Date | May 2016 | Marks available | 5 | Reference code | 16M.2.hl.TZ0.4 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Hence and Sketch | Question number | 4 | Adapted from | N/A |
Question
Consider the differential equation dydx=xy, where y≠0.
Find the general solution of the differential equation, expressing your answer in the form f(x, y)=c, where c is a constant.
(i) Hence find the particular solution passing through the points (1,±√2).
(ii) Sketch the graph of your solution and name the type of curve represented.
(i) Write down the particular solution passing through the points (1, ±1).
(ii) Give a geometrical interpretation of this solution in relation to part (b).
(i) Find the general solution of the differential equation dydx=xy+yx, where xy≠0.
(ii) Find the particular solution passing through the point (1, √2).
(iii) Sketch the particular solution.
(iv) The graph of the solution only contains points with |x|>a.
Find the exact value of a, a>0.
Markscheme
attempt to separate the variables M1
∫ydydxdx=∫xdx A1
Note: Accept ∫ydy=∫xdx.
obtain 12y2=12x2+ constant (⇒y2−x2=c) A1
[3 marks]
(i) substitute the coordinates for both points M1
(±√2)2−12=1
obtain y2−x2=1 or equivalent A1
(ii) A1A1
Note: A1 for general shape including two branches and symmetry;
A1 for values of the intercepts.
(rectangular) hyperbola A1
[5 marks]
(i) y2−x2=0 A1
(ii) the two straight lines y=±x A1
form the asymptotes to the hyperbola found above, or equivalent A1
[3 marks]
(i) the equation is homogeneous, so attempt to substitute y=vx M1
as a first step write dydx=xdvdx+v (A1)
then xdvdx+v=1v+v A1
attempt to solve the resulting separable equation M1
∫vdv=∫1xdx A1
obtain 12v2=ln|x|+ constant⇒y2=2x2ln|x|+cx2 A1
(ii) substituting the coordinates (M1)
obtain c=2⇒y2=2x2ln|x|+2x2 A1
(iii) A1
(iv) since y2>0 and x2≠0 R1
ln|x|>−1⇒|x|>e−1 A1
a=e−1 A1
Note: The R1 may be awarded for a correct reason leading to subsequent correct work.
[12 marks]
Examiners report
Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.
Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.
Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.
For part (d) most candidates used the correct solution method for a homogeneous differential equation. A few found the algebra hard going in finding the particular solution. Most approaches to the final part were unsatisfactory, with a lack of proper consideration of the inequalities in the question.