3. Ordinary Differential Equations

First-order first-degree differential equations

First-order first-degree ordinary differential equations are foundational pillars of calculus involving derivatives of highest order one and highest degree one, typically expressed in the general form $M(x,y)dx + N(x,y)dy = 0$.

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

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Definition and statement

Existence and Uniqueness Theorem (Picard-Lindelöf Theorem)

Let $f(x, y)$ be a real-valued continuous function in a closed rectangular domain $R = \{(x, y) : |x - x_0| \le a, |y - y_0| \le b\}$, and satisfy the Lipschitz condition with respect to $y$, i.e., $|f(x, y_1) - f(x, y_2)| \le L |y_1 - y_2|$ for all $(x, y_1), (x, y_2) \in R$. Then the differential equation $\frac{dy}{dx} = f(x, y)$ with initial condition $y(x_0) = y_0$ possesses a unique solution $y(x)$ on the interval $[x_0 - h, x_0 + h]$ for some $h > 0$.

📐 Geometric Interpretation

Geometrically, a first-order differential equation defines a slope field or direction field in the $xy$-plane, where each point $(x, y)$ associates a unique tangent line slope $m = f(x, y)$. The solution curve is an integral curve passing through the point $(x_0, y_0)$ such that at every point on the curve, its tangent matches the slope field.

⚙️ Algebraic & Structural Meaning

Algebraically, solving a first-order differential equation transforms a differential operator relation into an algebraic or transcendental relation involving an arbitrary constant of integration $C$, establishing the family of integral curves.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity of $f(x, y)$ in domain $R$	`Closed rectangular domain $R = [x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b]$`	Strictly Mandatory / Necessary	Peano's existence theorem may hold without uniqueness, or no solution may pass through the region if discontinuity causes a break in the vector field.
Lipschitz Continuity with respect to $y$	`Open or closed interval in $y$`	Strictly Mandatory for Uniqueness	If the Lipschitz condition fails (e.g., $\frac{\partial f}{\partial y}$ is unbounded), multiple solutions or infinitely many solutions may pass through the same initial point.
Initial Condition Specification	`Boundary point $(x_0, y_0)$`	Essential for particular solution	Without initial conditions, only the general solution containing the arbitrary constant $C$ can be found; particular curves cannot be pinned down.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Variable Separable Recognition

$$f(x)dx = g(y)dy \implies \int f(x)dx = \int g(y)dy + C$$

Rule: If the equation can be factored cleanly as $f_1(x)g_1(y)dx + f_2(x)g_2(y)dy = 0$, separate variables instantly by dividing by $g_1(y)f_2(x)$ and integrate directly.

Application: Frequently asked in UP TGT/PGT with logarithmic and trigonometric combinations.

Homogeneous Equation Substitution

$$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$

Rule: If $M(x,y)$ and $N(x,y)$ are homogeneous functions of the exact same degree, substitute $y = vx$ or $x = vy$ to convert the ODE into a variable-separable form in terms of $v$ and $x$.

Integrating Factor (I.F.) for Linear ODE

$$\frac{dy}{dx} + P(x)y = Q(x) \implies \text{I.F.} = e^{\int P(x)dx}$$

Rule: For standard linear form, immediately compute the integrating factor $e^{\int P(x)dx}$ and write the solution as $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$.

Bernoulli Equation Reduction

$$\frac{dy}{dx} + P(x)y = Q(x)y^n \implies \text{Substitute } v = y^{1-n}$$

Rule: Divide by $y^n$, substitute $v = y^{1-n}$, and transform the nonlinear Bernoulli equation into a standard linear differential equation in $v$.

Application: High-level UP PGT / LT Grade questions test this transformation step directly.

Exact Differential Equation Test

$$Mdx + Ndy = 0 \text{ is exact if } \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

Rule: Check exactness by partial differentiation; if exact, the general solution is $\int_{y \text{ const}} M dx + \int (\text{terms of } N \text{ free of } x) dy = C$.

Clairaut's Equation General Solution

$$y = x p + f(p) \implies y = cx + f(c)$$

Rule: For Clairaut's form $y = xp + f(p)$ where $p = \frac{dy}{dx}$, replace $p$ with arbitrary constant $c$ to instantly write the general solution without elimination steps.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Non-Linearity Mistake in Linear Equations

Common Pitfall: Candidates mistake equations containing terms like $y^2$, $\sin(y)$, or $e^y$ multiplied by $\frac{dy}{dx}$ as linear.

Counterexample: $x \frac{dy}{dx} + y^2 = 0$

Why it fails: This is a Bernoulli equation ($n=2$), not a linear differential equation, so standard linear integrating factor methods cannot be directly applied without substitution.

🛡️ Exam Defense: Check that the dependent variable $y$ and its derivative $\frac{dy}{dx}$ appear only to the first degree and are never multiplied together.

⚠️ COMMON ERROR #02

Ignoring Loss of Solutions during Division

Common Pitfall: Dividing both sides of a differential equation by an expression containing variables (like $y$ or $y-1$) without checking if that expression can be zero.

Counterexample: $(y-1)\frac{dy}{dx} = x

Why it fails: Dividing by $(y-1)$ assumes $y \neq 1$, thereby dropping the singular or constant solution $y = 1$ which satisfies the original differential equation.

🛡️ Exam Defense: Always factorize expressions completely and check if the factored terms set to zero form valid solutions before dividing.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Variable Separable	$f(x)dx = g(y)dy$	$\int f(x)dx = \int g(y)dy + C$	Isoclines have separable slope components	Variables $x$ and $y$ factor completely.
Homogeneous ODE	$M(x,y), N(x,y)$ homogeneous of same degree	$y = vx \implies \text{separable in } v, x$	Slope depends only on the ratio $y/x$	Powers of $x$ and $y$ add up uniformly in every term.
Linear ODE	First degree in $y$ and $\frac{dy}{dx}$	$\text{I.F.} = e^{\int P(x)dx}$	Superposition principle applies to solutions	Contains $P(x)y = Q(x)$ structure.
Exact ODE	$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$	$\int M dx + \int N_{free} dy = C$	Total differential of a potential function $u(x,y)=C$	Symmetric mixed partial derivatives match.

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the general solution of the differential equation $\frac{dy}{dx} + \frac{y}{x} = \log x$.

(A) $xy = \frac{x^2}{2}(\log x - \frac{1}{2}) + C$(B) $y = x \log x + C$(C) $y = \frac{\log x}{x} + C$(D) $xy = x^2 \log x + C$

📚 Method A: Formal Textbook Derivation

Step 1: Identify as linear ODE with $P(x) = \frac{1}{x}$ and $Q(x) = \log x$. Step 2: Calculate $\text{I.F.} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$. Step 3: Multiply by I.F. to get $\frac{d}{dx}(xy) = x \log x$. Step 4: Integrate by parts: $\int x \log x dx = \frac{x^2}{2}\log x - \frac{x^2}{4}$. Step 5: Divide by $x$ to obtain $y = \frac{x}{2}(\log x - \frac{1}{2}) + \frac{C}{x}$ or $xy = \frac{x^2}{2}(\log x - \frac{1}{2}) + C$.

⚡ Method B: 25-Second Exam Shortcut

Recognize linear form, calculate I.F. = $x$, integrate $x \log x$ instantly by standard calculus formulas, match option (A).

💡 Tip: Watch out for constant of integration placement when multiplying by $x$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The solution of the differential equation $x \frac{dy}{dx} - y = x^2 \sin x$ is:

(A) $y = -x \cos x + C x$(B) $y = x \sin x + C(C) $y = -x \sin x + C x$(D) $y = x \cos x + C x$

📚 Method A: Formal Textbook Derivation

Step 1: Divide by $x$ to standard form $\frac{dy}{dx} - \frac{1}{x}y = x \sin x$. Step 2: Here $P(x) = -\frac{1}{x}$, so $\text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$. Step 3: Multiply equation by $\frac{1}{x}$: $\frac{d}{dx}\left(\frac{y}{x}\right) = \sin x$. Step 4: Integrate both sides: $\frac{y}{x} = -\cos x + C \implies y = -x \cos x + Cx$.

⚡ Method B: 25-Second Exam Shortcut

Dividing by $x^2$ or adjusting integrating factor directly reveals total derivative of $\frac{y}{x}$.

💡 Tip: Check sign of $P(x)$ carefully during integrating factor exponentiation.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The integrating factor of the differential equation $(1 - x^2)\frac{dy}{dx} - xy = 1$ is:

A $\frac{1}{1-x^2}$

B $\sqrt{1-x^2}$

C $\frac{1}{\sqrt{1-x^2}}$

D $\log(1-x^2)$

Derivation & Analysis:

Standard form: $\frac{dy}{dx} - \frac{x}{1-x^2}y = \frac{1}{1-x^2}$. Here $P(x) = -\frac{x}{1-x^2}$. $\text{I.F.} = e^{\int -\frac{x}{1-x^2} dx} = e^{\frac{1}{2}\log(1-x^2)} = \sqrt{1-x^2}$. Wait, let's re-evaluate: $\int -\frac{x}{1-x^2}dx = \frac{1}{2}\log(1-x^2)$, so $e^{\frac{1}{2}\log(1-x^2)} = \sqrt{1-x^2}$. Option (C) is correct.

MCQ 2: The degree of the differential equation $\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^3 = x^2$ is:

A 1

B 3

C 2

D Not defined

Derivation & Analysis:

The highest order derivative is $\frac{dy}{dx}$ (order 1), and its highest exponent in the polynomial equation is 3. Hence, the degree is 3.

MCQ 3: The general solution of the differential equation $\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}$ is:

A $e^y = e^x + \frac{x^3}{3} + C$

B $e^y = e^x + x^3 + C$

C $e^y = e^{-x} + \frac{x^3}{3} + C$

D $e^{-y} = e^x + \frac{x^3}{3} + C$

Derivation & Analysis:

Rewrite as $\frac{dy}{dx} = e^{-y}(e^x + x^2) \implies e^y dy = (e^x + x^2) dx$. Integrating both sides gives $e^y = e^x + \frac{x^3}{3} + C$. Option (A) is correct.

MCQ 4: Which of the following differential equations is homogeneous?

A $(x^3+y^3)dx + xy^2 dy = 0$

B $(x^2+y)dx + (y^2+x)dy = 0$

C $(xy + y^2)dx - x^2 dy = 0$

D $(\sin x + \cos y)dx + (\cos x - \sin y)dy = 0$

Derivation & Analysis:

In option (C), every term in $M(x,y) = xy + y^2$ is of degree 2, and in $N(x,y) = -x^2$ is of degree 2. Thus, it is a homogeneous differential equation.

MCQ 5: The solution of $x \frac{dy}{dx} + y = x^3$ is:

A $xy = \frac{x^3}{3} + C$

B $4xy = x^4 + C$

C $y = \frac{x^2}{4} + \frac{C}{x}$

D $xy = \frac{x^4}{4} + C$

Derivation & Analysis:

Rewrite as $\frac{d}{dx}(xy) = x^3 \implies xy = \int x^3 dx = \frac{x^4}{4} + C$. Option (D) is correct.

MCQ 6: The differential equation of all circles passing through the origin and having their centers on the x-axis is:

A $(x^2-y^2)\frac{dy}{dx} = 2xy$

B $(x^2+y^2)\frac{dy}{dx} = 2xy$

C $(x^2-y^2)\frac{dy}{dx} = xy$

D $2(x^2-y^2)\frac{dy}{dx} = xy$

Derivation & Analysis:

Equation of circle: $(x-a)^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0 \implies 2ax = x^2 + y^2 \implies a = \frac{x^2+y^2}{2x}$. Differentiating with respect to $x$ yields $(x^2-y^2)\frac{dy}{dx} = 2xy$.

MCQ 7: The solution of the exact differential equation $(2x + y)dx + (x + 2y)dy = 0$ is:

A $x^2 + xy + y^2 = C$

B $2x^2 + xy + 2y^2 = C$

C $x^2 + 2xy + y^2 = C$

D $x^2 + \frac{1}{2}xy + y^2 = C$

Derivation & Analysis:

Integrating $M(x,y)$ with respect to $x$ keeping $y$ constant gives $\int (2x+y)dx = x^2 + xy$. Adding terms of $N(x,y)$ free of $x$ gives $\int 2y dy = y^2$. Thus, $x^2 + xy + y^2 = C$.

MCQ 8: The general solution of Clairaut's equation $y = x \frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2$ is:

A $y = cx + c^2$

B $y = cx - c^2$

C $y = c^2x + c$

D $y = x + c$

Derivation & Analysis:

Replacing $\frac{dy}{dx}$ by $c$ in Clairaut's form $y = xp + f(p)$ immediately yields the general solution $y = cx + c^2$.

MCQ 9: The integrating factor of $\frac{dy}{dx} + y \cot x = \csc x$ is:

A $\sin x$

B $\cos x$

C $\tan x$

D $-\sin x$

Derivation & Analysis:

Here $P(x) = \cot x$. $\text{I.F.} = e^{\int \cot x dx} = e^{\log(\sin x)} = \sin x$. Option (A) is correct.

Q10

MCQ 10: If $\frac{dy}{dx} = \frac{y}{x} + \tan\left(\frac{y}{x}\right)$, then the solution is:

A $\sin\left(\frac{y}{x}\right) = Cx$

B $\cos\left(\frac{y}{x}\right) = Cx$

C $\sin\left(\frac{y}{x}\right) = C$

D $\tan\left(\frac{y}{x}\right) = Cx$

Derivation & Analysis:

Substitute $y = vx \implies v + x\frac{dv}{dx} = v + \tan v \implies \cot v dv = \frac{dx}{x} \implies \log(\sin v) = \log x + \log C \implies \sin v = Cx \implies \sin\left(\frac{y}{x}\right) = Cx$.

Common questions

A homogeneous equation of first order has every term in $M(x,y)$ and $N(x,y)$ of identical total degree in $x$ and $y$, solvable via $y=vx$. A linear equation is of degree one in $y$ and $\frac{dy}{dx}$, where coefficients depend solely on $x$, solvable via integrating factor $e^{\int P(x)dx}$.

Making sign errors during integrating factor calculation, forgetting the arbitrary constant $C$, or losing singular solutions by premature division by variable expressions.

Yes, direct questions asking for either the general solution by substituting $p=c$ or the singular solution via $\frac{\partial}{\partial c}(F(x,y,c))=0$ frequently appear in PGT and LT Grade exams.

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