How Do You Solve a Homogeneous Differential Equation Using the Substitution y = vx?

Substitute $y=vx$, where $v$ is a function of $x$, and use the product-rule derivative $\frac{dy}{dx}=v+x\frac{dv}{dx}$. This converts the equation into a separable differential equation. Integrate it and then replace $v$ with $\frac{y}{x}$.

The Method

In IB Mathematics: Analysis and Approaches HL, a first-order homogeneous differential equation has the form

$$\frac{dy}{dx}=f\left(\frac{y}{x}\right).$$

The right-hand side depends only on the ratio $\frac{y}{x}$, not on $x$ and $y$ separately. This is HL-only content in AHL 5.18 and may appear on Paper 1, Paper 2, or Paper 3.

Set $y=vx$, where $v$ depends on $x$. Since $\frac{y}{x}=v$, differentiate using the :

$$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$

Substituting gives $v+x\frac{dv}{dx}=f(v)$. Rearrange:

$$\frac{1}{f(v)-v}\,dv=\frac{1}{x}\,dx.$$

Then integrate both sides, include an arbitrary constant $C$, and use $\int \frac{1}{x}\,dx=\ln|x|$.

This rearrangement assumes $f(v)-v\neq 0$. Before dividing, check whether $f(v)-v=0$ has constant solutions. Each valid root may produce an additional straight-line solution that division would otherwise remove. Retain the absolute value in unless a domain such as is specified.

Worked Example

Solve $\frac{dy}{dx}=1+\frac{y}{x}$. Let , so

$$v+x\frac{dv}{dx}=1+v.$$

The $v$-terms cancel, leaving $x\frac{dv}{dx}=1$. Hence

$$v=\ln|x|+C.$$

Since $v=\frac{y}{x}$, back-substitution produces

$$y=x\left(\ln|x|+C\right),\qquad x\neq 0.$$

In less direct questions, simplify (f(v)-v) before integrating and use partial fractions when required. Keep the answer implicit if rearranging for $y$ is impossible or unnecessary, unless the question asks otherwise.

Exam Technique

Show every stage: identify dependence on $\frac{y}{x}$, state the substitution, apply the product rule, separate, integrate, and back-substitute. The common misconception is writing $\frac{dy}{dx}=v$; this is incorrect because varies with , so omitting loses method marks. Also include and consider solutions lost through division.