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Homogeneous differential equations are one of three first-order types in JEE Main and Class 12, and the method is always the same substitution. Students still lose marks in three places: they misjudge whether an equation is homogeneous, they forget that dy/dx changes when y = vx, and they drop a solution when they divide. This post covers all three with four worked examples, none of them from the NCERT exercise set.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "solution of a homogeneous and linear differential equation" alongside separation of variables. The CBSE Mathematics curriculum for 2026-27 lists "solutions of homogeneous differential equations of first order and first degree". The method is section 9.4.2 of the NCERT Class 12 chapter on differential equations (Reprint 2026-27). Its historical note credits Leibniz with the method of separation of variables in 1691 and, "a year later", the method for homogeneous equations of the first order. Check your own exam year's syllabus.
How do you test whether an equation is homogeneous?
NCERT defines F(x, y) as homogeneous of degree n if F(λx, λy) = λⁿF(x, y) for every nonzero λ. The equation dy/dx = F(x, y) is homogeneous when F has degree zero, which means F can be written as g(y/x).
Quick checks:
(x² + y²)/(xy): replace x, y by λx, λy and the λ² cancels. Homogeneous.
y/x + sin(y/x): already a function of y/x. Homogeneous.
(x + y + 1)/(x − y): the constant 1 does not scale with λ. Not homogeneous.
(x³ + y²)/x²: terms of different degree in the numerator. Not homogeneous.
Why does y = vx work?
If y = vx, then dy/dx = v + x dv/dx by the product rule. The equation dy/dx = g(v) becomes
x dv/dx = g(v) − v
Now v and x separate: dv/[g(v) − v] = dx/x. Integrate, then replace v by y/x.
Worked example 1: a square root
Solve x dy/dx = y + √(x² + y²) for x > 0.
Divide by x: dy/dx = y/x + √(1 + (y/x)²), using x > 0. Put y = vx:
v + x dv/dx = v + √(1 + v²), so dv/√(1 + v²) = dx/x.
Integrate: ln(v + √(1 + v²)) = ln x + ln C, so v + √(1 + v²) = Cx. Multiply by x:
y + √(x² + y²) = Cx²
Worked example 2: a particular solution
Solve dy/dx = y/x + e^(y/x), given y = 0 when x = 1.
Put y = vx: x dv/dx = e^v, so e^(−v) dv = dx/x. Integrate: −e^(−v) = ln x + C.
At x = 1, v = 0: −1 = 0 + C, so C = −1. Then e^(−y/x) = 1 − ln x, which gives
y = −x ln(1 − ln x), for 1 − ln x > 0.
Check: at x = 1, ln 1 = 0, so y = −1 × ln 1 = 0.
Worked example 3: the solution you can lose
Solve x² dy/dx = y² + xy.
Divide by x²: dy/dx = (y/x)² + y/x. Put y = vx: x dv/dx = v². For v ≠ 0, dv/v² = dx/x, so −1/v = ln|x| + C, and
y = −x/(ln|x| + C)
Now go back to the step where you divided by v². If v = 0, then y = 0, and the original equation reads 0 = 0. So y = 0 is also a solution, and no value of C produces it. Whenever g(v) − v = 0 for some constant v₀, the line y = v₀x is a solution that dividing removes.
Worked example 4: when x = vy is easier
NCERT notes that if the equation is dx/dy = h(x/y), you can put x = vy instead. Solve dx/dy = x/y + e^(−x/y).
With x = vy, dx/dy = v + y dv/dy. Then y dv/dy = e^(−v), so e^v dv = dy/y. Integrate:
e^(x/y) = ln|y| + C
Common mistakes and how to avoid them
Writing dy/dx = dv/dx. Always v + x dv/dx.
Calling an equation with a constant term homogeneous. (x + y + 1)/(x − y) fails the λ test.
Applying the initial condition too early. Integrate first, then substitute to find C.
Losing the line solutions. Check where g(v) − v = 0.
Ignoring the domain. √(x² + y²)/x = √(1 + v²) only for x > 0.
To decide quickly between separable, homogeneous and linear types, use the checklist in our guide to differential equations for JEE. If you keep getting stuck after the substitution, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning at each step.
A practice routine
Write five right-hand sides and run the λ test on each.
Solve dy/dx = (x² + y²)/(xy) by y = vx and check by differentiating.
Redo Example 2 with y = 0 when x = e^(−1).
For dy/dx = y/x + tan(y/x), find any lines y = v₀x that are solutions.
Tonight, write the four-line method (test, substitute, separate, back-substitute) at the top of your differential equations notes.
