Differential Equations: Substitution \(u=xy^2\) — HCI 2025 H2 Math Prelim Paper 1
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Question
It is given that \(\left(2xy\,\dfrac{\mathrm{d}y}{\mathrm{d}x} + y^2\right)\cos x = \dfrac{1}{xy^2}\), where \(0 < x < \dfrac{\pi}{2}\) and \(y > 0\).
(a) Using the substitution \(u = xy^2\), show that the differential equation can be reduced to \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{\sec x}{u}\).
(b) Hence find the general solution to the differential equation.
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(a) With \(u = xy^2\), differentiate implicitly with respect to \(x\): \[\begin{aligned} \frac{\mathrm{d}u}{\mathrm{d}x} &= y^2 + x \cdot 2y \frac{\mathrm{d}y}{\mathrm{d}x}\\ &= y^2 + 2xy \frac{\mathrm{d}y}{\mathrm{d}x}. \end{aligned}\]
This is exactly the bracketed expression in the original DE. Substituting: \[\begin{aligned} \frac{\mathrm{d}u}{\mathrm{d}x} \cdot \cos x &= \frac{1}{xy^2}\\ &= \frac{1}{u}. \end{aligned}\]
Hence \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{u\cos x} = \dfrac{\sec x}{u}\). (Shown)
(b) Separating variables: \[ \int u \,\mathrm{d}u = \int \sec x \,\mathrm{d}x. \]
Integrating both sides: \[\begin{aligned} \tfrac{1}{2}u^2 &= \ln|\sec x + \tan x| + C_1. \end{aligned}\]
For \(0 < x < \tfrac{\pi}{2}\), both \(\sec x > 0\) and \(\tan x > 0\), so the modulus may be dropped. Substituting \(u = xy^2\) gives \(u^2 = x^2 y^4\), and multiplying through by \(2\) (absorbing \(2C_1\) into a new constant \(C\)): \[ \boxed{\, x^2 y^4 = 2\ln(\sec x + \tan x) + C \,} \]