Hint
Separate variables before integrating. Integrate to y²=kx²+a².
Worked solution
(a)
Separate as y dy=kx dx. Integration gives y²/2=kx²/2+C₀. The initial condition gives y²=kx²+a². Since y(0)>0, use the positive square root.
Answer: \(y=\sqrt{3 x^{2} + 4}\)
(b)
Differentiating the positive square root gives kx/√(kx²+a²)=kx/y. Its value at zero is a and its denominator is positive for every real x.
Answer: \(y'=kx/y\) with \(y(0)=2\).
(c)
Substitute the stated y into the squared relation, solve for x² and keep the positive x specified.
Answer: \(2\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 3 marks
- 1 mark: Separate and integrate both sides with a constant.
- 1 mark: Apply the initial value and select the positive branch.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not lose a factor of two when integrating y dy.
(b) · 2 marks
- 1 mark: Differentiate the explicit positive branch and verify both the differential equation and initial value.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: An implicit squared relation can contain an invalid branch.
(c) · 2 marks
- 1 mark: Use the squared solution relation and keep only the requested positive x.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not lose a factor of two when integrating y dy.
Total: 7 marks.