Question 3(e) Merit to Excellence

Opens with Simpson’s Rule on five measured offsets from a garden path, then a root quotient, a product-of-sines area, an exponential-quotient acceleration integrated to a velocity, and a differential equation that has to be factorised before it will separate.

Consider the differential equation (1 − x²)(1 + y)dy/dx + (1 − x)(1 − y²) = 0. Given that y = 0 when x = 2, find the value(s) of y when x = 6.

The 2023 paper, page 13 ↗ Marking schedule, page 6 ↗

Work it out

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

1. 1 Both brackets with squares are differences of two squares. Factorise them.

   1 − x² = (1 − x)(1 + x) and 1 − y² = (1 − y)(1 + y).

2. 2 Move the second term across and cancel what the two sides share.

   (1 − x)(1 + x)(1 + y)dy/dx = −(1 − x)(1 − y)(1 + y)
   Cancel (1 − x) and (1 + y): (1 + x)dy/dx = −(1 − y).

3. 3 Separate and integrate.

   int (1/(1 − y)) dy = −int (1/(1 + x)) dx
   −ln(1 − y) = −ln(1 + x) + c

4. 4 Use y = 0 when x = 2.

   −ln 1 = −ln 3 + c, so c = ln 3, and ln(1 − y) = ln(1 + x) − ln 3 = ln(1 + x)/3.

5. 5 Undo the logs, then put x = 6 in.

   1 − y = (1 + x)/3, so y = (2 − x)/3, and at x = 6, y = −4/3.

6. 6 The question says value(s). Is there another?

   No. With modulus signs, |1 − y| = (1 + x)/3 also allows y = 1 + (1 + x)/3, but that line gives y = 2 at x = 2, not 0, so it is not this solution.

What the marker wanted

• Achieved Either of the two integrals correct.
• Merit The solution of the differential equation, with its constant: −ln(1 − y) = −ln(1 + x) + ln 3.
• Excellence The value y = −4/3, with correct integrations, is E8; a solution with one minor error is E7.

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

How did it go?

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