All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
\frac{dz}{dx} = 3 - 2z

\frac{1}{3-2z} \frac{dz}{dx} = 1

\int \frac{1}{3-2z} dz = \int 1 dx

-\frac{1}{2} \mathrm{ln} |3-2z| = x + C

Since z > \frac{3}{2},

\mathrm{ln} ( 3 -2z )= -2x -2 C

3-2z = e^{-2x  -2C}

2z = 3 - e^{-2x -2c }

z = \frac{3}{2} - Ae^{-2x} where A=\frac{e^{-2C}}{2}

(ii)
\frac{dy}{dx} = z

\frac{dy}{dx} = \frac{3}{2} - Ae^{-2x}

y = \frac{3}{2}x + \frac{Ae^{-2x}}{2} + D

(iii)
\frac{d^{2}y}{dx^{2}} = 2Ae^{-2x}

\frac{d^{2}y}{dx^{2}}= 2(\frac{3}{2} - \frac{dy}{dx})

\frac{d^{2}y}{dx^{2}}= 3 - 2\frac{dy}{dx}

\therefore, a=-2, b = 3

(iv)
Consider A = 0, then y = \frac{3}{2} + D

Let D > 0, so y\frac{3}{2} + 1 and D=0, then y = \frac{3}{2}x

\therefore, \mathrm{they~are} y=\frac{3}{2}x ~\mathrm{and}~ y= \frac{3}{2} + 1

Graph for 10iv
Graph for 10iv

KS Comments:

This is fairly good question. The first part explores students’ abilities to relate a 3-variables differential equation. In (iv), they are expected to understand the question first and set the arbitrary constants to the necessary conditions. Don’t forget to use the graphing calculator to draw the graphs!

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