The variables $x$ and $y$ are related by the differential equation \[x^2 \frac{\mathrm{d}y}{\mat

Question

The variables $x$ and $y$ are related by the differential equation $$x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for}\hspace{0.5em} x > 0.$$ Using the substitution $y = ux^2$, for $x > 0$, show that the differential equation can be reduced to $$\frac{\mathrm{d}u}{\mathrm{d}x} = u(10 - u).$$[2] Hence show that the general solution for the differential equation $$x^2 \frac{\mathrm{d}y}{\mathrm{d}x} - 2xy = 10x^2y - y^2 \quad \text{for}\hspace{0.5em} x > 0$$ can be expressed in the form $y = \frac{10x^2}{Qe^{-10x} + 1}$, where $Q$ is an arbitrary constant.[6] Given that the graph of $y = \mathrm{f}(x)$ passes through the point with coordinates $\left(\frac{1}{10}, \frac{1}{10(1 - \mathrm{e}^{-1})}\right)$, find the equation of the graph of $y = \mathrm{f}(x)$.[2]

Accepted Answer

(i) $\dfrac{\mathrm du}{\mathrm dx}=u(10-u)$ (ii) $y=\dfrac{10x^2}{Q\mathrm e^{-10x}+1}$ (iii) $y=\dfrac{10x^2}{1-\mathrm e^{-10x}}$, $x>0$

2024 MI PRU2 P2 Q5

2024 MI PRU2 P2 Q5

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