2023 TJC_NYJC_VJC FM Prelim Paper 2 (Suggested Solution)

1 2023 TJC H2 FM Prelim Paper 2 (Suggested Solution) Question 1 (a) 2d ( 1)d y yx  Let 2f ( , ) ( 1)x y y Using the Euler Method with step size 0.5, nx ny d d y x 0 1.6 6.76 0.5 1.6 0.5 6.76 1.78    0.6084 1 1.78 0.5 0.6084 2.0842   Using Euler method, 2.8042 2.80y (3 s.f.) when 1x (b) The particular solution is 15 1 13xy   As seen in the diagram below, the tangent to the curve at 0.5x is steep which results in the first approximation to be already exceed the asymptotic value of y. (c) Using the improved Euler method, 0 0x  , 0 1.6y  , 0.5h , 2f ( , ) ( 1)x y y 0 0 0 0 0 0 10 f ( , ) f ( , f ( , )) 2 6.76 0.60841.6 0.5 2 0.2421 x y x h y h x yy y h         1 1 1 1 1 1 21 f ( , ) f ( , f ( , )) 2 x y x h y h x yy y h     0.441 (3 s.f.) The accuracy of the estimate can be improved by reducing the step size.

2 Question 2 2 1 1 1 1 (dividing throughout by ( 1)) 1 nn nn nn u n au n b n n uu a b n n nn v av b             1 1 11 1 1 1 1 1 ( l a)(i) 2 General solution, (2 ) Note that Thus 0 2 2 1000 1 1, (10000 )2 Particu ar solution, (1000 )2 (10000 )2 (1 0 2 0000 )2 n n nn n n nn n n vuv B b A b v b b u bb n u b b n v v b v A B v A B u A B B B b                              (a)(ii) This will eventually go to zero w hen 10000. The population of virus will first increase to a maximum value and then decrease to zero. b 1( 22b) nnvv  19998(2 ) 2n nun  Using GC: 1000000 0 when 7 i.e. after 1 week nun   Question 3 (i) (a) Characteristic equation: 0ab cd     2 ( )( ) 0 ( ) 0 a d bc a d ad bc            2( ) 4( ) 2 a d a d ad bc      2( ) 4 4(1 )(1 ) 2 a d a d ad d a       2( ) 4(1 ) 2 a d a d a d      2( ) 4( ) 4 2 a d a d a d     

3 2( 2) 2 a d a d    ( 2) 2 a d a d    2( ) 2 2 or 22 a d a d a d      1 or 1 (shown)ad   Or 1 ,1 a b a d c d a d            M Characteristic equation: 0MI  1 01 ad ad          1 1 0a d d a          2 10a d ad a d ad          2 10a d a d          1 1 0 ad     1 or 1 (shown)ad   1: a b x x c d y y               ( 1) 0 ( 1) ( 1) ( 1)( 1) 0 1// 1 a x by by a x d y a xcx d y xd ya                         1: ( 1)a b x xa d a d c d y y                   (1 ) 0 ( 1)(1 ) 0 1// 1 by d x by d x by bxcx a y x y                      The eigenvectors corresponding to 1 and 111 are and respectively.11 dad a           

4 Question 4 (a) a = 2, b + c = 6 Equation of top ellipse: 2 2 2 2 2