Homogeneous differential equations

A function $f(x, y)$ is homogeneous of degree $n$ if $f(tx, ty) = t^n f(x, y)$ for all $t > 0$. Examples: $x^2 + xy$ is homogeneous of degree $2$; $\dfrac{x^2 + y^2}{xy}$ is of degree $0$; $\dfrac{y}{x}$ is of degree $0$.

A first-order DE is called homogeneous when it can be written as $\frac{dy}{dx} = F\!\left(\frac{y}{x}\right),$ i.e. the right-hand side depends only on the ratio $y/x$. Equivalently, $\dfrac{dy}{dx} = \dfrac{P(x, y)}{Q(x, y)}$ where $P$ and $Q$ are homogeneous functions of the same degree.

The substitution

Let $y = v x$, so $v = y/x$. Then $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Substituting: $v + x\frac{dv}{dx} = F(v),$ which separates as $\frac{dv}{F(v) - v} = \frac{dx}{x}.$ Integrate, then substitute $v = y/x$ back to recover $y$ as a function of $x$ (or as an implicit relation).

If $F(v) = v$ identically, the equation is degenerate , every line $y = vx$ is a solution.

Recognition

A DE $\dfrac{dy}{dx} = \dfrac{P(x, y)}{Q(x, y)}$ is homogeneous when both $P$ and $Q$ are homogeneous of the same degree. Quick test: replace each $x$ by $tx$ and each $y$ by $ty$; the equation should be unchanged (the $t$'s cancel).

Examples:

• $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$ , homogeneous (degree 1 on top and bottom). Divide top and bottom by $x$: $\dfrac{dy}{dx} = \dfrac{1 + y/x}{1 - y/x}$.
• $\dfrac{dy}{dx} = \dfrac{xy}{x^2 - y^2}$ , homogeneous.
• $\dfrac{dy}{dx} = x + y$ , not homogeneous (the right side is degree-1 polynomial, but combined "scalarly" $tx + ty = t(x+y)$, so actually it scales by $t^1$, making it homogeneous? Yes , it is homogeneous of degree 1, but the DE form differs: we need $F(y/x)$. We have $dy/dx = x(1 + y/x)$, which has an extra $x$. Not in the required form.)

The clearer test for the DE: after dividing top and bottom by $x^n$ (the common degree), does the right side depend only on $y/x$? If yes, homogeneous. The example $\dfrac{dy}{dx} = x + y$ becomes $\dfrac{dy}{dx} = x(1 + y/x)$ , still has bare $x$, so not homogeneous.

Worked examples

Example 1. Solve $\dfrac{dy}{dx} = \dfrac{x + y}{x}$.

Rewrite: $\dfrac{dy}{dx} = 1 + \dfrac{y}{x}$. Substitute $y = vx$: $v + x v' = 1 + v$, so $x v' = 1$, i.e. $dv = \dfrac{dx}{x}$. Integrate: $v = \ln|x| + C$. Restore: $\dfrac{y}{x} = \ln|x| + C$, so $y = x(\ln|x| + C)$.

Example 2. Solve $\dfrac{dy}{dx} = \dfrac{y - x}{y + x}$.

Divide numerator and denominator by $x$: $\dfrac{dy}{dx} = \dfrac{(y/x) - 1}{(y/x) + 1}$. Let $v = y/x$: $v + xv' = \dfrac{v - 1}{v + 1}$. So $x v' = \dfrac{v - 1}{v + 1} - v = \dfrac{(v - 1) - v(v + 1)}{v + 1} = \dfrac{-v^2 - 1}{v + 1}$. Separate: $\dfrac{(v + 1)\,dv}{-(v^2 + 1)} = \dfrac{dx}{x}$. Integrate: $-\dfrac{1}{2}\ln(v^2 + 1) - \tan^{-1}v = \ln|x| + C_1$. Restore: combine the $\ln$ terms. (The implicit form is $\tan^{-1}(y/x) + \dfrac{1}{2}\ln\left(\dfrac{x^2 + y^2}{x^2}\right) + \ln|x| = C_2$, or equivalently $\tan^{-1}(y/x) + \dfrac{1}{2}\ln(x^2 + y^2) = C$.)

Example 3. Solve $x \dfrac{dy}{dx} = y + x \tan(y/x)$.

Divide by $x$: $\dfrac{dy}{dx} = \dfrac{y}{x} + \tan(y/x)$. Substitute $v = y/x$: $v + xv' = v + \tan v$, so $x v' = \tan v$, separating to $\cot v\,dv = \dfrac{dx}{x}$. Integrate: $\ln|\sin v| = \ln|x| + C_1$, so $\sin v = Cx$. Restore: $\sin(y/x) = Cx$.

Example 4. Solve $(x^2 + y^2)\,dx - 2xy\,dy = 0$.

Rewrite: $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}$. Divide top and bottom by $x^2$: $\dfrac{dy}{dx} = \dfrac{1 + v^2}{2v}$ where $v = y/x$. Substitute $y = vx$: $v + xv' = \dfrac{1 + v^2}{2v}$, so $xv' = \dfrac{1 + v^2}{2v} - v = \dfrac{1 + v^2 - 2v^2}{2v} = \dfrac{1 - v^2}{2v}$. Separate: $\dfrac{2v\,dv}{1 - v^2} = \dfrac{dx}{x}$. Integrate: $-\ln|1 - v^2| = \ln|x| + C_1$. So $|1 - v^2| \cdot |x| = C$, i.e. $x(1 - y^2/x^2) = C$, giving $x^2 - y^2 = Cx$. (Hyperbola family.)

Example 5. Solve $\dfrac{dy}{dx} = \dfrac{y}{x} + \cos^2(y/x)$, $y(1) = \pi/4$.

Substitute $v = y/x$: $v + xv' = v + \cos^2 v$, so $xv' = \cos^2 v$. Separate: $\sec^2 v\,dv = \dfrac{dx}{x}$. Integrate: $\tan v = \ln|x| + C$. Apply IC: at $x = 1$, $v = \pi/4$, so $\tan(\pi/4) = 0 + C$, giving $C = 1$. Particular: $\tan(y/x) = \ln|x| + 1$.

Example 6. Solve $(x - y)\,dy = (x + y)\,dx$.

Rewrite: $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$. Substitute: $v + xv' = \dfrac{1 + v}{1 - v}$. So $x v' = \dfrac{1 + v}{1 - v} - v = \dfrac{1 + v - v(1 - v)}{1 - v} = \dfrac{1 + v^2}{1 - v}$. Separate: $\dfrac{1 - v}{1 + v^2}\,dv = \dfrac{dx}{x}$. Integrate: $\tan^{-1}v - \dfrac{1}{2}\ln(1 + v^2) = \ln|x| + C_1$. Restore: $\tan^{-1}(y/x) - \dfrac{1}{2}\ln(1 + y^2/x^2) = \ln|x| + C_1$, simplifying to $\tan^{-1}(y/x) = \ln\sqrt{x^2 + y^2} + C$.

Try it yourself

1. $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$
2. $\dfrac{dy}{dx} = \dfrac{y(x - y)}{x(x + y)}$
3. $x \dfrac{dy}{dx} = y(\log y - \log x + 1)$
4. $\dfrac{dy}{dx} = \dfrac{y}{x} - \cos^2(y/x)$
5. $(x^2 - y^2)\dfrac{dy}{dx} = 2xy$
6. $x \cos(y/x)\dfrac{dy}{dx} = y \cos(y/x) + x$
7. $\dfrac{dy}{dx} = \dfrac{y^2}{x^2 - xy}$
8. $(x^2 + xy)\,dy = (x^2 + y^2)\,dx$
9. $\dfrac{dy}{dx} = \dfrac{xy}{x^2 - y^2}$
10. $y(x + y)\,dx + (x + 2y)\,dy = 0$ , verify homogeneity (each term degree 2).
11. $(y^2 - 2xy)\,dx = (x^2 - 2xy)\,dy$
12. $\dfrac{dy}{dx} = \dfrac{y}{x} + \sin(y/x)$, $y(1) = 0$.
13. $x \dfrac{dy}{dx} - y = \sqrt{x^2 + y^2}$
14. Show that $\dfrac{dy}{dx} = \dfrac{ax + by}{cx + dy}$ is always homogeneous.

Pitfalls and tricks

• Check homogeneity first. Both sides of $P/Q$ must have the same degree.
• The substitution is $y = vx$, not $v = xy$. Also valid: $x = vy$ if it gives cleaner algebra.
• After integration, substitute $v = y/x$ back. The answer should be in terms of $x$ and $y$.
• Watch for degenerate cases where $F(v) = v$ , no separation happens; instead $\dfrac{dv}{dx} = 0$, so $v$ is constant.
• Combine logarithms cleanly. $\ln$-form answers often simplify with $\ln|x| + \ln|y/x| = \ln|y|$ etc.