IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Homogeneous Differential Equations

A homogeneous first-order DE has the form dy/dx = f(y/x) — the RHS depends only on the ratio y/x. These aren’t separable as written, but one substitution — v = y/x — turns them INTO a separable DE. Then it’s the technique from the previous note.

📘 What you need to know

Recognise the form: rearrange the DE until the RHS depends only on y/x (no lonely x or lonely y). Trick: divide top and bottom by the highest power of x.
The substitution: v = y/x ⟺ y = vx. Then dy/dx = v + x · dv/dx (product rule).
After substitution: the equation becomes x · dv/dx = f(v) − v, which IS separable in v and x.
Apply separation of variables from the previous note: get v‘s on one side, x‘s on the other, integrate, add +c.
Back-substitute v = y/x at the end to convert the solution in (v, x) back to (x, y).
Quick homogeneity test: replace x → kx and y → ky in the RHS. If the k cancels out (scale-invariant), the equation is homogeneous.
Implicit solutions are fine — many homogeneous DEs don’t have a clean y = f(x) form.

The substitution that does the magic

Key substitution — turns homogeneous into separable v = yx ⟹ y = vx ⟹ dydx = v + x · dvdx

The product rule gives dy/dx = d/dx(vx) = v·1 + x·dv/dx. Substitute this AND v = y/x into the homogeneous DE; the result is a separable DE in the new variables.

The substitution v = y/x turns a homogeneous DE into a separable one in (v, x). The “v + x dv/dx” comes from the product rule on y = vx. After solving the separable equation for v(x), back-substitute to recover y.

Spotting a homogeneous DE

✓ Homogeneous — RHS only depends on y/x

dy/dx = f(y/x)

Examples: (x+y)/x = 1 + y/x; (x²+y²)/(xy) = x/y + y/x; y/x + (y/x)². RHS scale-invariant under (x, y) → (kx, ky).

✗ Not homogeneous — extra x or y terms

dy/dx = (something not just y/x)

Examples: x + y²/x (the y² isn’t scaled by x²); x + y + 1 (constant breaks symmetry). The k doesn’t cancel under the scale test.

The scale-test trick: in the RHS, replace every x with kx and every y with ky. If every k cancels out, the RHS depends only on y/x — homogeneous. If a stray k remains, it isn’t.

🧭 Recipe — solving a homogeneous DE

Rearrange to dy/dx = … and confirm the RHS depends only on y/x. Divide top and bottom by the highest power of x to make this obvious.
Substitute v = y/x: replace every y/x with v, and replace dy/dx with v + x·dv/dx.
Simplify to separable form: x·dv/dx = f(v) − v. Use separation of variables on this.
Integrate both sides. Use partial fractions, substitution, or parts as needed.
Back-substitute v = y/x to express the solution in terms of x and y. Apply any initial condition to find the constant.

Worked examples

WE 1

Show an equation is homogeneous

Show that xy · dy/dx = x² + y² is a homogeneous differential equation.

Step 1 — isolate dy/dx dy/dx = (x² + y²)/(xy) Step 2 — divide top and bottom by x² dy/dx = (1 + y²/x²)/(y/x) = (1 + (y/x)²)/(y/x) = 1/(y/x) + (y/x) Step 3 — the RHS is a function of y/x alone dy/dx = f(y/x) where f(v) = 1/v + v Homogeneous ✓ — substitute v = y/x to solve. the “divide by highest power of x” move is the standard technique to expose the y/x structure. If you end up with a function of only y/x, the equation is homogeneous.

WE 2

General solution — simplest case

Find the general solution of dy/dx = (x + y)/x.

Step 1 — rewrite as function of y/x dy/dx = 1 + y/x = 1 + v (where v = y/x) Step 2 — substitute dy/dx = v + x dv/dx v + x dv/dx = 1 + v x dv/dx = 1 Step 3 — separate and integrate dv = (1/x) dx v = ln|x| + C Step 4 — back-substitute v = y/x y/x = ln|x| + C y = x ln|x| + Cx y = x ln|x| + Cx (general solution) notice how cleanly the v terms cancel: v + x dv/dx = 1 + v ⟹ x dv/dx = 1. The two v’s on each side disappear. This is a common pattern in homogeneous DEs — the substitution simplifies the algebra a lot.

WE 3

Particular solution with initial condition

Solve 2xy · dy/dx = x² + y² given y(1) = 2.

Step 1 — rewrite dy/dx = (x² + y²)/(2xy) = (1 + (y/x)²)/(2 y/x) = (1 + v²)/(2v) Step 2 — substitute v + x dv/dx = (1 + v²)/(2v) x dv/dx = (1 + v²)/(2v) − v = (1 + v² − 2v²)/(2v) = (1 − v²)/(2v) Step 3 — separate and integrate 2v/(1 − v²) dv = (1/x) dx LHS: let u = 1 − v², du = −2v dv, ∫ −du/u = −ln|u| −ln|1 − v²| = ln|x| + C |1 − v²| = A/|x| ⟹ 1 − v² = A/x (drop modulus) Step 4 — back-substitute v = y/x 1 − y²/x² = A/x Multiply by x²: x² − y² = Ax Step 5 — apply IC y(1) = 2 1² − 2² = A·1 ⟹ A = −3 x² − y² = −3x ⟹ y² = x² + 3x y² = x² + 3x ⟹ y = √(x² + 3x) verify: y(1) = √4 = 2 ✓. The substitution u = 1 − v² in step 3 is just inverse chain rule — standard from Further Integration.

WE 4

Recognizing NON-homogeneous

Show that dy/dx = (x + y²)/x is NOT a homogeneous differential equation.

Step 1 — apply the scale test: replace x → kx, y → ky RHS(kx, ky) = (kx + (ky)²)/(kx) = (kx + k²y²)/(kx) = 1 + (k·y²/x) Step 2 — compare with the original RHS RHS(x, y) = (x + y²)/x = 1 + y²/x RHS(kx, ky) = 1 + k(y²/x) Step 3 — the k does NOT cancel RHS depends on k → not scale-invariant dy/dx = (x+y²)/x is NOT homogeneous. The substitution v = y/x wouldn’t simplify it. the scale-test is the quickest way to check. If you can simplify the RHS to a pure function of y/x, it’s homogeneous; if a stray x or y term refuses to combine, it isn’t.

WE 5

v² in the separable equation

Solve x² · dy/dx = xy + y² given y(1) = 1. Express your answer as y = f(x).

Step 1 — rewrite dy/dx = (xy + y²)/x² = y/x + (y/x)² = v + v² Step 2 — substitute v + x dv/dx = v + v² x dv/dx = v² Step 3 — separate and integrate (1/v²) dv = (1/x) dx −1/v = ln|x| + C v = −1/(ln|x| + C) Step 4 — back-substitute v = y/x y/x = −1/(ln|x| + C) y = −x/(ln|x| + C) Step 5 — apply IC y(1) = 1 1 = −1/(ln 1 + C) = −1/C ⟹ C = −1 y = −x/(ln|x| − 1) = x/(1 − ln|x|) y = x1 − ln|x| verify: y(1) = 1/(1 − 0) = 1 ✓. Note: the solution blows up when ln|x| = 1, i.e., x = e. This vertical asymptote at x = e is a real feature of the DE, not an error.

WE 6

Implicit answer — when y = f(x) isn’t clean

Find the general solution of dy/dx = (y − x)/(y + x), expressing your answer in implicit form.

Step 1 — rewrite Divide top and bottom by x: dy/dx = (y/x − 1)/(y/x + 1) = (v − 1)/(v + 1) Step 2 — substitute v + x dv/dx = (v − 1)/(v + 1) x dv/dx = (v − 1)/(v + 1) − v = (v − 1 − v(v + 1))/(v + 1) = (v − 1 − v² − v)/(v + 1) = −(1 + v²)/(v + 1) Step 3 — separate (v + 1)/(1 + v²) dv = −(1/x) dx Split LHS: v/(1+v²) + 1/(1+v²) Step 4 — integrate ∫ v/(1+v²) dv = (1/2) ln(1+v²) [u = 1+v², du = 2v dv] ∫ 1/(1+v²) dv = arctan(v) (1/2) ln(1+v²) + arctan(v) = −ln|x| + C Step 5 — back-substitute v = y/x and simplify (1/2) ln(1 + y²/x²) + arctan(y/x) = −ln|x| + C (1/2) ln((x² + y²)/x²) = (1/2) ln(x²+y²) − ln|x| ⟹ (1/2) ln(x²+y²) − ln|x| + arctan(y/x) = −ln|x| + C The −ln|x| cancels on both sides: arctan(y/x) + 12 ln(x² + y²) = C no clean y = f(x) form exists here — the answer stays implicit. Geometrically the level curves are spirals in the plane. When a question says “leave your answer in implicit form” or doesn’t specify “in the form y = f(x)”, don’t waste time forcing the rearrangement.

💡 Top tips

Divide by the highest power of x in numerator AND denominator to expose the y/x structure. This is the standard “convert to homogeneous form” move.
The product rule is automatic: y = vx ⟹ dy/dx = v + x dv/dx. Don’t forget the v term — it’s a common slip.
The v terms often cancel after substitution, leaving a clean separable equation in v and x. If they don’t cancel cleanly, double-check your algebra.
Apply separation of variables from the previous note — same toolkit, just in (v, x) instead of (y, x).
Back-substitute LAST. Do all the integration in (v, x), then convert at the end with v = y/x.

⚠ Common mistakes

Wrong derivative formula: writing dy/dx = x dv/dx (forgetting the v). The correct formula is dy/dx = v + x dv/dx by the product rule on y = vx.
Forgetting to back-substitute: the answer must be in terms of x and y, not v and x. Once you’ve found v(x), replace v with y/x.
Applying the method to a non-homogeneous DE: if the RHS has a lonely x or y term that breaks scale-invariance, the substitution won’t reduce to separable form. Test first.
Not simplifying f(v) − v: this step is where the magic happens. Combine into a single fraction so the separation is clean.
Forcing y = f(x) when the implicit form is final. arctan and log terms often refuse to combine — leave the answer implicit and you’ll save time.

Up next: Integrating Factor. The third analytical technique — for DEs in the standard form dy/dx + P(x)y = Q(x) (linear in y). You multiply both sides by an “integrating factor” e^{∫P(x) dx}, which makes the LHS into an exact derivative. Then one integration finishes the job.

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