UG Applied Mathematics Booster Test 3 - Formation of Differential Equations -

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UG Applied Mathematics Booster Test 3 - 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 3 - Formation of Differential Equations

1 / 20

QUESTION 1 OF 20

If the functional relation is
\(log⁡y=cx,\)
how is the constant \(c\) most efficiently and algebraically eliminated to form the differential equation?

Answer Review

1 If the functional relation is
\(log⁡y=cx,\)
how is the constant \(c\) most efficiently and algebraically eliminated to form the differential equation?

A Differentiate to \(\frac{y^{'}}{y}=c\), then substitute \(c=\frac{\log\,y}{x}\)to directly get \(xy^{'}=ylog⁡y\).

B Multiply the entire equation by \(e^{x}\).

C Differentiate twice with respect to \(x\).

D Set \(c=0\) unconditionally.

💡 Short Explanation

Differentiate the given relation once. Express the arbitrary constant \(c\) from the original equation. Substitute this value into the differentiated equation. This eliminates the constant and yields the required differential equation.

✅ Correct Answer Explanation

The given family of curves is \(log⁡y=cx.\) Differentiating both sides with respect to \(x\), \(\frac{1}{y}\frac{dy}{dx}=c.\) From the original equation, \(c=\frac{\log\,y}{x}.\) Substituting this into the differentiated equation, \(\frac{y^{'}}{y}=\frac{\log\,y}{x}.\) Multiplying both sides by \(xy\), \(xy^{'}=ylog⁡y.\) Thus, the arbitrary constant \(c\) is eliminated efficiently, giving the required differential equation. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because multiplying by \(e^{x}\)does not eliminate the arbitrary constant \(c\).
Option C) Incorrect because there is only one arbitrary constant, so only one differentiation is required.
Option D) Incorrect because the arbitrary constant cannot be assumed to be zero unless explicitly given.

🎯 Strategy

Differentiate and Substitute

Application:

Differentiate the given relation.
Express the arbitrary constant from the original equation.
Substitute it into the differentiated equation to eliminate the parameter.

Final Logic:

\(log⁡y=cx\)
↓
\(\frac{y^{'}}{y}=c\)
↓
\(c=\frac{\log\,y}{x}\)
↓
\(xy^{'}=ylog⁡y.\)
Therefore, Option A is correct.

🧠 Memory Trick

"DESE: Differentiate, Express, Substitute, Eliminate."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants

2 Match the complex families of curves to their corresponding differential equations.

List I	List II
1. \(y=ae^{2x}+be^{-x}\)	a. \(y^{''}-4y^{'}+4y=0\)
2. \(y=ce^{-x^{3}}\)	b. \(y^{''}+y=0\)
3. \(y=(a+bx)e^{2x}\)	c. \(y^{''}-y^{'}-2y=0\)
4. \(y=asin⁡x+bcos⁡x\)	d. \(y^{'}+3x^{2}y=0\)

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

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

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

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

💡 Short Explanation

Exponential solution families correspond to linear differential equations with constant coefficients. A single exponential function produces a first-order differential equation. The repeated exponential factor gives a repeated-root differential equation. Sine and cosine together correspond to the harmonic differential equation.

✅ Correct Answer Explanation

Each family of curves corresponds to a standard differential equation: 1 → c: \(y=ae^{2x}+be^{-x}\) has characteristic roots \(2\) and \(-1\). Hence the characteristic equation is \((m-2)(m+1)=0\) which gives \(y^{''}-y^{'}-2y=0.\) 2 → d: For \(y=ce^{-x^{3}},\) differentiating gives \(y^{'}=-3x^{2}y,\) or \(y^{'}+3x^{2}y=0.\) 3 → a: The family \(y=(a+bx)e^{2x}\) corresponds to a repeated root \(m=2\), giving \(\left(m-2)^{2}=0\right,\) which expands to \(y^{''}-4y^{'}+4y=0.\) 4 → b: The family \(y=asin⁡x+bcos⁡x\) satisfies the harmonic differential equation \(y^{''}+y=0.\) Thus, the correct matching is \(1\rightarrow c,2\rightarrow d,3\rightarrow a,4\rightarrow b.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A) Incorrect because it mismatches all four standard solution–equation pairs.
Option B) Incorrect because exponential, repeated-root, and trigonometric families are paired with incorrect differential equations.
Option C) Incorrect because only the repeated-root family is correctly matched, while the remaining pairs are incorrect.

🎯 Strategy

Pattern Recognition of Standard Differential Equation Forms

Application:

Identify the characteristic roots for exponential solution families.
Recognize repeated-root solutions.
Differentiate single exponential functions directly.
Recall the standard harmonic solution involving sine and cosine.

Final Logic:

Distinct Exponential → Characteristic Equation
↓
Repeated Exponential → Repeated Root DE
↓
Single Exponential → First-Order DE
↓
Sine & Cosine → Harmonic DE
Hence, Option D is correct.

🧠 Memory Trick

"Distinct Exp → Characteristic Roots | Single Exp → First Order | Repeated Exp → Repeated Root | Sin & Cos → Harmonic."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Standard Families of Curves)

3 Forming the differential equation for all circles with fixed radius \(r\) and varying centres strictly on the x-axis
\(\left(x-h)^{2}+y^{2}=r^{2}\right.\)
(where \(r\) is constant and \(h\) is an arbitrary parameter). Which statements are true?
Statements:
A. There is exactly 1 independent parameter (h).
B. The resulting differential equation is of order 1.
C. Differentiating once yields
2(x-h)+2yy^'=0,
meaning
(x-h)=-yy^'.
D. Substituting back yields the differential equation
y^2 (y^' )^2+y^2=r^2.

A A, B, D

B A, B

C A, B, C

D A, B, C, D

💡 Short Explanation

The family of circles contains one arbitrary parameter, \(h\). One differentiation is sufficient to eliminate this parameter. The differentiated equation gives \((x-h)=-yy^{'}\). Substituting this result into the original equation produces the required first-order differential equation.

✅ Correct Answer Explanation

Let's evaluate each statement: Statement A is correct because the family of circles contains only one arbitrary parameter, namely \(h\). Statement B is correct because eliminating one arbitrary parameter requires only one differentiation, resulting in a first-order differential equation. Statement C is correct because differentiating \(\left(x-h)^{2}+y^{2}=r^{2}\right.\) gives \(2(x-h)+2yy^{'}=0,\) which simplifies to \((x-h)=-yy^{'}.\) Statement D is correct because substituting \((x-h)=-yy^{'}\) into the original equation gives \(\left(-yy^{'})^{2}+y^{2}=r^{2}\right,\) or \(y^{2}(y^{'})^{2}+y^{2}=r^{2}.\) Therefore, all four statements are correct, making Option D the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because Statement C is also correct.
Option B) Incorrect because Statements C and D are also correct.
Option C) Incorrect because Statement D is also correct.

🎯 Strategy

Parameter Counting Rule

Application:

Count the number of arbitrary parameters.
Differentiate once to eliminate the parameter.
Substitute back into the original equation to obtain the differential equation.

Final Logic:

One Parameter
↓
One Differentiation
↓
Parameter Eliminated
↓
First-Order Differential Equation
Hence, all statements are true.

🧠 Memory Trick

"One Shift → One Parameter → One Differentiation → One DE."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Family of Circles)

4 For circles passing through the origin with centres uniquely on the y-axis, the equation is
\(x^{2}+y^{2}-2by=0.\)
Identify the INCORRECT statement in forming its differential equation.

A Differentiating gives 2x+2yy^'-2by^'=0.

B The parameter \(2b\) isolates to (2x+2yy^')/y^' .

C Substituting 2bback yields (x^2-y^2)y^'+2xy=0 instead of x^2-y^2+2xyy^'=0.

D The differential equation order is definitively 2 because radius and centre act as parameters.

💡 Short Explanation

The family contains only one arbitrary parameter, \(b\). One differentiation is sufficient to eliminate it. The resulting differential equation is first order. Therefore, the statement claiming a second-order differential equation is incorrect.

✅ Correct Answer Explanation

The given family of circles is \(x^{2}+y^{2}-2by=0,\) where \(b\) is the only arbitrary parameter. Differentiating with respect to \(x\), \(2x+2yy^{'}-2by^{'}=0,\) so Statement A is correct. Rearranging, \(2by^{'}=2x+2yy^{'}\) or \(2b=\frac{2x+2yy^{'}}{y^{'}},\) so Statement B is correct. Substituting this value of \(2b\) into the original equation, \(x^{2}+y^{2}-y\left(\frac{2x+2yy^{'}}{y^{'}}\right)=0,\) which simplifies to \((x^{2}-y^{2})y^{'}+2xy=0.\) Hence, Statement C is also correct. However, Statement D is incorrect because the family contains only one arbitrary parameter (\(b\)). The radius is not an independent parameter; it depends on the centre since the circle passes through the origin. Therefore, only one differentiation is required, and the resulting differential equation is first order, not second order. Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the differentiation is performed correctly.
Option B) Incorrect because the parameter \(2b\) is correctly isolated from the differentiated equation.
Option C) Incorrect because substituting the value of \(2b\) correctly gives
\((x^{2}-y^{2})y^{'}+2xy=0.\)

🎯 Strategy

Parameter Counting Rule

Application:

Determine the number of independent arbitrary parameters before deciding the order of the differential equation.

Final Logic:

One Independent Parameter (\(b\))
↓
One Differentiation
↓
First-Order Differential Equation
Therefore, Statement D is incorrect.

🧠 Memory Trick

"One Parameter → One Order."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Circles Passing Through the Origin)

5 In a mixture of tangent lines to the standard parabola \(x^{2}=4ay\), if the family of lines is given by
\(y=mx+\frac{a}{m},\)
where \(m\) is the arbitrary parameter, what is the resulting non-linear differential equation?

A \(y=xy^{'}+\frac{a}{y^{'}}\)

B \(y=xy^{'}+a\)

C \(y^{'}=m\)

D \(y=\frac{x}{y^{'}}\)

💡 Short Explanation

Differentiate the given family once. Replace the arbitrary parameter \(m\) by \(y^{'}\). Substitute this value into the original equation. This eliminates the parameter and produces the required non-linear differential equation.

✅ Correct Answer Explanation

The given family of tangent lines is \(y=mx+\frac{a}{m},\) where \(m\) is the only arbitrary parameter. Differentiating with respect to \(x\), \(\frac{dy}{dx}=m.\) Thus, \(m=y^{'}.\) Substituting this into the original family, \(y=xy^{'}+\frac{a}{y^{'}}.\) This equation contains no arbitrary parameter and is therefore the required differential equation. Since the derivative appears in the denominator, the equation is non-linear. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the term \(\frac{a}{m}\)becomes \(\frac{a}{y^{'}}\), not simply \(a\).
Option C) Incorrect because \(y^{'}=m\) is only the intermediate result obtained after differentiation; the parameter has not yet been eliminated.
Option D) Incorrect because it does not result from substituting \(m=y^{'}\)into the original family of curves.

🎯 Strategy

Differentiate and Substitute

Application:

Differentiate the given family.
Replace the arbitrary parameter using the derivative.
Substitute back into the original equation to eliminate the parameter.

Final Logic:

\(y=mx+\frac{a}{m}\)
↓
\(y^{'}=m\)
↓
\(m=y^{'}\)
↓
\(y=xy^{'}+\frac{a}{y^{'}}.\)
Hence, Option A is correct.

🧠 Memory Trick

"Slope becomes \(y^{'}\)."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants

6 The family of lines having an x-intercept \(a\) and y-intercept \(b\) is
\(\frac{x}{a}+\frac{y}{b}=1.\)
If \(a\) and \(b\) are entirely independent parameters, what is the exact geometric constraint inherently shown by its defined differential equation
\(y^{''}=0\)

A The slope is continuously changing.

B The rate of change of the slope is rigorously zero, meaning it's a straight line family.

C It implicitly forms a circle boundary.

D The axis intercepts must theoretically be equal.

💡 Short Explanation

The differential equation is \(y^{''}=0\). This means the second derivative is zero everywhere. Hence, the slope remains constant. Therefore, the family represents straight lines.

✅ Correct Answer Explanation

The family \(\frac{x}{a}+\frac{y}{b}=1\) contains two independent parameters, \(a\) and \(b\). Eliminating these parameters leads to the differential equation \(y^{''}=0.\) The equation \(y^{''}=0\) implies that the second derivative of \(y\) with respect to \(x\) is zero, meaning the slope \(\frac{dy}{dx}\) is constant. A constant slope is the defining property of a straight line. Therefore, the differential equation represents the entire family of straight lines, making Option B correct.

❓ Why Other Options Are Incorrect

Option A) Incorrect because a continuously changing slope would require \(y^{''}\neq 0\).
Option C) Incorrect because circles satisfy second-order differential equations different from \(y^{''}=0\).
Option D) Incorrect because there is no requirement that the intercepts \(a\) and \(b\) be equal; they are independent arbitrary parameters.

🎯 Strategy

Geometric Interpretation of Differential Equations

Application:

Interpret the meaning of the second derivative rather than solving the equation.

Final Logic:

\(y^{''}=0\)
↓
Slope is Constant
↓
Graph is a Straight Line
↓
Option B is correct.

🧠 Memory Trick

"\(y^{''}=0\) ⇒ Constant Slope."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (Straight Line Family)

7 Like computing sequential decay steps, for the family
\(y=Ae^{kx},\)
where \(k\) is a fixed constant and \(A\) is the sole arbitrary parameter, the first derivative gives
\(y^{'}=kAe^{kx}.\)
What is the exact differential equation?

A \(y^{'}=ky\)

B \(y^{'}=k^{2}y\)

C \(y^{''}=ky\)

D \(y^{'}=Ay\)

💡 Short Explanation

Differentiate the given family once. Replace \(Ae^{kx}\)by \(y\). This eliminates the arbitrary constant \(A\). The resulting differential equation is \(y^{'}=ky\).

✅ Correct Answer Explanation

The given family of curves is \(y=Ae^{kx},\) where \(A\) is the only arbitrary parameter and \(k\) is a fixed constant. Differentiating with respect to \(x\), \(y^{'}=kAe^{kx}.\) Since \(Ae^{kx}=y,\) substituting into the differentiated equation gives \(y^{'}=ky.\) This equation contains no arbitrary parameter and is therefore the required differential equation. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because differentiating \(Ae^{kx}\)once introduces only one factor of \(k\), not \(k^{2}\).
Option C) Incorrect because only one arbitrary parameter is present, so one differentiation is sufficient; the required equation is first order.
Option D) Incorrect because \(A\) is the arbitrary parameter that must be eliminated from the differential equation.

🎯 Strategy

Differentiate and Eliminate the Arbitrary Constant

Application:

Differentiate the family once.
Replace the original expression by \(y\).
Eliminate the arbitrary constant \(A\).

Final Logic:

\(y=Ae^{kx}\)
↓
\(y^{'}=kAe^{kx}\)
↓
\(Ae^{kx}=y\)
↓
\(y^{'}=ky\)
Hence, Option A is correct.

🧠 Memory Trick

"\(Ae^{kx}\) becomes \(y\)."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (Exponential Family)

8 Verifying the steps for
\(y=(a+bx)e^{2x},\)
the first derivative is
\(y^{'}=2y+be^{2x}.\)
Differentiating again gives
\(y^{''}=2y^{'}+2be^{2x}.\)
To permanently eliminate \(b\), we substitute
\(be^{2x}=y^{'}-2y\)
into the second equation. The result is:

A \(y^{''}-4y^{'}+4y=0\)

B \(y^{''}+4y=0\)

C \(y^{''}-2y^{'}=0\)

D \(y^{''}-4y^{'}-4y=0\)

💡 Short Explanation

Differentiate the given family twice. Express \(be^{2x}\)from the first derivative. Substitute this expression into the second derivative. Rearranging gives the required second-order differential equation.

✅ Correct Answer Explanation

The given family is \(y=(a+bx)e^{2x}.\) Its first derivative is \(y^{'}=2y+be^{2x}.\) Hence, \(be^{2x}=y^{'}-2y.\) Differentiating again, \(y^{''}=2y^{'}+2be^{2x}.\) Substituting \(be^{2x}=y^{'}-2y,\) gives \(y^{''}=2y^{'}+2(y^{'}-2y).\) Simplifying, \(y^{''}=4y^{'}-4y.\) Rearranging, \(y^{''}-4y^{'}+4y=0.\) Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the \(y^{'}\)term cannot disappear after substitution.
Option C) Incorrect because it ignores the contribution of the \(y\) term obtained during substitution.
Option D) Incorrect because the sign of the \(4y\) term is incorrect; it should be positive after rearranging to one side.

🎯 Strategy

Differentiate Twice and Substitute

Application:

Differentiate the family twice.
Isolate the remaining arbitrary-parameter term.
Substitute it into the second derivative.
Rearrange to eliminate the parameter completely.

Final Logic:

\(y^{'}\)
↓
\(be^{2x}=y^{'}-2y\)
↓
Substitute into \(y^{''}\)
↓
\(y^{''}-4y^{'}+4y=0.\)
Hence, Option A is correct.

🧠 Memory Trick

"DISR: Differentiate, Isolate, Substitute, Rearrange."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (Repeated Root Exponential Family)

9 What is the probability that the general differential equation for the conic
\(ax^{2}+2hxy+by^{2}+2gx+2fy+c=0\)
(containing 5 independent ratios) has a final order of exactly 5?

A 0

B 0.5

C 0.75

D 1

💡 Short Explanation

A general conic has five independent arbitrary constants (ratios). The order of the differential equation equals the number of independent arbitrary parameters eliminated. Therefore, the resulting differential equation is always of order 5.

✅ Correct Answer Explanation

The general equation of a conic is \(ax^{2}+2hxy+by^{2}+2gx+2fy+c=0.\) Since multiplying all coefficients by the same non-zero constant does not change the curve, there are five independent ratios among the six coefficients. According to the principle of forming differential equations: Order of the differential equation = Number of independent arbitrary parameters. Hence, \(5 independent parameters⟹Order =5.\) Therefore, the probability that the resulting differential equation has order exactly 5 is certain, i.e., \(P=1.\) Hence, Option D is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the order is certainly 5, not impossible.
Option B) Incorrect because the order is not determined by chance; it is fixed by the number of independent parameters.
Option C) Incorrect because there is no uncertainty involved. The order is always 5.

🎯 Strategy

Parameter Counting Rule

Application:

Count the number of independent arbitrary parameters in the family of curves.

Final Logic:

5 Independent Ratios
↓
5 Arbitrary Parameters
↓
5 Differentiations
↓
Order = 5
↓
Probability = 1
Hence, Option D is correct.

🧠 Memory Trick

"Five Ratios → Fifth Order."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (General Conic Family)

10 For
\(y=e^{x}(Acos⁡x+Bsin⁡x),\)
treating the trigonometric components like orthogonal vectors during differentiation gives
\(y^{'}-y=e^{x}(-Asin⁡x+Bcos⁡x).\)
Differentiating again yields
\(y^{''}-y^{'}=e^{x}(-Acos⁡x-Bsin⁡x)+e^{x}(-Asin⁡x+Bcos⁡x).\)
Substituting the previously obtained relations gives:

A \(y^{''}-2y^{'}+2y=0\)

B \(y^{''}+2y^{'}+2y=0\)

C \(y^{''}-y^{'}+y=0\)

D \(y^{''}=0\)

💡 Short Explanation

Differentiate the family twice. Use the first derivative relation to eliminate the arbitrary constants. Substitute into the second derivative. Rearranging gives the required second-order differential equation.

✅ Correct Answer Explanation

The given family is \(y=e^{x}(Acos⁡x+Bsin⁡x).\) First differentiation gives \(y^{'}-y=e^{x}(-Asin⁡x+Bcos⁡x).\) Differentiating again, \(y^{''}-y^{'}=e^{x}(-Acos⁡x-Bsin⁡x)+e^{x}(-Asin⁡x+Bcos⁡x).\) Notice that \(e^{x}(-Acos⁡x-Bsin⁡x)=-y,\) and \(e^{x}(-Asin⁡x+Bcos⁡x)=y^{'}-y.\) Substituting these relations, \(y^{''}-y^{'}=-y+(y^{'}-y).\) Simplifying, \(y^{''}-y^{'}=y^{'}-2y,\) which becomes \(y^{''}-2y^{'}+2y=0.\) Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the signs of both the \(y^{'}\)and \(y\) terms are incorrect after substitution.
Option C) Incorrect because it omits one factor of \(y^{'}\)and does not satisfy the given family of solutions.
Option D) Incorrect because the given family contains two arbitrary constants, requiring a second-order differential equation rather than \(y^{''}=0\).

🎯 Strategy

Differentiate Twice and Substitute

Application:

Differentiate the family twice.
Express the exponential–trigonometric terms using previously obtained relations.
Eliminate the arbitrary constants by substitution.

Final Logic:

\(y\)
↓
\(y^{'}-y\)
↓
Differentiate Again
↓
Substitute
↓
\(y^{''}-2y^{'}+2y=0.\)
Hence, Option A is correct.

🧠 Memory Trick

"Exp–Trig ⇒ \(2, -2, +2\)."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (Exponential–Trigonometric Family)

11 The differential equation for parabolas with vertex at origin and axis along x-axis is derived from y² = 4ax. The area-related differential form 2yy' = 4a leads to substitution 4a = 2yy'. The differential equation is:

A y = 2x y'

B x = 2y y'

C y² = x y'

D y' = 2xy

💡 Short Explanation

Differentiate parabola Eliminate parameter a Substitute back

✅ Correct Answer Explanation

y² = 4ax Differentiate → 2yy' = 4a From original → 4a = y²/x Substitute → y² = 2xyy' → simplifies to y = 2x y'

❓ Why Other Options Are Incorrect

B → wrong rearrangement
C → incorrect structure
D → not derived form

🎯 Strategy

Substitution elimination

Final Logic: remove a using original equation

🧠 Memory Trick

"Parabola → y = 2x y'"

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

12 For the structural family
\(x^{2}=4ay,\)
differentiating gives
\(2x=4ay^{'}.\)
Dividing
\(x^{2}=4ay\)
directly by
\(2x=4ay^{'}\)
yields
\(\frac{x}{2}=\frac{y}{y^{'}}.\)
This uniquely simplifies without integration to:

A \(xy^{'}=2y\)

B \(xy^{'}=y\)

C \(yy^{'}=2x\)

D \(x^{2}y^{'}=2y\)

💡 Short Explanation

Differentiate the family once. Divide the original equation by the differentiated equation. Simplify the resulting expression. Rearranging gives the required differential equation.

✅ Correct Answer Explanation

The given family is \(x^{2}=4ay.\) Differentiating with respect to \(x\), \(2x=4ay^{'}.\) Now divide the original equation by the differentiated equation: \(\frac{x^{2}}{2x}=\frac{4ay}{4ay^{'}}.\) This simplifies to \(\frac{x}{2}=\frac{y}{y^{'}}.\) Cross-multiplying, \(xy^{'}=2y.\) Thus, the arbitrary parameter \(a\) is eliminated, giving the required differential equation. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because cross-multiplication gives \(xy^{'}=2y\), not \(xy^{'}=y\).
Option C) Incorrect because it incorrectly interchanges the variables during simplification.
Option D) Incorrect because an extra factor of \(x\) is introduced, which does not arise from the algebraic manipulation.

🎯 Strategy

Divide and Simplify

Application:

Differentiate the given family.
Divide the original equation by the differentiated equation.
Cross-multiply to eliminate the arbitrary parameter.

Final Logic:

\(x^{2}=4ay\)
↓
\(2x=4ay^{'}\)
↓
\(\frac{x}{2}=\frac{y}{y^{'}}\)
↓
\(xy^{'}=2y.\)
Hence, Option A is correct.

🧠 Memory Trick

"Parabola in \(x\)⇒ \(xy^{'}=2y\)."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Family of Parabolas)

13 For the hyperbola
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\)
differentiating twice gives
\(xy y^{''}+x(y^{'})^{2}-yy^{'}=0.\)
What happens analytically if the family were ellipses
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
instead?

A The signs in the final differential equation change completely.

B The resulting differential equation is algebraically identical.

C The order of the equation strictly reduces to 1.

D It morphs into a non-polynomial differential equation.

💡 Short Explanation

Both the ellipse and hyperbola families contain two arbitrary parameters. Eliminating these parameters requires the same differentiation process. After elimination, the sign difference disappears. Hence, both families produce the same differential equation.

✅ Correct Answer Explanation

The hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) both contain the arbitrary parameters \(a\) and \(b\). When the parameters are eliminated through successive differentiation, the intermediate sign difference between the two equations is removed during the algebraic elimination process. Consequently, both families lead to the same second-order differential equation: \(xy y^{''}+x(y^{'})^{2}-yy^{'}=0.\) Thus, the resulting differential equation is algebraically identical for both families. Hence, Option B is the correct answer

❓ Why Other Options Are Incorrect

Option A) Incorrect because the sign difference in the original family disappears after eliminating the arbitrary parameters.
Option C) Incorrect because both families contain two arbitrary parameters, so the resulting differential equation remains second order.
Option D) Incorrect because the elimination process yields a polynomial differential equation, not a non-polynomial one.

🎯 Strategy

Parameter Elimination Analysis

Application:

Compare the elimination process for both conic families rather than the original equations.

Final Logic:

Hyperbola
↓
Differentiate
↓
Eliminate \(a,b\)
↓
Second-Order DE
↓
Ellipse follows the same elimination
↓
Identical Differential Equation
Hence, Option B is correct.

🧠 Memory Trick

"± disappears after elimination."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Conic Families by Eliminating Arbitrary Constants)

14 For the equation of a circle
\(x^{2}+y^{2}=r^{2}\)
centered at the origin, if \(r\) varies, the family covers the entire plane. The differential equation
\(x+yy^{'}=0\)
is of order 1.
If the circle's centre was instead moving along the x-axis with arbitrary radius, the differential equation's order would definitively be:

A 1

B 2

C 3

D 4

💡 Short Explanation

A circle centered at the origin has only one arbitrary parameter (\(r\)). A circle with its centre moving along the x-axis and arbitrary radius has two arbitrary parameters. The order of the differential equation equals the number of independent arbitrary parameters. Therefore, the required differential equation is of second order.

✅ Correct Answer Explanation

The family \(x^{2}+y^{2}=r^{2}\) contains only one arbitrary parameter, namely the radius \(r\). Hence, eliminating this parameter requires one differentiation, giving a first-order differential equation. If the centre is allowed to move along the x-axis, the family becomes \(\left(x-h)^{2}+y^{2}=r^{2}\right,\) where \(h\)= x-coordinate of the centre (arbitrary), \(r\)= radius (arbitrary). Thus, the family contains two independent arbitrary parameters. According to the rule for forming differential equations, Order of the differential equation = Number of independent arbitrary parameters. Therefore, \(2 parameters ⟹ Second-order differential equation.\) Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because the family has two independent arbitrary parameters, not one.
Option C) Incorrect because there are only two arbitrary parameters, so a third-order differential equation is unnecessary.
Option D) Incorrect because the order cannot exceed the number of independent arbitrary parameters.

🎯 Strategy

Parameter Counting Rule

Application:

Count the independent arbitrary parameters before determining the order of the differential equation.

Final Logic:

\(\left(x-h)^{2}+y^{2}=r^{2}\right.\)
↓
Parameters: \(h, r\)
↓
2 Independent Parameters
↓
2 Differentiations
↓
\(Order =2\)
Hence, Option B is correct.

🧠 Memory Trick

"Parameters = Order."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (Family of Circles)

15 For the specific single circle
\(x^{2}+y^{2}=25\)
(where the radius is strictly fixed at \(5\) and not arbitrary), differentiating once gives
\(x+yy^{'}=0.\)
This identical differential equation represents:

A Only the curve \(x^{2}+y^{2}=25\)

B All concentric circles of the form \(x^{2}+y^{2}=a^{2}\)

C Straight lines spanning from the origin

D Parabolas oriented along the axes

💡 Short Explanation

Differentiation removes the constant radius. The resulting differential equation no longer contains the value \(5\). Hence, it represents the entire family of concentric circles. A differential equation describes a family of curves, not a single curve.

✅ Correct Answer Explanation

The given circle is \(x^{2}+y^{2}=25.\) Differentiating with respect to \(x\), \(2x+2yy^{'}=0,\) or \(x+yy^{'}=0.\) Notice that the constant \(25\) disappears during differentiation. Consequently, the differential equation does not retain information about the specific radius \(5\). Integrating the differential equation, \(x+yy^{'}=0,\) gives \(x^{2}+y^{2}=C,\) where \(C\) is an arbitrary constant. Writing \(C=a^{2}\), \(x^{2}+y^{2}=a^{2}.\) Thus, the differential equation represents all concentric circles centered at the origin, not just the single circle of radius \(5\). Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A) Incorrect because differentiation eliminates the fixed constant \(25\), so the differential equation represents a family rather than one specific circle.
Option C) Incorrect because \(x+yy^{'}=0\) is not the differential equation of a family of straight lines.
Option D) Incorrect because parabolas satisfy different differential equations and do not arise from this relation.

🎯 Strategy

Interpret the General Solution

Application:

Differentiate the given curve and observe that the fixed constant disappears. Then identify the family obtained by integrating the differential equation.

Final Logic:

\(x^{2}+y^{2}=25\)
↓
Differentiate
↓
\(x+yy^{'}=0\)
↓
Integrate
↓
\(x^{2}+y^{2}=C=a^{2}\)
↓
Family of Concentric Circles
Hence, Option B is correct.

🧠 Memory Trick

"One Circle becomes All Circles."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (Family of Circles)

16 For the complex family of circles passing through the origin and having their centres explicitly on the y-axis, the initial equation is
\(x^{2}+y^{2}-2ay=0.\)
Algebraically eliminating the arbitrary parameter \(a\) yields:

A \((x^{2}-y^{2})y^{'}-2xy=0\)

B \((x^{2}-y^{2})-2xyy^{'}=0\)

C \(x^{2}+y^{2}-2xyy^{'}=0\)

D \(x^{2}y^{'}+y^{2}=0\)

💡 Short Explanation

Differentiate the given family once. Express the arbitrary parameter \(a\) in terms of \(x\), \(y\), and \(y^{'}\). Substitute this expression into the original equation. Simplifying gives the required first-order differential equation.

✅ Correct Answer Explanation

The given family of circles is \(x^{2}+y^{2}-2ay=0,\) where \(a\) is the only arbitrary parameter. Differentiating with respect to \(x\), \(2x+2yy^{'}-2ay^{'}=0.\) Hence, \(2a=\frac{2x+2yy^{'}}{y^{'}}.\) Substituting this value into the original equation, \(x^{2}+y^{2}-y\left(\frac{2x+2yy^{'}}{y^{'}}\right)=0.\) Multiplying throughout by \(y^{'}\), \(x^{2}y^{'}+y^{2}y^{'}-2xy-2y^{2}y^{'}=0.\) Combining like terms, \(x^{2}y^{'}-y^{2}y^{'}-2xy=0.\) Factoring, \((x^{2}-y^{2})y^{'}-2xy=0.\) Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the derivative should multiply the entire factor \(\left(x^{2},\ y^{2}\right)\), not only the term \(2xy\).
Option C) Incorrect because it fails to eliminate the parameter correctly and retains an incorrect algebraic form.
Option D) Incorrect because it does not result from substituting the value of \(a\) into the original equation.

🎯 Strategy

Differentiate and Substitute

Application:

Differentiate the given family.
Express the arbitrary parameter from the differentiated equation.
Substitute back into the original equation and simplify.

Final Logic:

\(x^{2}+y^{2}-2ay=0\)
↓
Differentiate
↓
Express \(a\)
↓
Substitute
↓
Simplify
↓
\((x^{2}-y^{2})y^{'}-2xy=0.\)
Hence, Option A is correct

🧠 Memory Trick

"DESF: Differentiate, Express, Substitute, Factor.

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations by Eliminating Arbitrary Constants (Family of Circles Passing Through the Origin)

📄 Passage

The formation of a differential equation involves translating a geometric family of curves into a relation between coordinates and their derivatives. For instance, consider the family of curves
y=e^2x (a+bx),

which contains two arbitrary constants aand b. Differentiating once with respect to xusing the product rule yields
y^'=2e^2x (a+bx)+be^2x.

Notice that the term
e^2x (a+bx)

is exactly the original function y. Substituting yback into the derivative simplifies the equation to
y^'=2y+be^2x.

To eliminate the remaining parameter b, we differentiate a second time to get
y^''=2y^'+2be^2x.

By algebraic substitution, we replace
be^2x=y^'-2y

to obtain the final second-order differential equation free of any arbitrary constants

Form the differential equation for ellipses with foci on the x-axis and their centre fixed at the origin
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.\)
After differentiating implicitly twice, the correct arrangement is:

A \(xy y^{''}+x(y^{'})^{2}-yy^{'}=0\)

B \(yy^{''}+x(y^{'})^{2}=0\)

C \(xy^{''}+yy^{'}=0\)

D \(x^{2}y^{''}+y^{2}y^{'}=0\)

💡 Short Explanation

The ellipse contains two arbitrary parameters, \(a\) and \(b\). Eliminating them requires two differentiations. The resulting equation is a second-order differential equation. The final simplified form is \(xy y^{''}+x(y^{'})^{2}-yy^{'}=0\).

✅ Correct Answer Explanation

The family of ellipses \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) contains two arbitrary constants, \(a\) and \(b\). Differentiating the equation twice and eliminating these parameters gives the differential equation \(xy y^{''}+x(y^{'})^{2}-yy^{'}=0.\) This equation contains no arbitrary constants and therefore completely represents the given family of ellipses. Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because it omits the essential \(xy\) and \(-yy^{'}\)terms.
Option C) Incorrect because it is not the differential equation obtained after eliminating both arbitrary parameters.
Option D) Incorrect because the powers of \(x\) and \(y\) do not arise in the elimination process.

🎯 Strategy

Parameter Elimination

Application:

Differentiate twice and eliminate the two arbitrary parameters \(a\) and \(b\).

Final Logic:

Two Parameters
↓
Two Differentiations
↓
Parameter Elimination
↓
\(xy y^{''}+x(y^{'})^{2}-yy^{'}=0.\)

🧠 Memory Trick

"Ellipse → Two Parameters → Two Differentiations."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Ellipse Family)

📄 Passage

The simplified differential equation representing the family of parabolas having their vertices exactly at the origin and foci on the positive direction of the x-axis
\(y^{2}=4ax\)
is mathematically:

A \(2x\frac{dy}{dx}-y=0\)

B \(x\frac{dy}{dx}-2y=0\)

C \(y\frac{dy}{dx}-2x=0\)

D \(2y\frac{dy}{dx}-x=0\)

💡 Short Explanation

Differentiate the parabola equation once. Eliminate the parameter \(a\). Rearrange the equation. The required differential equation is \(2xy^{'}-y=0\).

✅ Correct Answer Explanation

The family of parabolas is \(y^{2}=4ax.\) Differentiating, \(2yy^{'}=4a.\) Substituting \(4a=2yy^{'}\) into the original equation, \(y^{2}=2xyy^{'}.\) Dividing by \(y\), \(y=2xy^{'}.\) Rearranging, \(2xy^{'}-y=0.\) Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the coefficients of \(x\) and \(y\) are interchanged.
Option C) Incorrect because it incorrectly places the derivative with \(y\).
Option D) Incorrect because differentiation does not produce this relation.

🎯 Strategy

Differentiate and Substitute

Application:

Differentiate once and substitute the value of the arbitrary parameter.

Final Logic:

\(y^{2}=4ax\)
↓
Differentiate
↓
Substitute
↓
\(2xy^{'}-y=0.\)

🧠 Memory Trick

"Parabola ⇒ \(2xy^{'}=y\)."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Parabola Family)

📄 Passage

According strictly to the passage's underlying substitution principles, if the curve is defined as
y=ae^2x,
where ais the only arbitrary constant, what is its fully simplified differential equation?

A \(y^{'}=2y\)

B \(y^{'}=a\)

C \(y^{'}=e^{2x}\)

D \(y^{''}=4y\)

💡 Short Explanation

Differentiate the equation once. Replace \(ae^{2x}\)by \(y\). The arbitrary constant disappears. The resulting equation is first order.

✅ Correct Answer Explanation

Given \(y=ae^{2x},\) differentiate to obtain \(y^{'}=2ae^{2x}.\) Since \(ae^{2x}=y,\) substituting gives \(y^{'}=2y.\) Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the arbitrary constant must be eliminated.
Option C) Incorrect because the derivative depends on both \(a\) and \(e^{2x}\).
Option D) Incorrect because only one arbitrary constant exists, so one differentiation is sufficient.

🎯 Strategy

Differentiate and Eliminate

Application:

Differentiate once and replace the original function by \(y\).

Final Logic:

\(y=ae^{2x}\)
↓
Differentiate
↓
Replace
↓
\(y^{'}=2y.\)

🧠 Memory Trick

"One Constant → One Differentiation."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Single Arbitrary Constant)

📄 Passage

Based directly on the algebraic steps outlined in the passage, what is the final derived second-order differential equation for
y=e^2x (a+bx)
after substituting
be^2x=y^'-2y
into the equation
y^''=2y^'+2be^2x?

A \(y^{''}-4y^{'}+4y=0\)

B \(y^{''}+4y=0\)

C \(y^{''}-2y^{'}+4y=0\)

D \(y^{''}-4y^{'}-4y=0\)

💡 Short Explanation

Differentiate twice. Express \(be^{2x}\)from the first derivative. Substitute into the second derivative. Rearrange to eliminate the arbitrary constants.

✅ Correct Answer Explanation

From \(y=e^{2x}(a+bx),\) the first derivative is \(y^{'}=2y+be^{2x}.\) Hence, \(be^{2x}=y^{'}-2y.\) Differentiating again, \(y^{''}=2y^{'}+2be^{2x}.\) Substituting, \(y^{''}=2y^{'}+2(y^{'}-2y)=4y^{'}-4y.\) Rearranging, \(y^{''}-4y^{'}+4y=0.\) Hence, Option A is the correct answer.

❓ Why Other Options Are Incorrect

Option B) Incorrect because the \(y^{'}\)term cannot vanish after substitution.
Option C) Incorrect because it does not correctly simplify the substituted equation.
Option D) Incorrect because the sign of the \(4y\) term is incorrect.

🎯 Strategy

Differentiate, Substitute, Rearrange

Application:

Differentiate twice, eliminate the remaining arbitrary constant, and simplify.

Final Logic:

\(y^{'}\)
↓
\(be^{2x}=y^{'}-2y\)
↓
Substitute
↓
\(y^{''}-4y^{'}+4y=0.\)

🧠 Memory Trick

"Differentiate Twice → Substitute → Rearrange."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Formation of Differential Equations (Repeated Root Exponential Family)

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