Date | May 2013 | Marks available | 10 | Reference code | 13M.3ca.hl.TZ0.2 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ0 |
Command term | Hence, Show that, and Solve | Question number | 2 | Adapted from | N/A |
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = {\cos ^2}x\), given that y = 2 when x = 0.
Use Euler’s method with a step length of 0.1 to find an approximation to the value of y when x = 0.3.
(i) Show that the integrating factor for solving the differential equation is \(\sec x\).
(ii) Hence solve the differential equation, giving your answer in the form \(y = f(x)\).
Markscheme
use of \(y \to y + \frac{{h{\text{d}}y}}{{{\text{d}}x}}\) (M1)
Note: Award A1 for \(y(0.1)\) and A1 for \(y(0.2)\)
\(y(0.3) = 2.23\) A2
[5 marks]
(i) \({\text{IF}} = {{\text{e}}^{\left( {\int {\tan x{\text{d}}x} } \right)}}\) (M1)
\({\text{IF}} = {{\text{e}}^{\left( {\int {\frac{{\sin x}}{{\cos x}}{\text{d}}x} } \right)}}\) (M1)
Note: Only one of the two (M1) above may be implied.
\( = {{\text{e}}^{( - \ln \cos x)}}{\text{ (or }}{{\text{e}}^{(\ln \sec x)}})\) A1
\( = \sec x\) AG
(ii) multiplying by the IF (M1)
\(\sec x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x\tan x = \cos x\) (A1)
\(\frac{{\text{d}}}{{{\text{d}}x}}(y\sec x) = \cos x\) (A1)
\(y\sec x = \sin x + c\) A1A1
putting \(x = 0,{\text{ }}y = 2 \Rightarrow c = 2\)
\(y = \cos x(\sin x + 2)\) A1
[10 marks]
Examiners report
Most candidates knew Euler’s method and were able to apply it to the differential equation to answer part (a). Some candidates who knew Euler’s method completed one iteration too many to arrive at an incorrect answer. Surprisingly few candidates were able to efficiently use their GDCs to answer this question and this led to many final answers that were incorrect due to rounding errors.
Most candidates were able to correctly derive the Integration Factor in part (b) but some lost marks due to not showing all the steps that would be expected in a “show that” question. The differential equation was solved correctly by a significant number of candidates but there were errors when candidates multiplied by \(\sec x\) before the inclusion of the arbitrary constant.