Why is the degree of a differential equation like $\sin\left(\frac{dy}{dx}\right) + y = 0$ considered undefined?

Degree is defined as the highest exponent of the highest-order derivative, provided the differential equation is a polynomial in its derivatives. When derivatives are inside non-polynomial functions like $\sin$, $\cos$, or $e^x$, it cannot be expanded into a finite polynomial in terms of $\frac{dy}{dx}$. Thus, order is $1$, but degree is undefined.

How do I decide whether to form a linear differential equation in $x$ or $y$?

Examine the powers of $y$ and $x$ in the given equation. If $y$ and $\frac{dy}{dx}$ appear linearly (degree $1$) without multiplying each other, use $\frac{dy}{dx} + P(x)y = Q(x)$. If $x$ and $\frac{dx}{dy}$ appear linearly while $y$ has powers or complex terms, invert the derivatives to form $\frac{dx}{dy} + P(y)x = Q(y)$.

What is the fastest way to solve homogeneous differential equations in competitive exams?

First check if the equation can be expressed as $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$ and substitute $y = vx$. Alternatively, for expressions involving $x \, dx + y \, dy$ or $x \, dy - y \, dx$, try using exact differentials such as $d(xy) = x \, dy + y \, dx$ or $d\left(\frac{y}{x}\right) = \frac{x \, dy - y \, dx}{x^2}$ to bypass lengthy substitutions.

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Key Formulas & High-Yield Equations

First-Order Linear Differential Equation (in $y$)

$\frac{dy}{dx} + P(x)y = Q(x)$

Solution is given by

y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C

, where

IF = e^{\int P(x) \, dx}

. Valid strictly when

P(x)

and

Q(x)

are continuous functions of

x

alone.

First-Order Linear Differential Equation (in $x$)

$\frac{dx}{dy} + P(y)x = Q(y)$

Solution is given by

x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C

, where

IF = e^{\int P(y) \, dy}

. Valid strictly when

P(y)

and

Q(y)

are continuous functions of

y

alone.

Homogeneous Differential Equation

$\frac{dy}{dx} = f\left(\frac{y}{x}\right)$

Solved by substitution

y = vx

, converting the equation to separable form

v + x \frac{dv}{dx} = f(v)

. Requires

x \neq 0

on the given domain.

Bernoulli's Differential Equation

$\frac{dy}{dx} + P(x)y = Q(x)y^n$

Reducible to linear form by dividing by

y^n

and substituting

z = y^{1-n}

. Valid for any real index

n \neq 0, 1

Variable Separable Form

$f(x) \, dx = g(y) \, dy$

Integrate directly as

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

. Requires

g(y) \neq 0

throughout the domain.

Exact Differential Equation Condition

$M(x,y) \, dx + N(x,y) \, dy = 0 \iff \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Guarantees existence of potential function

U(x,y) = C

such that

dU = M \, dx + N \, dy

. Valid for continuously differentiable functions

M

and

N

Solved High-Yield Sample Questions

Practice Mode

Question #1JEE Main 2024

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

Solution Locked

Step-by-Step Solution & Concept

Step 1

Identify the differential equation as a first-order linear equation $\frac{dy}{dx} + P(x)y = Q(x)$ , where $P(x) = \tan x$ and $Q(x) = \sec x$ .

Step 2

Compute the Integrating Factor $IF = e^{\int \tan x \, dx} = e^{\ln|\sec x|} = \sec x$ .

Step 3

Apply the solution formula $y \cdot (IF) = \int Q(x) \cdot (IF) \, dx + C$ , yielding $y \sec x = \int \sec^2 x \, dx + C = \tan x + C$ .

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Question #2JEE Main 2023

If $y = y(x)$ satisfies the differential equation $x \frac{dy}{dx} + 2y = x^2$ with initial condition $y(1) = 1$ , then find the value of $y(2)$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Convert to standard linear differential equation form by dividing by $x$ ( $x \neq 0$ ): $\frac{dy}{dx} + \frac{2}{x}y = x$ , giving $P(x) = \frac{2}{x}$ and $Q(x) = x$ .

Step 2

Calculate $IF = e^{\int \frac{2}{x} \, dx} = e^{2 \ln x} = x^2$ . Multiply through to obtain $y x^2 = \int x \cdot x^2 \, dx + C = \frac{x^4}{4} + C$ .

Step 3

Apply $y(1) = 1$ : $1 \cdot (1)^2 = \frac{1}{4} + C \implies C = \frac{3}{4}$ . Thus $y(x) = \frac{x^2}{4} + \frac{3}{4x^2}$ . Substituting $x = 2$ gives $y(2) = \frac{4}{4} + \frac{3}{16} = \frac{19}{16}$ .

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Mastering Differential Equations: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

First-Order Linear Differential Equation (in $y$)

First-Order Linear Differential Equation (in $x$)

Homogeneous Differential Equation

Bernoulli's Differential Equation

Variable Separable Form

Exact Differential Equation Condition

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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