Rates of change and differential equations — Question 277

ANSWER

(a) Horizontal slopes at $y=2$; slope $-2$ at $y=4$. (b) $y=2-2e^{-x}$ approaches $y=2$ from below as x increases. (c) $y=2-2e^{-x}$.

Worked solution

(a) The derivative depends only on y, so each row has a constant slope. Use 0 on $y=2$ and $-2$ on $y=4$.

(b) The curve passes through the initial point with slope 2, increases and is concave down. It does not cross the equilibrium line $y=2$.

(c) Separation gives $-\ln|2-y|=x+C$. On the branch through $(0,0)$, $2-y=Ke^{-x}$ and $K=2$.

Equivalent justified methods accepted; respect any requested proof method.

Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.

Part a: Draw horizontal segments on $y=2$.

[1 mark]

Part a: Draw slope-$-2$ segments on $y=4$.

[1 mark]

Part b: Draw the increasing concave-down curve through $(0,0)$.

[1 mark]

Part b: Indicate the asymptote and approach from below.

[1 mark]

Part c: Separate and integrate correctly.

[1 mark]

Part c: Apply the initial condition and express y.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.