DHS Euler's & Improved Euler's Method (9649) (Revision Solutions)

Numerical Methods – Euler’s Method NUMERICAL METHODS [FM] EULER’S METHOD (SOLUTIONS) 1 A solution to the differential equation d 2d y xyx  has y = 1 when x = 1. (i) Use two iterations of Euler method of step size 0.5 to estimate the value of y when x = 2. [3] (ii) Explain whether you would expect this value to be an under-estimate or an over-estimate of the true value. [2] (iii) Explain why it is usually better to improve accuracy by using the improved Euler method rather than by simply using smaller step sizes in Euler method. [1] [TJC/FM/2018/P1//Q2] [Solution] (i) d 2d y xyx  Given x1 = 1, y1 = 1 and h = 0.5 When x2 = 1.5,  1 21 d 2( 1 ) ( 1 )2 . 5d 10 . 5 xx yyy h x        When x3 = 2,  2 32 d 2 (1.5)(2.5) 4.47 (correct to 3 s.f.)d 2.5 0.5 xx yyy h x        (ii) xn d d nxx y x  1 3 1.5 3.936 2 2 2(4.46825) = 4.989 Since d d y x is increasing over [1, 2], the value is an under-estimate. Alternatively, 2 2 d1 d 1 dd 22 yy xyxx yx   H2 Double Math

Numerical Methods – Euler’s Method Since x > 0, y > 0 and d 0d y x  , we have 2 2 d 0d y x  Thus the curve is concave upwards (iii) It is usually more efficient to use the improved Euler method as the convergence is faster and so requires fewer computations or lesser computing time.

Numerical Methods – Euler’s Method 2 The differential equation 2d tan 1,d y yxx  where 1 when 1,yx is to be solved numerically. (i) Carry out two steps of Euler’s method with step length 0.1 to es timate the value of y when 1.2x , giving your answer to 4 decimal places. [3] (ii) The method in part (i) is now replaced by the improved Euler method. The estimate obtained is 2.0156, given to 4 decimal places . State, with a reason, whethe r this estimate and the one found in part (i) are likely to be overestimates or underestimates of the actual value of y when 1.2x . [2] (iii) Explain why it would be inappropriate to continue the process in part (i) to estimate the value of y when 1.6.x [2] [VJC/FM/2019/P2/Q3] [Solution] 1 (i) 22 00 d 1t a n , 1 ,1 , f ( , ) 1t a nd y yx xy x y yxx     , By Euler’s Method with h = 0.1: 1 2 10 . 1 f ( 1 , 1 ) 10 . 1 ( 1t a n 1 ) 1.255741 1.25574 0.1f(1.1,1.255741) 1.665560 y y        the value of y when x = 1.2 is 1.6656 (4d.p) . (ii) The improved Euler method usually gives a better estimate than the Euler method. Since 2.0156 > 1.6656, it is likely that the actual value of y is bigger than these estimates. Hence both estimates are likely to underestimate the actual value. (iii) 2f( , ) 1 tanx yy x is discontinuous at 1.572x  .

Numerical Methods – Euler’s Method 3 Determine the maximum number of iterations, using the Euler Method, with step size 0.01 on the differential equation 2 d 1,d yx yx  where 0y when 1x , such that the error between the actual value of y and the approximated value of y is less than 0.001. [6] [NYJC/FM/2018/P1//Q2] [Solution] 2 2 2 1 1 11 1 1ln |1 | 1e d 1 d 1e d d1 x x yx dyyx yc x y yx yy d A A xx x y           Given 0, 1yx , 1 1 e 1 ee 01 A A     11 1 e1ee 1xxy    Using Euler Method, 1 1 2 10.01(1 0.01( ) ,11) n nn yyy n n     , 0 0y 

Numerical Methods – Euler’s Method n ny actualy Difference 20 0.182334 0.181360 0.000974 21 0.190545 0.189525 0.00102 From GC, max no of iterations = 20 U(n): euler V(n): exact W(n): difference

Numerical Methods – Euler’s Method 4. The variables x and y are related by the differential equation d f( , )d y x yx  . (i) Taking the initial value as 00()yx y  , explain with the aid of a diagram, how the Euler method can be applied once on the different ial equation to approximate the solution at 0x xh . [3] Given that f, yxy x y x and 00(, )xy  (1.5, 2). (ii) Apply the following methods with a step size 0.5 to estimate y at 2.5x . (a) Euler method [2] (b) Improved Euler method [3] (iii) Comment on the accuracy of the estimates found in part (ii). [2] (iv) State one advantage and one disadvanta ge in using the improved Euler method compared to the Euler method. [2] [RVHS/FM/2018/P1/Q9] [Solution] (i) d f,d y x yx  ,  00yx y  Let 10x xh At the initial point  00,x y , the tangent line has slope  0 00 d f,d xx y xyx   .  00f,khx y Hence the solution at 0x xh is approximated by 10 0 0 f,yyh x y x y  00,x y  11,x y 0x 10x xh h solution curve slope =  00f, x y k

Numerical Methods – Euler’s Method (ii)(a) h = 0.5 x y  f,yhx y 1.5 2 0.833333 2 2.83333 2.12500 2.5 4.95833 (2.5) 4.96y  (ii)(b) h = 0.5 x y  f, x y u y x   1.5 2 1.66667 2.83333 2.95833 2 3.47917 5.21875 6.08854 9.00234 2.5 7.98034 (2.5) 7.98y  (iii) As the gradient of the solution curve is increasing, the estimate found by the improved Euler method is more accurate than that found by the Euler method because gradient used in improved Euler method is adjusted by taking mean of gradients at xi and xi + h, as the iteration goes. However, noticing that the gradient is increasing very rapidly from x = 2 to x = 2.5, the accuracy of both method is likely to be poor (as seen also from the difference in the two estimates). (iv) Advantage: - Estimate is more accurate - Estimate is numerically stable Disadvantage: - Require large computation time - Error estimation is not easily done

Numerical Methods – Euler’s Method 5 A particle moves along a straight line which passes through a fixed point O. It is acted on by two resistive forces, one of which is proportional to its displacement x from O while the other is proportional to its speed v. As a result, the motion of the particle is governed by the differential equation d 72 4 .d vvx vx   Given that 121 when 0,vx estimate the value of v when 1x using (i) one iteration of Euler’s Formula, [2] (ii) one iteration of the improved Euler formula. [2] Hence, explain why v is approximately a linear function of for 0 1.xx  [2] By considering the values of for 0 1,x xv  use the given differential equation to find an expression for this linear function. [2] [VJC/FM/2018/P1//Q5] [Solution] d7 24 f( , )d vx x vx v   (a) 00 0, 121:xv h = 1 One iteration of Euler formula: 10 0 0 (1) f( , )vv x v 7(0)121 24 97121    (b) One iteration of improved Euler formula: u1 = 97, h = 1 10 0 0 1 1 1 f( , ) f( , )2vv x v x u  17 ( 1 )121 24 2429 7 96.9639... 97.0           The two estimates are approximately equal. This means that the tangent at 00(,)x v used in the Euler formula has a gradient that is almost equal to the average of the gradients at 00(,)x v and 1(1, )u . In other words, tangents at 00(,)x v and 1(1, )u have about the same gradient. This suggests that v is approximately a linear function of x for 01 .x