Differential Equations: Substitution \(u=y/x\) — DHS 2025 H2 Math Prelim Paper 1

Let \(u = \dfrac{y}{x}\), so \(y = ux\). Differentiating with respect to \(x\): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}u}{\mathrm{d}x}\,x + u \quad \cdots (1) \]

From the given equation, \(2xy\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^2 - y^2\): \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^2 - y^2}{2xy} = \frac{x^2 - u^2x^2}{2ux^2} = \frac{1-u^2}{2u} \quad \cdots (2) \]

Equating (1) and (2): \[\begin{aligned} \frac{\mathrm{d}u}{\mathrm{d}x}\,x + u &= \frac{1-u^2}{2u} \\ \frac{\mathrm{d}u}{\mathrm{d}x}\,x &= \frac{1-u^2}{2u} - u = \frac{1-3u^2}{2u} \end{aligned}\]

Therefore \(\dfrac{2u}{1-3u^2}\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{x}\) (shown).

Separating variables: \[\begin{aligned} \int \frac{2u}{1-3u^2}\,\mathrm{d}u &= \int \frac{1}{x}\,\mathrm{d}x \\ -\frac{1}{3}\int \frac{-6u}{1-3u^2}\,\mathrm{d}u &= \int \frac{1}{x}\,\mathrm{d}x \\ -\frac{1}{3}\ln|1-3u^2| &= \ln x + c \quad (\because x > 0) \\ \ln|1-3u^2| &= -3\ln x - 3c \\ 1 - 3u^2 &= \pm e^{-3c} \cdot e^{\ln x^{-3}} = \frac{A}{x^3} \quad (A = \pm e^{-3c}) \\ 3u^2 &= 1 - \frac{A}{x^3} \end{aligned}\]

Substituting \(u = \dfrac{y}{x}\): \[ 3\left(\frac{y}{x}\right)^2 = 1 - \frac{A}{x^3} \implies y^2 = \frac{x^2}{3}\left(1 - \frac{A}{x^3}\right) \]

Since \(y < 0\):