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JEE Main Maths Differential Equations 2027: Order, Degree, Separable, Homogeneous & Linear

Order and degree of a differential equation, separation of variables, homogeneous equations and the linear first-order equation with integrating factor, taught with worked examples.

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October 1, 2026

JEE Main Maths Differential Equations 2027: Order, Degree, Separable, Homogeneous & Linear

The JEE Main syllabus keeps this chapter tight: only three solving methods are named. That is good news, because it means every question is really asking you to spot which of the three types you are looking at, and then apply a fixed recipe.

Before solving a differential equation, ask which of the three recipes it fits. The recipe decides everything that follows.

Chapter at a Glance

SnapshotDetail
NTA unitUnit 9 of 14: Differential Equations
Priority (trend-based)Moderate
Typical question styleSolving differential equations by the three named methods, order-and-degree MCQs
Best first stepIdentify the type first: separable, homogeneous, or linear

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

  • Ordinary differential equations, their order and degree
  • Solution of a differential equation by the method of separation of variables
  • Solution of a homogeneous differential equation
  • Solution of a linear differential equation of the type dydx+p(x)y=q(x)\dfrac{dy}{dx} + p(x)y = q(x)

Master These Topics

1. Order and Degree

The order is the order of the highest derivative present. The degree is the power of the highest-order derivative, once the equation is written as a polynomial in derivatives (free of roots and fractional powers of derivatives).

Worked example: (d2ydx2)3+(dydx)2+y=0\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + y = 0 has order 2 (from d2y/dx2d^2y/dx^2) and degree 3 (its power).

Trap: If the equation contains a derivative inside a square root or a trigonometric function, it must first be rearranged into polynomial form before the degree can be read off.

2. Separation of Variables

If the equation can be written as dydx=f(x)g(y)\dfrac{dy}{dx} = f(x)g(y), separate the variables and integrate both sides.

For $\dfrac{dy}{dx} = y$, the slope at every point equals its height, so the solution curve through $(0,1)$ is $y = e^x$.
For $\dfrac{dy}{dx} = y$, the slope at every point equals its height, so the solution curve through $(0,1)$ is $y = e^x$.

Worked example: dydx=y\dfrac{dy}{dx} = y. Separating gives dyy=dx\dfrac{dy}{y} = dx, so ln⁡∣y∣=x+C\ln\lvert y \rvert = x + C, giving y=Aexy = Ae^x. If the curve passes through (0,1)(0, 1), then A=1A = 1 and y=exy = e^x.

3. Homogeneous Differential Equations

A first-order equation is homogeneous if it can be written as dydx=f ⁣(yx)\dfrac{dy}{dx} = f\!\left(\dfrac{y}{x}\right). Substitute y=vxy = vx, so dydx=v+xdvdx\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}, which turns the equation into a separable one in vv and xx.

Worked example: dydx=x2+y2xy\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}. Dividing through by x2x^2 gives dydx=1+(y/x)2y/x\dfrac{dy}{dx} = \dfrac{1 + (y/x)^2}{y/x}. With y=vxy = vx, this becomes v+xdvdx=1+v2vv + x\dfrac{dv}{dx} = \dfrac{1 + v^2}{v}, so xdvdx=1vx\dfrac{dv}{dx} = \dfrac{1}{v}. Separating and integrating gives v2=2ln⁡∣x∣+Cv^2 = 2\ln\lvert x \rvert + C, and substituting back v=y/xv = y/x gives the solution in terms of xx and yy.

4. Linear Differential Equations

For dydx+P(x)y=Q(x)\dfrac{dy}{dx} + P(x)y = Q(x), the integrating factor is

I.F.=e∫P(x) dx\text{I.F.} = e^{\int P(x)\,dx}

and the solution is y⋅I.F.=∫Q(x)⋅I.F. dx+Cy \cdot \text{I.F.} = \displaystyle\int Q(x) \cdot \text{I.F.}\,dx + C.

Worked example: dydx+yx=x\dfrac{dy}{dx} + \dfrac{y}{x} = x. Here P=1/xP = 1/x, so I.F.=e∫dx/x=eln⁡x=x\text{I.F.} = e^{\int dx/x} = e^{\ln x} = x. Then

xy=∫x⋅x dx=x33+C⟹y=x23+Cxxy = \int x \cdot x \,dx = \frac{x^3}{3} + C \quad \Longrightarrow \quad y = \frac{x^2}{3} + \frac{C}{x}

Trap: The equation must be in the exact form dydx+P(x)y=Q(x)\dfrac{dy}{dx} + P(x)y = Q(x), with the coefficient of dydx\dfrac{dy}{dx} equal to 1, before reading off P(x)P(x).

Common Traps to Avoid

  • Reading off the degree before clearing radicals or fractional powers of derivatives.
  • Forgetting the substitution dydx=v+xdvdx\dfrac{dy}{dx} = v + x\dfrac{dv}{dx} when solving a homogeneous equation.
  • Using the wrong sign or forgetting a constant when finding the integrating factor.
  • Not dividing through to make the coefficient of dy/dxdy/dx equal to 1 before identifying P(x)P(x).

60-Second Revision Sheet

  • Order: highest derivative present; degree: its power once polynomial in derivatives
  • Separable: dydx=f(x)g(y)\dfrac{dy}{dx} = f(x)g(y), separate and integrate
  • Homogeneous: substitute y=vxy = vx
  • Linear: I.F.=e∫P dx\text{I.F.} = e^{\int P\,dx}, solution y⋅I.F.=∫Q⋅I.F. dx+Cy \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\,dx + C

Your Study Plan

  1. Day 1: order, degree and separable equations.
  2. Day 2: homogeneous equations with the v=y/xv = y/x substitution.
  3. Day 3: linear equations and integrating factor practice.
  4. Day 4: mixed identification (which type is it?) and a timed set.

Practice Differential Equations Questions Free → (opens in a new tab)


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Frequently Asked Questions

How do I know which method to use for a given differential equation?

Check first whether the variables separate directly. If not, see whether it depends only on y/xy/x (homogeneous). If it is linear in yy and dy/dxdy/dx, use the integrating factor method.

What is the integrating factor?

For dydx+P(x)y=Q(x)\dfrac{dy}{dx} + P(x)y = Q(x), it is e∫P(x) dxe^{\int P(x)\,dx}, a multiplier that turns the left side into the derivative of a simple product.

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