Rates of change and differential equations — Question 278

ANSWER

(a) A positive even curve with minimum $(0,e^{-1/2})$. (b) $y=e^{(x^2-1)/2}$; minimum $e^{-1/2}$. (c) $y\equiv0$.

Worked solution

(a) The field has negative slopes for $x<0,y>0$ and positive slopes for $x>0,y>0$. The solution falls then rises, remaining positive. Its algebraic expression depends on $x^2$, giving reflection symmetry about the y-axis.

(b) For this positive branch, $dy/y=x\,dx$. Integrate to obtain $\ln y=x^2/2+C$. From $y(-1)=1$, $C=-1/2$. The exponent is least when $x=0$, so the minimum is $e^{-1/2}$.

(c) The zero function satisfies $y\prime=xy$ but not this initial condition. It belongs to the differential equation's solution family and is excluded by dividing by y.

Equivalent justified methods accepted; respect any requested proof method.

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

Part a: Sketch the positive curve through the initial point with a minimum at x=0.

[1 mark]

Part a: Indicate and justify the reflected shape.

[1 mark]

Part b: Separate and integrate.

[1 mark]

Part b: Use the initial condition.

[1 mark]

Part b: Identify the exact global minimum on the interval.

[1 mark]

Part c: Identify the equilibrium solution and distinguish it from this IVP.

[1 mark]

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