Question 1(e) Merit to Excellence

Five independent parts, no shared scenario: a standard integral, a differential equation with a boundary condition, a pair of definite integrals that pin down two constants, a trig area, and a differential equation you have to rearrange before you can separate it.

Consider the differential equation y − xy − (1 + x)dy/dx = 0. Given that y = 3 when x = 0, find the value(s) of y when x = 2.

The 2025 paper, page 5 ↗ Marking schedule, page 2 ↗

Work it out

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

1. 1 Get the derivative on its own first. What do the two y terms share?

   y − xy = (1 + x)dy/dx, and y − xy = y(1 − x), so
   dy/dx = (y(1 − x))/(1 + x)

2. 2 Separate the variables: y with dy on one side, x with dx on the other.

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

3. 3 The right-hand fraction is top-heavy: its top has the same degree as its bottom. Divide before integrating.

   1 − x = −(1 + x) + 2, so (1 − x)/(1 + x) = −1 + 2/(1 + x).

4. 4 Integrate both sides, with one constant.

   ln y = −x + 2 ln(1 + x) + c
   No modulus signs are needed: starting from y = 3 the solution stays positive, and 1 + x > 0 from x = 0 to x = 2.

5. 5 Use y = 3 when x = 0.

   ln 3 = 0 + 2 ln 1 + c, so c = ln 3.

6. 6 Now put x = 2 in, and undo the log.

   ln y = −2 + 2 ln 3 + ln 3 = 3 ln 3 − 2
   y = e^{3 ln 3 − 2} = e^{ln 27}/e² = 27/e² ≈ 3.654

What the marker wanted

• Achieved One of the two integrations correct.
• Merit Both integrations correct, and the value of c found.
• Excellence Reaching ln y = 3 ln 3 − 2 is E7; the value of y itself, 27/e² = 3.654, is E8.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

How did it go?

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