3. Ordinary Differential Equations

Singular solutions

A singular solution of a differential equation is a solution that cannot be obtained from the general solution by assigning specific values to the arbitrary constants, representing the envelope of the family of integral curves.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Singular Solutions & Discriminant Loci

Given a differential equation expressed in Clairaut's form $$y = x p + f(p)$$ where $p = \frac{dy}{dx}$, the general solution is $y = cx + f(c)$. The singular solution is obtained by eliminating $p$ between $f(p) + x f'(p) = 0$ and $y = x p + f(p)$, forming the envelope of the family of straight lines.

📐 Geometric Interpretation

Geometrically, the general solution represents a family of tangent lines to a single curve known as the envelope. The singular solution is this envelope itself, where every point on the envelope is touched by some member of the family of integral curves, and at each point, the singular solution shares a common tangent with a specific particular solution.

⚙️ Algebraic & Structural Meaning

Algebraically, the singular solution corresponds to the locus where the partial derivative of the differential equation with respect to the derivative parameter vanishes (the p-discriminant locus), resulting in a solution that violates uniqueness theorems (such as Picard-Lindelöf theorem) due to failure of Lipschitz continuity.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-linearity in Derivative (Degree > 1)	`Domain of $x$ and $p$`	Strictly Mandatory	First-order linear differential equations cannot possess singular solutions because their general solutions contain the complete parameter space and uniqueness holds universally.
Smoothness and Differentiability of $f(p)$	`Open interval of $p$`	Strictly Mandatory	Discontinuities or non-differentiable points in $f(p)$ fail to generate a valid envelope, leading to false singular loci or cuspidal edges.
Non-zero Second Derivative ($f''(p) \neq 0$)	`Parameter domain`	Essential for Envelope formation	If $f''(p) = 0$, the family reduces to a parallel system of lines with no envelope, meaning no singular solution exists.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Clairaut's Equation Direct Substitution

$$ \text{If } y = xp + f(p), \text{ replace } p \text{ by constant } c \text{ for general solution, and eliminate } p \text{ via } x + f'(p) = 0 \text{ for singular solution.} $$

Rule: Directly substitute $c$ for $p$ in the equation for general solution; differentiate partially w.r.t $p$ and set to zero to find the singular solution instantly without solving complicated integrals.

Application: Directly tested in UP PGT and TGT Mathematics for equations like $y = xp + a \sqrt{1 + p^2}$.

c-Discriminant Loci (CDL) Method

$$ \text{Given } F(x, y, c) = 0, \text{ eliminate } c \text{ from } F(x, y, c) = 0 \text{ and } \frac{\partial F}{\partial c} = 0. $$

Rule: For a general solution polynomial in $c$, find the discriminant with respect to $c$. The factors of the discriminant yield the singular solution, node locus, or cusp locus.

Application: Frequently asked in KVS PGT and DSSSB PGT to classify the nature of discriminant loci.

p-Discriminant Equation (PDE) Rule

$$ \text{Given } f(x, y, p) = 0, \text{ eliminate } p \text{ from } f(x, y, p) = 0 \text{ and } \frac{\partial f}{\partial p} = 0. $$

Rule: The p-discriminant equation provides the envelope (singular solution), taclocus, and cusp-locus in a single sweep. Check which factor satisfies the differential equation.

Degree 2 in $p$ (Quadratic in $p$)

$$ A(x,y)p^2 + B(x,y)p + C(x,y) = 0 \implies B^2 - 4AC = 0 $$

Rule: For any differential equation quadratic in $p$, the singular solution is obtained simply by setting the discriminant of the quadratic equation to zero: $B^2 - 4AC = 0$.

Constant Coefficient Envelopes

$$ (y - c x)^2 = R(1 + c^2) \implies x^2 + y^2 = R^2 $$

Rule: Recognize standard geometric forms. Circles and parabolas appearing as envelopes can be written down directly from the radius or focal distance parameters.

Application: Standard geometric ODE problems in UP PGT.

Extraneous Loci Elimination

$$ \text{Test: Does } \frac{dy}{dx} \text{ match original ODE?} $$

Rule: Discriminants often produce extraneous loci (cusp-loci, tac-loci). Always verify that the derived singular solution actually satisfies the original differential equation.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming p-Discriminant is Always a Singular Solution

Common Pitfall: Candidates blindly assume every factor of the p-discriminant is a singular solution.

Counterexample: Tac-loci or cusp-loci that satisfy $\frac{\partial f}{\partial p} = 0$ and $f = 0$ but fail to satisfy $\frac{dy}{dx} = p$.

Why it fails: The p-discriminant contains not only the envelope (singular solution) but also the tac-locus and cusp-locus, which may not be solutions to the differential equation.

🛡️ Exam Defense: Always substitute the derived locus back into the original differential equation to verify it is a valid solution.

⚠️ COMMON ERROR #02

Confusing General Solution Constant with Singular Solution

Common Pitfall: Trying to obtain the singular solution by substituting a specific numerical value for $c$ in the general solution.

Counterexample: For $y = cx + c^2$, setting $c = 1$ gives $y = x + 1$ (a particular solution, not singular).

Why it fails: Singular solutions cannot be derived by putting $c = \text{constant}$; they represent envelopes and are obtained via discriminant elimination or derivative w.r.t parameter.

🛡️ Exam Defense: If an option gives a solution with no arbitrary constants but it can be formed by setting $c = \text{constant}$, it is NOT a singular solution.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
General Solution	Contains essential arbitrary constant $c$	$y = cx + f(c)$	Family of integral curves (straight lines/curves)	Includes parameter $c$
Singular Solution	Envelope of family, violates uniqueness	Eliminate $p$ from $f = 0, \frac{\partial f}{\partial p} = 0$	Envelope touching all family curves	No arbitrary constant, satisfies ODE
c-Discriminant Locus (CDL)	Polynomial in $c$	Discriminant w.r.t $c = 0$	Envelope, node locus, cusp locus	Derived from general solution
p-Discriminant Locus (PDE)	Polynomial in $p$	Discriminant w.r.t $p = 0$	Envelope, tac-locus, cusp-locus	Derived directly from ODE

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the singular solution of the differential equation $y = xp + p^2$.

(A) $4y + x^2 = 0$(B) $y + 4x^2 = 0$(C) $y^2 + x = 0$(D) $4y - x^2 = 0$

📚 Method A: Formal Textbook Derivation

Step 1: The equation is in Clairaut's form $y = xp + f(p)$ where $f(p) = p^2$. Step 2: Differentiate with respect to $p$: $x + 2p = 0 \implies p = -x/2$. Step 3: Substitute $p = -x/2$ into the original equation: $y = x(-x/2) + (-x/2)^2 = -x^2/2 + x^2/4 = -x^2/4$. Step 4: Simplify to get $4y + x^2 = 0$.

⚡ Method B: 25-Second Exam Shortcut

Use quadratic in $p$ discriminant rule: $p^2 + xp - y = 0$. Discriminant $B^2 - 4AC = x^2 - 4(1)(-y) = x^2 + 4y = 0 \implies 4y + x^2 = 0$. Takes 10 seconds.

💡 Tip: Recognize Clairaut's form immediately and use the quadratic discriminant shortcut if degree in $p$ is 2.

CANONICAL PROBLEM 02 Dual-Method Breakdown

What is the singular solution of the differential equation $xp^2 - yp + a = 0$?

(A) $y^2 = 4ax$(B) $xy = a$(C) $y^2 = 2ax$(D) $x^2 = 4ay$

📚 Method A: Formal Textbook Derivation

Step 1: Given $xp^2 - yp + a = 0$, quadratic in $p$. Step 2: Write p-discriminant: $B^2 - 4AC = (-y)^2 - 4(x)(a) = 0$. Step 3: $y^2 - 4ax = 0 \implies y^2 = 4ax$. Step 4: Verify that $y^2 = 4ax$ satisfies the original differential equation.

⚡ Method B: 25-Second Exam Shortcut

Quadratic in $p$: $A = x, B = -y, C = a$. Set $B^2 - 4AC = y^2 - 4ax = 0 \implies y^2 = 4ax$.

💡 Tip: For equations of the form $xp^2 - yp + \text{constant} = 0$, the p-discriminant yields the standard parabola envelope.

Practice questions

Choose an answer to check it and see the solution:

The singular solution of the differential equation $y = xp + \sqrt{a^2 p^2 + b^2}$ is given by:

A $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

B $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

C $x^2 + y^2 = a^2 + b^2$

D $ax^2 + by^2 = 1$

Derivation & Analysis:

Given $y = xp + \sqrt{a^2p^2 + b^2}$ (Clairaut's form). Differentiating w.r.t $p$: $x + \frac{a^2p}{\sqrt{a^2p^2 + b^2}} = 0 \implies x^2(a^2p^2 + b^2) = a^4p^2 \implies p^2 = \frac{b^2 x^2}{a^2(a^2 - x^2)}$. Substituting $p$ back yields the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

A singular solution of a differential equation is characterized geometrically as:

A A normal to the family of curves

B The envelope of the family of integral curves

C An asymptote to the integral curves

D The orthogonal trajectory

Derivation & Analysis:

By definition, the singular solution represents the envelope of the family of curves represented by the general solution.

The differential equation $p^2 - (x+y)p + xy = 0$ can be factored into:

A $(p-x)(p-y) = 0$

B $(p+x)(p+y) = 0$

C $(p-1)(p-xy) = 0$

D $(xp-1)(yp-1) = 0$

Derivation & Analysis:

Factorizing the quadratic expression in $p$: $p^2 - xp - yp + xy = p(p-x) - y(p-x) = (p-x)(p-y) = 0$. Thus the equations are $p = x$ and $p = y$.

Which of the following differential equations possesses a singular solution?

A $\frac{dy}{dx} + y = x$

B $x \frac{dy}{dx} - y = 0$

C $(y - xp)^2 = 1 + p^2$

D $\frac{dy}{dx} = \frac{y}{x}$

Derivation & Analysis:

Linear differential equations of the first order (options A, B, D) do not possess singular solutions. Clairaut's-type or higher degree non-linear equations (option C) possess singular solutions.

If the p-discriminant of a differential equation has a repeated factor, that factor corresponds to:

A General solution only

B Singular solution or tac-locus/cusp-locus

C Only orthogonal trajectories

D Particular solution with zero constant

Derivation & Analysis:

The p-discriminant locus contains the envelope (singular solution), tac-locus, and cusp-locus.

The singular solution of $y = xp - p^3$ is:

A $27y^2 = 4x^3$

B $4y^2 = 27x^3$

C $y^3 = 27x^2$

D $27y^3 = 4x^2$

Derivation & Analysis:

Given $y = xp - p^3$. Differentiating w.r.t $p$: $x - 3p^2 = 0 \implies p^2 = x/3 \implies p = \sqrt{x/3}$. Substitute into ODE: $y = x(\sqrt{x/3}) - (x/3)^{3/2} = \frac{2x\sqrt{x}}{3\sqrt{3}}$. Squaring both sides: $y^2 = \frac{4x^3}{27} \implies 27y^2 = 4x^3$.

Clairaut's equation is represented in standard form as:

A $y = xp + f(p)$

B $y = mx + c$

C $f(x, y, p) = 0$

D $p^2 + f(x)p + g(x) = 0$

Derivation & Analysis:

Clairaut's differential equation is strictly defined as $y = xp + f(p)$ where $p = \frac{dy}{dx}$.

If the general solution of a differential equation is $y = c x + c^2$, its singular solution is:

A $4y + x^2 = 0$

B $4y = x^2$

C $y = x^2$

D $y^2 = 4x$

Derivation & Analysis:

Given $F(x,y,c) = c^2 + cx - y = 0$. Differentiating w.r.t $c$: $\frac{\partial F}{\partial c} = 2c + x = 0 \implies c = -x/2$. Substituting $c = -x/2$ into general solution: $y = (-x/2)x + (-x/2)^2 = -x^2/4 \implies 4y + x^2 = 0$.

What is the order and degree of the differential equation from which a singular solution is typically derived?

A First order, degree 1

B First order, degree greater than 1

C Second order, degree 1

D Second order, degree 2

Derivation & Analysis:

Singular solutions arise predominantly from first-order differential equations of degree greater than 1 (such as Clairaut's or general non-linear equations in $p$).

Q10

If a singular solution satisfies the differential equation but cannot be derived from the general solution for any real value of the arbitrary constant, it is because:

A It represents the envelope where the family curves degenerate

B The Lipschitz condition fails on the envelope

C The general solution is incomplete

D Both A and B

Derivation & Analysis:

The singular solution acts as an envelope where uniqueness theorems fail (Lipschitz condition breakdown) and the parameter cannot span the envelope.

Common questions

No. First-order linear differential equations (e.g., $\frac{dy}{dx} + P(x)y = Q(x)$) always have a general solution that includes all possible integral curves through any given point where coefficients are continuous. Thus, uniqueness holds and no singular solutions exist.

Methods like the p-discriminant or c-discriminant often yield extraneous loci such as cusp-loci or tac-loci, which satisfy $\frac{\partial f}{\partial p} = 0$ but are not actual solutions to the differential equation. Verification ensures mathematical validity.

Do not solve via lengthy elimination. Simply replace $p$ with $c$ in the given equation for the general solution, and for the objective options, look for the discriminant elimination result or use the quadratic formula shortcut if degree in $p$ is 2.

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