Question 1(d) Merit

A linear-plus-reciprocal integral, a velocity in km hr⁻¹ integrated to a distance from a point 3 km away, an area between a root and a parabola, a separable equation, and an area bounded by three different curves at once.

Consider the differential equation dy/dx = y(2x − 3x²). Given that y = 1 when x = 2, find the value(s) of y when x = 1.

The 2023 paper, page 3 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Separate the variables.

   int (1/y) dy = int (2x − 3x²)dx

2. 2 Integrate both sides.

   ln y = x² − x³ + c

3. 3 Use y = 1 when x = 2. What is ln 1?

   0 = 4 − 8 + c, so c = 4.

4. 4 Now x = 1, and undo the log.

   ln y = 1 − 1 + 4 = 4, so y = e⁴ ≈ 54.6.

What the marker wanted

• Achieved Reaching ln y = x² − x³ + c.
• Merit The value y = e⁴ = 54.6, from correct integration.

In our words, from NZQA’s 2023 marking schedule.

How did it go?

Got it moves it further out. Review brings it back tomorrow.