on:

A \(x, y,\)and \(A\) only

B \(x, y, y^{'},\)and \(y^{''}\)

C Integrals of \(y\)

D Constants \(A\) and \(B\)

💡 Short Explanation

Since the solution contains two independent arbitrary constants (\(A\) and \(B\)), it must be differentiated twice. After eliminating the constants, the resulting differential equation involves only \(x\), \(y\), \(y^{'}\), and \(y^{''}\).

✅ Correct Answer Explanation

The given family of curves is \(y=Ae^{3x}+Be^{-2x},\) where \(A\) and \(B\) are two arbitrary constants. Differentiate once: \(y^{'}=3Ae^{3x}-2Be^{-2x}.\) Differentiate again: \(y^{''}=9Ae^{3x}+4Be^{-2x}.\) The three equations, \(y,y^{'},y^{''},\) are then used to eliminate the arbitrary constants \(A\) and \(B\). The resulting differential equation contains only \(x\), \(y\), \(y^{'}\), and \(y^{''}\). Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
Incorrect because the arbitrary constant A should be eliminated.
The final differential equation cannot contain arbitrary constants.
Option B)
✅ Correct.
After eliminating both constants, the differential equation depends only on \(x\), \(y\), \(y^{'}\), and \(y^{''}\).
Option C)
Incorrect because forming a differential equation requires differentiation, not integration.
Option D)
Incorrect because the purpose of forming a differential equation is to eliminate the arbitrary constants.

🎯 Strategy

Used

Formation of Differential Equations by Eliminating Arbitrary Constants

Application:

1. Count the number of arbitrary constants.
2. Differentiate the equation the same number of times.
3. Eliminate the constants.
4. Express the final equation only in terms of \(x\), \(y\), and the required derivatives.

🧠 Memory Trick

The final differential equation never contains arbitrary constants.

📚 Chapter: Differential Equations

2 Match the family of curves to their fully resolved differential equation form.

List I	List II
1. — Circles centered at origin \(\left(x^{2}+y^{2}=a^{2}\right)\)	a. — \(y=2xy^{'}\)
2. — Straight lines \(\left(y=mx+c\right)\)	b. — \(x+yy^{'}=0\)
3. — Parabolas along x-axis \(\left(y^{2},\ 4ax\right)\)	c. — \(xy y^{''}+x(y^{'})^{2}-yy^{'}=0\)
4. — Central conics \(\left(\frac{x^{2}}{a^{2}}\pm \frac{y^{2}}{b^{2}}=1\right)\)	d. — \(y^{''}=0\)

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

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

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

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

💡 Short Explanation

Each family of curves has a standard differential equation obtained by eliminating its arbitrary constant(s). Match each curve with its corresponding differential equation.

✅ Correct Answer Explanation

1. Circles centered at the origin \(x^{2}+y^{2}=a^{2}\) Differentiating gives \(2x+2y\frac{dy}{dx}=0\) or \(x+y\frac{dy}{dx}=0.\) Hence, 1 → b 2. Straight lines \(y=mx+c\) Differentiate once: \(\frac{dy}{dx}=m.\) Differentiate again: \(\frac{d^{2}y}{dx^{2}}=0.\) Hence, 2 → d 3. Parabolas along the x-axis \(y^{2}=4ax\) Eliminating the constant \(a\) gives \(y=2x\frac{dy}{dx}.\) Hence, 3 → a 4. Central conics \(\frac{x^{2}}{a^{2}}\pm \frac{y^{2}}{b^{2}}=1\) Eliminating the two arbitrary constants requires two differentiations, giving \(xy\frac{d^{2}y}{dx^{2}}+x{\left(\frac{dy}{dx}\right)}^{2}-y\frac{dy}{dx}=0.\) Hence, 4 → c Therefore, the correct matching is: 1 → b 2 → d 3 → a 4 → c Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A
Correct. Every family of curves is matched with its correct differential equation.
Option B
Incorrect because circles do not correspond to
\(\frac{d^{2}y}{dx^{2}}=0,\)
and straight lines do not correspond to
\(x+y\frac{dy}{dx}=0.\)
Option C
Incorrect because the differential equations are mismatched with the corresponding families of curves.
Option D
Incorrect because the parabola and central conic equations are incorrectly matched.

🎯 Strategy

Used

Identify the Standard Family of Curves and Its Differential Equation
Recall the standard equation of each family.
Differentiate and eliminate arbitrary constant(s).
Match the resulting differential equation with the given family.

🧠 Memory Trick

Mnemonic: Circle → Line → Parabola → Conic (CLPC).

📚 Chapter: Differential Equations

3 For the circle equation x² + y² + 2gx + 2fy + c = 0 where the centre (-g, -f) is completely fixed but the radius varies (making 'c' the only parameter), which statements are true?
1. Differentiating the equation exactly once eliminates the parameter c.
2. The resulting differential equation is of order 1.
3. The unsimplified differential equation is 2x + 2yy' + 2g + 2fy' = 0.
4. The differential equation is of order 3.

A 1, 2, 4

B 1, 2

C 1, 2, 3

D 1, 2, 3, 4

💡 Short Explanation

One parameter (c) First differentiation removes it First-order DE formed

✅ Correct Answer Explanation

Since only c varies, differentiating once eliminates it. Thus: 1 true, 2 true, 3 true. 4 false because order is not 3.

❓ Why Other Options Are Incorrect

D → incorrect order claim
A/B → incomplete sets

🎯 Strategy

Elimination of single parameter

Final Logic: 1 parameter → 1st order DE

🧠 Memory Trick

"Only c varies → one derivative enough"

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

4 Identify the INCORRECT step in eliminating the fixed radius 'a' from the generic circle equation x² + (y - b)² = a² (assuming 'b' is a fixed known constant):

A Differentiating once gives 2x + 2(y - b)y' = 0.

B The constant 'a' vanishes immediately upon the first differentiation.

C The resulting differential equation has an order of 2.

D The final simplified DE is x + (y - b)y' = 0.

💡 Short Explanation

One constant only First derivative sufficient First-order DE only

✅ Correct Answer Explanation

Only parameter a exists → one differentiation removes it → order is 1, not 2.

❓ Why Other Options Are Incorrect

A → correct differentiation
B → correct elimination
D → correct simplification

🎯 Strategy

Order–parameter rule

Final Logic: 1 constant → first order

🧠 Memory Trick

"1 constant → order 1"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Radius Elimination

5 In the mixture of lines y = mx + c, if the intercept 'c' is strictly fixed at 5, we have y = mx + 5. The differential equation generated by eliminating the slope parameter 'm' is:

A y = xy' + 5

B y' = 5

C y'' = 0

D x y' = y

💡 Short Explanation

Differentiate once Replace slope m Substitute back

✅ Correct Answer Explanation

y = mx + 5 y' = m Substitute → y = xy' + 5

❓ Why Other Options Are Incorrect

B → incorrect constant relation
C → unrelated
D → misses constant 5

🎯 Strategy

Substitution method

Final Logic: m = y'

🧠 Memory Trick

"m → y' replacement"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Straight Line Family

6 Constraining the slope m = 1 in the family y = mx + c gives the parallel line family y = x + c. What is the specific differential equation eliminating the intercept parameter 'c'?

A y' = 1

B y' = x

C y'' = 1

D y = x

💡 Short Explanation

Fixed slope No elimination needed First derivative constant

✅ Correct Answer Explanation

Since slope is 1, derivative of y is always 1 → y' = 1.

❓ Why Other Options Are Incorrect

B → incorrect dependence
C → wrong order
D → original equation

🎯 Strategy

Direct slope interpretation

Final Logic: slope = derivative

🧠 Memory Trick

"Fixed slope → fixed derivative"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Straight Line Constraint

7 Sequentially calculating the derivative of
\(y=ce^{x}\)
once yields
\(y^{'}=ce^{x}.\)
What is the final differential equation after substituting \(c\) back?

A y^'=y

B y^'=c

C y^'=e^x

D y^''=y

💡 Short Explanation

Differentiate the given equation and substitute the original expression for \(ce^{x}\). This eliminates the arbitrary constant \(c\), giving the required differential equation.

✅ Correct Answer Explanation

Given, \(y=ce^{x},\) where \(c\) is an arbitrary constant. Differentiate with respect to \(x\): \(\frac{dy}{dx}=ce^{x}.\) From the original equation, \(ce^{x}=y.\) Substituting this into the derivative, \(\frac{dy}{dx}=y.\) Thus, the differential equation after eliminating the arbitrary constant is \(\frac{dy}{dx}=y.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
Correct.
Substituting \(ce^{x}=y\) into
\(y^{'}=ce^{x}\)
gives
\(y^{'}=y.\)
Option B)
Incorrect because the derivative of
\(ce^{x}\)
is \(ce^{x}\), not simply the constant \(c\)
Option C)
Incorrect because the arbitrary constant \(c\) has been omitted.
Option D)
Incorrect because the question asks for the differential equation after one differentiation. Although
\(y^{''}=y\)
is also true for this function, it is not the required equation.

🎯 Strategy

Used

Formation of Differential Equation by Eliminating an Arbitrary Constant

Application:

1. Differentiate the given family of curves.
2. Use the original equation to substitute for the arbitrary constant.
3. Eliminate the constant.
4. Express the final differential equation in terms of \(x\), \(y\), and derivatives only.

🧠 Memory Trick

\(y^{'}=y.\)

📚 Chapter: Differential Equations

8 Similar to averaging twice, differentiating y = a cos x + b sin x twice sequentially yields y'' = -a cos x - b sin x. The resulting simplified differential equation is:

A y'' - y = 0

B y'' + y = 0

C y'' = ab

D y' + y = 0

💡 Short Explanation

Two differentiations Constants cancel Harmonic equation

✅ Correct Answer Explanation

y'' = -a cos x - b sin x = -y → y'' + y = 0

❓ Why Other Options Are Incorrect

A → wrong sign
C → meaningless
D → incorrect form

🎯 Strategy

Standard harmonic pattern

Final Logic: sin/cos → second derivative negative

🧠 Memory Trick

"sin/cos → minus after 2nd derivative"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Harmonic Functions

9 What is the mathematical probability that the general solution polynomial y = c₁x + c₂x² + c₃x³ requires exactly a 3rd order differential equation to completely eliminate the three independent arbitrary constants?

A 0

B 0.5

C 0.75

D 1

💡 Short Explanation

Two constants Deterministic rule Second-order DE

✅ Correct Answer Explanation

Two constants always produce second-order DE, so probability is certain (1), not 3rd order but rule certainty is 1.

❓ Why Other Options Are Incorrect

A/B/C → probabilistic incorrectness

🎯 Strategy

Rule certainty

Final Logic: deterministic outcome

🧠 Memory Trick

"n constants → fixed order"

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

10 Vector-like repeated differentiation of
\(y=e^{2x}(a+bx)\)
requires the product rule. The first derivative evaluates to
\(y^{'}=2e^{2x}(a+bx)+be^{2x}.\)
This algebraically simplifies to:
\(y^{'}=2y+be^{2x}\)
\(y^{'}=2y\)
\(y^{'}=y+b\)
\(y^{'}=ae^{2x}\)

💡 Short Explanation

Differentiate the product using the Product Rule and substitute \(y=e^{2x}(a+bx)\) to simplify the expression.

✅ Correct Answer Explanation

Given, \(y=e^{2x}(a+bx).\) Apply the Product Rule: \(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx},\) where \(u=e^{2x},v=a+bx.\) Differentiate each factor: \(\frac{du}{dx}=2e^{2x},\frac{dv}{dx}=b.\) Therefore, \(y^{'}=2e^{2x}(a+bx)+be^{2x}.\) Since \(e^{2x}(a+bx)=y,\) substituting gives \(y^{'}=2y+be^{2x}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
Correct.
Replacing
\(e^{2x}(a+bx)\)
with \(y\) gives
\(y^{'}=2y+be^{2x}.\)
Option B)
Incorrect because it ignores the derivative of
\((a+bx),\)
which contributes the additional term
\(be^{2x}.\)
Option C)
Incorrect because the derivative is not obtained by simply adding \(b\) to \(y\).
Option D)
Incorrect because differentiation of the product produces two terms, not just
\(ae^{2x}.\)

🎯 Strategy

Used

Product Rule + Substitution

Application:

1. Identify the product of two functions.
2. Apply the Product Rule.
3. Differentiate each factor.
4. Substitute the original expression for \(y\).
5. Simplify the result.

🧠 Memory Trick

\(e^{2x}(a+bx)=y.\)

📚 Chapter: Differential Equations

11 To find the differential equation of parabolas with vertex at origin and symmetric about the x-axis (y² = 4ax), we differentiate to get 2yy' = 4a. Substituting 4a back gives the area-related slope relation:

A y² = 2xyy'

B y² = xyy'

C 2y² = xyy'

D y = xy'

💡 Short Explanation

Differentiate parabola Eliminate parameter a Substitute back

✅ Correct Answer Explanation

From y² = 4ax → differentiate: 2yy' = 4a From original: 4a = y²/x Substitute → 2yy' = y²/x → y² = 2xyy'

❓ Why Other Options Are Incorrect

B → missing factor 2
C → incorrect scaling
D → incomplete relation

🎯 Strategy

Substitution → eliminate a using original equation

Final Logic: express DE purely in x, y, y'

🧠 Memory Trick

"y² parabola → 2xyy'"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Parabola (x-axis form)

12 For parabolas x² = 4ay (axis along y-axis), eliminating 'a' yields 2x = 4ay'. Substituting 4a = x²/y into this yields the integration-free differential equation:

A x y' - 2y = 0

B 2x y' - y = 0

C x y' + 2y = 0

D x² y' - y = 0

💡 Short Explanation

Differentiate equation Eliminate parameter a Simplify

✅ Correct Answer Explanation

x² = 4ay Differentiate → 2x = 4a y' From original: 4a = x²/y Substitute → x y' = 2y → x y' - 2y = 0

❓ Why Other Options Are Incorrect

B → wrong coefficient placement
C → wrong sign
D → incorrect structure

🎯 Strategy

Substitution elimination of parameter a

Final Logic: reduce to first-order DE

🧠 Memory Trick

"x² parabola → x y' = 2y"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Parabola

13 For the hyperbola family x²/a² - y²/b² = 1, separating variables after the first derivative gives x/y = (a²/b²)y'. Differentiating again implicitly eliminates the ratio. What specific differentiation rule is essential here?

A Quotient Rule on x/y

B Logarithmic differentiation

C Partial differentiation

D Integration by parts

💡 Short Explanation

Ratio form appears Requires quotient rule Eliminates constants

✅ Correct Answer Explanation

Hyperbola manipulation produces x/y type expressions, requiring quotient rule for correct differentiation and elimination of constants.

❓ Why Other Options Are Incorrect

B → not required
C → irrelevant
D → unrelated

🎯 Strategy

Rule identification → choose correct differentiation tool

Final Logic: ratio → quotient rule

🧠 Memory Trick

"Ratio → Quotient Rule"

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

14 The differential equation xy y'' + x(y')² - y y' = 0 represents central conics. It applies identically to both ellipses and hyperbolas centered at the origin because:

A The arbitrary constants a² and b² are eliminated in exactly the same algebraic manner.

B Ellipses and hyperbolas inherently share the same eccentricity.

C The equation degree is strictly zero.

D They both pass identically through the origin.

💡 Short Explanation

Same elimination process Constants cancel similarly Unified DE form

✅ Correct Answer Explanation

Both conics reduce to identical second-order DE because elimination of a² and b² follows the same algebraic structure.

❓ Why Other Options Are Incorrect

B → false property
C → incorrect concept
D → irrelevant

🎯 Strategy

Structural comparison → identify common elimination pattern

Final Logic: same algebraic reduction

🧠 Memory Trick

"Different curves → same DE"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Conic Sections DE

15 The family of concentric circles at the origin is x² + y² = a². Its differential equation x + y y' = 0 geometrically implies that:

A The normal to the curve always passes through the origin.

B The tangent is strictly parallel to the x-axis.

C The radius is mathematically infinite.

D The geometric curve is a straight line.

💡 Short Explanation

Circle centered at origin Normal passes through center Geometric property

✅ Correct Answer Explanation

For x² + y² = a², the normal passes through the centre (origin), hence x + y y' = 0 implies normal alignment.

❓ Why Other Options Are Incorrect

B → not always true
C → meaningless
D → incorrect

🎯 Strategy

Geometric interpretation of DE

Final Logic: normal = radius direction

🧠 Memory Trick

"Circle → normal through centre"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Circle Geometry

16 Circles with their centre strictly on the x-axis and passing through the origin have the equation x² + y² = 2ax. Differentiating once yields 2x + 2yy' = 2a. Substituting 2a back gives the final differential equation:

A y² - x² - 2xyy' = 0

B x² - y² + 2xyy' = 0

C x² + y² + 2xyy' = 0

D x + yy' = 0

💡 Short Explanation

Differentiate Eliminate parameter Rearrange

✅ Correct Answer Explanation

Differentiate and substitute to eliminate a → final symmetric quadratic DE.

❓ Why Other Options Are Incorrect

B → wrong sign
C → incorrect form
D → unrelated

🎯 Strategy

Substitution + elimination

Final Logic: combine original + derivative

🧠 Memory Trick

"Circle through origin → mixed form"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Circle Passing Through Origin

📄 Passage

When finding the differential equation for families of conic sections, the number of arbitrary constants dictates the order. The equation of the family of ellipses having their foci on the y-axis and centre at the origin is x²/b² + y²/a² = 1. Since there are two arbitrary constants, 'a' and 'b', we must differentiate twice. The first differentiation yields 2x/b² + (2y/a²)(dy/dx) = 0. Rearranging gives (y y') / x = -a²/b². Since the right side is a constant, a second differentiation using the quotient rule on the left side will completely eliminate the constants, resulting in a second-order differential equation. Similarly, a family of parabolas with a fixed vertex but varying focus requires one differentiation.

Question:
According to the passage, why is the quotient rule applied to the term (y y') / x during the second differentiation?

A To safely integrate the function over bounds.

B Because the right side -a²/b² is a constant, and differentiating it yields zero, eliminating the parameters.

C To geometrically convert the ellipse into a hyperbola.

D To manually reduce the order from 2 to 1.Options:

💡 Short Explanation

RHS constant Derivative becomes zero Parameters eliminated

✅ Correct Answer Explanation

Since RHS is constant, differentiation yields zero; quotient rule ensures correct elimination of variables in LHS.

❓ Why Other Options Are Incorrect

A → unrelated
C → false
D → incorrect concept

🎯 Strategy

Constant differentiation rule

Final Logic: constant → 0

🧠 Memory Trick

"Constant → vanish"

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

📄 Passage

Question:
Based explicitly on the passage, a family of parabolas with a fixed vertex (e.g., origin) and varying focus has how many arbitrary constants, and what is its DE's order?

A 2 constants, Order 2

B 1 constant, Order 1

C 0 constants, Order 0

D 3 constants, Order 3

💡 Short Explanation

One parameter One differentiation First order DE

✅ Correct Answer Explanation

Only focus varies → one constant → one differentiation → first-order DE.

❓ Why Other Options Are Incorrect

A → extra constants
C → incorrect
D → unnecessary

🎯 Strategy

Parameter count rule

Final Logic: 1 parameter → order 1

🧠 Memory Trick

"1 parameter → 1 DE"

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

📄 Passage

Question:
A curve family defined by y = c sin x has exactly one parameter. Its resulting differential equation is

A y' = y cot x

B y' = c cos x

C y'' = -y

D y' = y tan x

💡 Short Explanation

Differentiate the given equation to eliminate the arbitrary constant \(c\). A second differentiation removes the trigonometric factor, giving the differential equation \(y^{''}=-y.\)

✅ Correct Answer Explanation

Given, \(y=csin⁡x,\) where \(c\) is the arbitrary constant. Differentiate once: \(y^{'}=ccos⁡x.\) Differentiate again: \(y^{''}=-csin⁡x.\) Since \(y=csin⁡x,\) substitute this into the second derivative: \(y^{''}=-y.\) Thus, the required differential equation is \(y^{''}=-y.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A)
\(y^{'}=ycot⁡x\)
Incorrect as the final differential equation.
From the first derivative,
\(y^{'}=ccos⁡x\)
and
\(c=\frac{y}{\sin\,x},\)
giving
\(y^{'}=ycot⁡x.\)
However, this still contains the trigonometric function \(\cot\,x\). The standard differential equation obtained after eliminating the parameter completely is
\(y^{''}=-y.\)
Option B)
\(y^{'}=ccos⁡x\)
Incorrect because the arbitrary constant \(c\) has not been eliminated.
Option C)
\(y^{''}=-y\)
Correct.
After two differentiations and substitution, the equation contains only \(y\) and its derivatives.
Option D)
\(y^{'}=ytan⁡x\)
Incorrect because
\(\frac{\cos\,x}{\sin\,x}=cot⁡x,\)
not \(\tan\,x\).

🎯 Strategy

Used

Formation of Differential Equations by Eliminating Arbitrary Constants

Application:

1. Differentiate the given family of curves.
2. Continue differentiating until the arbitrary constant is eliminated.
3. Substitute the original expression wherever possible.
4. Write the final differential equation only in terms of \(x\), \(y\), and derivatives.

🧠 Memory Trick

\(y^{''}=-y.\)

📚 Chapter: Differential Equations

📄 Passage

Question:
The family of functions given by y = c₁x + c₂ has exactly two independent parameters. Its final simplified differential equation is: A) y'' = 0 B) y' = c₁ C) x y'' = y' D) y'' = y

A y'' = 0

B y' = c₁

C x y'' = y'

D y'' = y

💡 Short Explanation

Linear function Two constants eliminated Second derivative zero

✅ Correct Answer Explanation

y = c₁x + c₂ → y' = c₁ → y'' = 0 removes all constants.

❓ Why Other Options Are Incorrect

B → still contains constant
C → incorrect relation
D → unrelated

🎯 Strategy

Direct differentiation

Final Logic: line → second derivative zero

🧠 Memory Trick

"Line → y'' = 0"

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Straight Line Family

UG Applied Mathematics Booster Test 2 - Formation of Differential Equations

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