Hint
The substitution \(\displaystyle v= \frac{1}{y}\) applies only on an interval with \(\displaystyle y\ne 0\). Invert the initial value when changing variables.
Worked solution
(a)
\(\displaystyle v(0)= \frac{1}{y(0)}= 11\)
Answer: \(11\).
Answer: \(11\)
(b)
\(\displaystyle C= v(0)- \frac{9}{2}= \frac{13}{2}\)
Answer: \(\displaystyle \frac{13}{2}\).
Answer: \(\displaystyle \frac{13}{2}\)
(c)
\(\displaystyle y= \frac{1}{2C+ \frac{9}{2}}= \frac{2}{35}\)
Answer: \(\displaystyle \frac{2}{35}\).
Answer: \(\displaystyle \frac{2}{35}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Invert the initial value when changing variables.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: v′ is \(\displaystyle -\frac{y\prime}{y^2}\), not \(\displaystyle \frac{y\prime}{y}\).
(b) · 2 marks
- 1 mark: Apply the initial condition to the transformed linear equation.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Transform the initial value as well as the equation.
(c) · 2 marks
- 1 mark: Use \(\displaystyle e^{ax}=2\) and invert v.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: v′ is \(\displaystyle -\frac{y\prime}{y^2}\), not \(\displaystyle \frac{y\prime}{y}\).
Total: 6 marks.