Differential Equations — Complete Guide with Solved Examples

What Are Differential Equations — And Why Should You Care?

A differential equation is any equation that contains a derivative. That’s it. When you write $\frac{dy}{dx} = 2x$ , you’re not giving the answer — you’re describing a relationship between how $y$ changes and what $x$ is. Solving the equation means finding the original $y$ .

Think of it this way: velocity is $\frac{dx}{dt}$ . If someone tells you “a particle’s velocity equals twice its position,” they’ve given you a differential equation: $\frac{dx}{dt} = 2x$ . Your job is to find where the particle actually is at any time $t$ .

This chapter carries serious weightage — 4-8 marks in CBSE Class 12 boards, and 1-2 questions in JEE Main almost every year. The good news: the types are well-defined, the methods are systematic, and with practice you can score full marks reliably.

Key Terms and Definitions

Order of a differential equation: the highest derivative present.

$\frac{dy}{dx} + y = x^2$ → Order 1
$\frac{d^2y}{dx^2} + 3\frac{dy}{dx} = 0$ → Order 2

Degree of a differential equation: the power of the highest-order derivative, after clearing all fractions and radicals involving derivatives.

$\left(\frac{dy}{dx}\right)^3 + y = 0$ → Degree 3
$\frac{d^2y}{dx^2} + \sin\left(\frac{dy}{dx}\right) = 0$ → Degree not defined (can’t be expressed as a polynomial in derivatives)

Many students confuse order and degree. Order = which derivative is highest. Degree = what power that derivative is raised to. They’re asking two different things.

General Solution: contains arbitrary constants (as many as the order).

Particular Solution: obtained by substituting given initial conditions to find the constants.

Formation of a differential equation: given a family of curves (with $n$ arbitrary constants), differentiate $n$ times and eliminate the constants.

Methods for Solving Differential Equations

Class 12 syllabus covers three main types. Let’s handle each one cleanly.

Method 1: Variable Separable

When to use: You can write the equation as $f(y)\, dy = g(x)\, dx$ — i.e., all $y$ -terms on one side, all $x$ -terms on the other.

How to solve:

Rearrange to separate variables
Integrate both sides
Add a single constant $C$ (one constant, not two)
Simplify

\frac{dy}{dx} = \frac{f(x)}{g(y)} \implies g(y)\,dy = f(x)\,dx \implies \int g(y)\,dy = \int f(x)\,dx + C

Worked Example:

Solve $\frac{dy}{dx} = \frac{y^2 + 1}{x^2 + 1}$

\frac{dy}{y^2+1} = \frac{dx}{x^2+1}

\int \frac{dy}{y^2+1} = \int \frac{dx}{x^2+1}

\tan^{-1}y = \tan^{-1}x + C

Done. This is the general solution.

Method 2: Homogeneous Differential Equations

When to use: The equation has the form $\frac{dy}{dx} = F\!\left(\frac{y}{x}\right)$ — meaning, if you replace $x \to kx$ and $y \to ky$ , the $k$ cancels out completely.

Substitution: Put $y = vx$ , so $\frac{dy}{dx} = v + x\frac{dv}{dx}$ .

This converts the homogeneous equation into a variable separable one.

Quick check for homogeneous: if every term in numerator and denominator has the same total degree, it’s homogeneous. Example: $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$ — each term has degree 2 on top, degree 2 on bottom. Homogeneous.

Worked Example:

Solve $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$

Put $y = vx$ :

v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x \cdot vx} = \frac{1 + v^2}{2v}

x\frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2 - 2v^2}{2v} = \frac{1-v^2}{2v}

Separate variables:

\frac{2v\,dv}{1-v^2} = \frac{dx}{x}

-\ln|1-v^2| = \ln|x| + C_1

\ln|x| + \ln|1-v^2| = -C_1 = K

x(1-v^2) = A

Back-substitute $v = y/x$ :

x\!\left(1 - \frac{y^2}{x^2}\right) = A \implies x^2 - y^2 = Ax

Method 3: Linear Differential Equations

Standard form: $\frac{dy}{dx} + P(x)\cdot y = Q(x)$

This is the most heavily tested type in both CBSE and JEE Main.

The Integrating Factor (IF):

\text{IF} = e^{\int P\,dx}

Multiply both sides by IF. The left side magically becomes $\frac{d}{dx}(y \cdot \text{IF})$ .

y \cdot e^{\int P\,dx} = \int Q \cdot e^{\int P\,dx}\,dx + C

Why does this work? The product rule says $\frac{d}{dx}(u \cdot v) = u'v + uv'$ . The IF is specifically chosen so that the left side becomes a product rule derivative — making the whole thing integrable directly.

Worked Example:

Solve $\frac{dy}{dx} + \frac{2y}{x} = x^2$

Here $P = \frac{2}{x}$ , $Q = x^2$

\text{IF} = e^{\int \frac{2}{x}dx} = e^{2\ln x} = x^2

Multiply through by $x^2$ :

\frac{d}{dx}(x^2 \cdot y) = x^2 \cdot x^2 = x^4

x^2 y = \int x^4\,dx = \frac{x^5}{5} + C

y = \frac{x^3}{5} + \frac{C}{x^2}

The equation can also appear as $\frac{dx}{dy} + P(y)\cdot x = Q(y)$ — treating $x$ as the dependent variable and $y$ as independent. Same method, same IF formula, just swap the roles.

Solved Examples (Graded)

Easy — CBSE Level

Find the differential equation representing the family of curves $y = ax + a^2$ , where $a$ is an arbitrary constant.

Differentiate: $\frac{dy}{dx} = a$

Substitute back: $y = x\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2$

This is Clairaut’s equation — a classic CBSE 2-marker.

Medium — JEE Main Level

Solve: $x\frac{dy}{dx} - y = \sqrt{x^2 + y^2}$ , given $y(1) = 0$ .

Rearrange: $\frac{dy}{dx} = \frac{y + \sqrt{x^2+y^2}}{x} = \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}$

This is homogeneous. Put $y = vx$ :

v + x\frac{dv}{dx} = v + \sqrt{1+v^2}

x\frac{dv}{dx} = \sqrt{1+v^2}

\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}

\ln\!\left|v + \sqrt{1+v^2}\right| = \ln|x| + C

At $x=1$ , $y=0 \Rightarrow v=0$ : $\ln(0 + 1) = \ln 1 + C \Rightarrow C = 0$

\ln\!\left|\frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}}\right| = \ln x

y + \sqrt{x^2+y^2} = x^2

Hard — JEE Advanced Level

Solve: $(1+y^2)\,dx = (\tan^{-1}y - x)\,dy$

Rewrite as $\frac{dx}{dy}$ :

\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1+y^2}

\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}

This is linear in $x$ (dependent variable), $y$ (independent variable).

$P = \frac{1}{1+y^2}$ , so $\text{IF} = e^{\tan^{-1}y}$

x \cdot e^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2} \cdot e^{\tan^{-1}y}\,dy

Let $t = \tan^{-1}y$ , $dt = \frac{dy}{1+y^2}$ :

= \int t \cdot e^t\,dt = e^t(t-1) + C = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C

x = (\tan^{-1}y - 1) + Ce^{-\tan^{-1}y}

This exact problem type — linear DE with $x$ as dependent variable — appeared in JEE Main 2023 (January session). Recognizing when to flip and treat $x$ as dependent saves 3-4 minutes under exam pressure.

Exam-Specific Tips

CBSE Class 12 Boards

The marking scheme for a 5-mark DE question is typically: 1 mark for identifying type, 1 mark for correct substitution, 2 marks for integration, 1 mark for final answer. Even if you can’t complete the problem, the substitution step is worth marks — always show it.

Standard questions: formation (2 marks), variable separable (3 marks), linear DE (5 marks). The linear DE is almost always on the paper.

JEE Main

1-2 questions per paper, mostly from linear DE and homogeneous type. Questions often involve finding the particular solution — so read the initial condition carefully and substitute correctly. The answer choices help: if you get a clean form matching one option, you’re right.

JEE Main 2024 Shift 2 (January) had: “If $\frac{dy}{dx} + 2y\tan x = \sin x$ , $y\!\left(\frac{\pi}{3}\right) = 0$ , find the maximum value of $y$ .” Classic linear DE + particular solution + optimization combo. Be ready for this multi-step pattern.

JEE Advanced

Expect non-standard forms — DEs mixed with inequalities, DEs where the solution curve has geometric properties, or systems of DEs. Here, the setup is the hard part. Once you identify the type, the mechanics are the same.

Common Mistakes to Avoid

Mistake 1: Degree of a DE involving $\sin\left(\frac{dy}{dx}\right)$

Students write degree = 1 because they see $\frac{dy}{dx}$ once. Wrong. Degree is only defined when the DE is a polynomial in derivatives. $\sin\left(\frac{dy}{dx}\right)$ is transcendental — degree is not defined. This is a guaranteed CBSE MCQ trap.

Mistake 2: Adding two constants after integration

When you integrate both sides of $f(y)\,dy = g(x)\,dx$ , write $C$ only on one side. Writing $C_1$ on the left and $C_2$ on the right, then combining as $C_1 - C_2 = C$ , is technically fine but wastes time. Add $C$ to the right side from the start.

Mistake 3: Forgetting to back-substitute in homogeneous equations

After solving for $v$ , the final answer must be in terms of $x$ and $y$ . Leaving $v$ in the answer gets you zero marks for the final step in boards, and a wrong MCQ option in JEE.

Mistake 4: Misidentifying the linear DE form

The standard form is $\frac{dy}{dx} + Py = Q$ where $P$ and $Q$ are functions of $x$ only. If $P$ or $Q$ contain $y$ , it’s not linear (it might be Bernoulli’s equation, which is out of CBSE but appears in JEE). Always check this before applying the IF formula.

Mistake 5: Using IF = $e^{\int P\,dx}$ without simplifying $\int P\,dx$ first

If $P = \frac{2}{x}$ , then $\int P\,dx = 2\ln x = \ln x^2$ , so $\text{IF} = e^{\ln x^2} = x^2$ . Students who write $\text{IF} = e^{2\ln x}$ and leave it there often make multiplication errors in the next step. Always simplify the IF to a clean form.

Practice Questions

Q1. Find the order and degree of: $\left(\frac{d^3y}{dx^3}\right)^2 + \frac{d^2y}{dx^2} = \sin x$

Order = 3 (highest derivative is $\frac{d^3y}{dx^3}$ ). Degree = 2 (that derivative is squared). The $\sin x$ on the right doesn’t affect either — it involves only $x$ , not derivatives.

Q2. Solve: $\frac{dy}{dx} = e^{x+y}$

Separate: $e^{-y}\,dy = e^x\,dx$

Integrate: $-e^{-y} = e^x + C$

Or equivalently: $e^x + e^{-y} = K$ (where $K = -C$ )

Q3. Form the DE for the family of circles $x^2 + y^2 = r^2$ .

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

So: $x + y\frac{dy}{dx} = 0$ , or $\frac{dy}{dx} = -\frac{x}{y}$

(One arbitrary constant $r$ → order 1 DE, as expected.)

Q4. Solve: $(x^2 - y^2)\,dx + 2xy\,dy = 0$

Rewrite: $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$

Check homogeneity: both terms have degree 2. Homogeneous. Put $y = vx$ :

v + x\frac{dv}{dx} = \frac{v^2 - 1}{2v}

x\frac{dv}{dx} = \frac{v^2-1}{2v} - v = \frac{v^2-1-2v^2}{2v} = \frac{-1-v^2}{2v}

\frac{2v\,dv}{1+v^2} = -\frac{dx}{x}

\ln(1+v^2) = -\ln x + \ln K

x(1+v^2) = K

Back-substitute: $x\!\left(1 + \frac{y^2}{x^2}\right) = K \Rightarrow x^2 + y^2 = Kx$

Q5. Solve: $\frac{dy}{dx} + y\cot x = 2\cos x$

Linear DE. $P = \cot x$ , $Q = 2\cos x$ .

$\text{IF} = e^{\int \cot x\,dx} = e^{\ln|\sin x|} = \sin x$

y\sin x = \int 2\cos x \cdot \sin x\,dx = \int \sin 2x\,dx = -\frac{\cos 2x}{2} + C

y = \frac{-\cos 2x}{2\sin x} + \frac{C}{\sin x}

Q6. Find the particular solution of $\frac{dy}{dx} = \frac{y}{x} + \tan\!\left(\frac{y}{x}\right)$ , given $y = \frac{\pi}{2}$ when $x = 1$ .

Homogeneous. Put $y = vx$ :

v + x\frac{dv}{dx} = v + \tan v

x\frac{dv}{dx} = \tan v

\cot v\,dv = \frac{dx}{x}

\ln|\sin v| = \ln|x| + C

\sin v = Ax

Back-substitute $v = y/x$ : $\sin\!\left(\frac{y}{x}\right) = Ax$

At $x=1$ , $y=\frac{\pi}{2}$ : $\sin\!\left(\frac{\pi}{2}\right) = A(1) \Rightarrow A = 1$

Particular solution: $\sin\!\left(\frac{y}{x}\right) = x$

Q7. Solve: $\frac{dy}{dx} - 3y = \sin 2x$ (JEE Main pattern)

Linear DE. $P = -3$ , $Q = \sin 2x$ .

$\text{IF} = e^{-3x}$

ye^{-3x} = \int e^{-3x}\sin 2x\,dx

Use the standard result $\int e^{ax}\sin bx\,dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2}$ :

\int e^{-3x}\sin 2x\,dx = \frac{e^{-3x}(-3\sin 2x - 2\cos 2x)}{9+4} = \frac{e^{-3x}(-3\sin 2x - 2\cos 2x)}{13}

y = \frac{-3\sin 2x - 2\cos 2x}{13} + Ce^{3x}

Q8. Form the DE for parabolas with vertex at origin and axis along the positive $x$ -axis.

Such parabolas have equation $y^2 = 4ax$ (one arbitrary constant $a$ ).

Differentiate: $2y\frac{dy}{dx} = 4a$ , so $a = \frac{y}{2}\frac{dy}{dx}$

Substitute back into $y^2 = 4ax$ :

y^2 = 4 \cdot \frac{y}{2}\frac{dy}{dx} \cdot x = 2xy\frac{dy}{dx}

y = 2x\frac{dy}{dx}

Or equivalently: $\frac{dy}{dx} = \frac{y}{2x}$

FAQs

What is the difference between a general solution and a particular solution?

The general solution contains arbitrary constants (as many as the order of the DE). A particular solution is obtained by substituting specific initial or boundary conditions — it’s one specific curve from the family represented by the general solution.

How do I identify which method to use?

Check in this order: (1) Can I separate variables cleanly? If yes, use variable separable. (2) Does every term have the same total degree? If yes, it’s homogeneous. (3) Is it in the form $\frac{dy}{dx} + Py = Q$ with $P, Q$ functions of $x$ only? If yes, linear DE with IF.

What does “degree not defined” mean?

When a DE contains derivatives inside transcendental functions like $\sin$ , $\log$ , $e^{(\cdot)}$ , it cannot be expressed as a polynomial in derivatives. Degree is only defined for polynomial DEs. This is a 1-mark conceptual question in CBSE — know it cold.

Is Bernoulli’s equation in the Class 12 syllabus?

No, it’s not in the CBSE/NCERT syllabus for Class 12. It appears in some JEE coaching modules, but JEE Main questions on DEs are typically solvable using the three standard methods above.

Why does multiplying by the integrating factor work?

The IF $\mu = e^{\int P\,dx}$ is chosen precisely so that $\mu' = P\mu$ . This makes $\frac{d}{dx}(\mu y) = \mu y' + \mu' y = \mu(y' + Py) = \mu Q$ . So after multiplying, the left side is always a perfect derivative — which is immediately integrable.

How many questions come from differential equations in JEE Main?

Typically 1-2 questions per paper. With the shift to a numerical answer format for some questions, you might be asked to find a specific value (like the value of $y$ at $x = 2$ ) rather than the full solution. Practice substituting the initial condition and evaluating.

Can the arbitrary constant appear in different forms in the final answer?

Yes. $\ln|y| = x + C$ is equivalent to $y = Ae^x$ (where $A = e^C$ ). Both are correct general solutions — they represent the same family of curves. In CBSE boards, either form is accepted; in JEE, the answer options will guide you to the expected form.

What is an exact differential equation?

An equation $M\,dx + N\,dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ . This is not in the CBSE Class 12 syllabus but appears in engineering mathematics (B.Tech). For boards and JEE, focus on the three methods above.