| Date | May 2008 | Marks available | 14 | Reference code | 08M.3ca.hl.TZ2.3 |
| Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ2 |
| Command term | Show and Solve | Question number | 3 | Adapted from | N/A |
Question
A curve that passes through the point (1, 2) is defined by the differential equation
\[\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x(1 + {x^2} - y){\text{ }}.\]
(a) (i) Use Euler’s method to get an approximate value of y when x = 1.3 , taking steps of 0.1. Show intermediate steps to four decimal places in a table.
(ii) How can a more accurate answer be obtained using Euler’s method?
(b) Solve the differential equation giving your answer in the form y = f(x) .
Markscheme
(a)
(i) \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x(1 + {x^2} - y)\)
Note: Award A2 for complete table.
Award A1 for a reasonable attempt.
\(f(1.3) = 2.14\,\,\,\,\,{\text{(accept 2.141)}}\) A1
(ii) Decrease the step size A1
[5 marks]
(b) \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x(1 + {x^2} - y)\)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2xy = 2x(1 + {x^2})\) M1
Integrating factor is \({{\text{e}}^{\int {2x{\text{d}}x} }} = {{\text{e}}^{{x^2}}}\) M1A1
So, \({{\text{e}}^{{x^2}}}y = \int {(2x} {{\text{e}}^{{x^2}}} + 2x{{\text{e}}^{{x^2}}}{x^2}){\text{d}}x\) A1
\( = {{\text{e}}^{{x^2}}} + {x^2}{{\text{e}}^{{x^2}}} - \int {2x{{\text{e}}^{{x^2}}}{\text{d}}x} \) M1A1
\( = {{\text{e}}^{{x^2}}} + {x^2}{{\text{e}}^{{x^2}}} - {{\text{e}}^{{x^2}}} + k\)
\( = {x^2}{{\text{e}}^{{x^2}}} + k\) A1
\(y = {x^2} + k{{\text{e}}^{ - {x^2}}}\)
\(x = 1,{\text{ }}y = 2 \to 2 = 1 + k{{\text{e}}^{ - 1}}\) M1
\(k = {\text{e}}\)
\(y = {x^2} + {{\text{e}}^{1 - {x^2}}}\) A1
[9 marks]
Total [14 marks]
Examiners report
Some incomplete tables spoiled what were often otherwise good solutions. Although the intermediate steps were asked to four decimal places the answer was not and the usual degree of IB accuracy was expected.
Some candidates surprisingly could not solve what was a fairly easy differential equation in part (b).
