N2018 P1 Q3

N2018 P1 Q3

Junior College 2
7 marks
Free
  1. It is given that \(x\frac{\mathrm dy}{\mathrm dx}=2y-6\). Using the substitution \(y=ux^2\), show that the differential equation can be transformed to \(\frac{\mathrm du}{\mathrm dx}=\mathrm{f}(x)\), where the function \(\mathrm{f}(x)\) is to be found.[3]
  2. Hence, given that \(y=2\) when \(x=1\), solve the differential equation \(x\frac{\mathrm dy}{\mathrm dx}=2y-6\), to find \(y\) in terms of \(x\).[4]

Default solution

(i) \(y=ux^2\Rightarrow \frac{\mathrm dy}{\mathrm dx}=x^2\frac{\mathrm du}{\mathrm dx}+2ux\).

\(x^3\frac{\mathrm du}{\mathrm dx}+2ux^2=2ux^2-6\Rightarrow\mathrm{f}(x)=\frac{\mathrm du}{\mathrm dx}=-\frac6{x^3}\).

(ii) \(u=\int-6x^{-3}\,\mathrm dx=3x^{-2}+C\), so \(y=3+Cx^2\).

\(x=1,\ y=2:\quad C=-1\Rightarrow y=3-x^2\).

Answer:(i) \(\mathrm{f}(x)=-\frac6{x^3}\); (ii) \(y=3-x^2\)

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