Question

Use the improved Euler's method to obtain a four-decimal approximation of the indicated value. First use h = 0.1 and then use h = 0.05. y = xy + Vy, 7(0) = 5; y(0.5) y(0.5) Ch 0.1) Y(0.5) (h = 0.05) Need Help? Read it Talk to a Tutor

Answer 1

Improved Euler's Method: Given, y'= xy + V y(0) = 5 We have to find the value of y(0.5) using step size of h=0.1 and h=0.05.

h The Improved Euler method states that Yn+1 = yn + f (xr, yn) + f(In+1, Ūn+1)), where În+1 = yn +h. f (In, yn) and Xn+1 = n + h. We have that h 1 10 20 = 0, y = 5, f(x,y) = xy +VY Step 1. 1 1 21 = 30 + h = 0+ 10 10 ỹ1 = yo +h. f (x0,y)=5+h. f(0,5) = (5) + 1 10 (15) V5 10 + 5 h Y1 = yo + ž (f(xo, yo) + f (x1,ịc)) (s(0,5) +s(TO 5)) = 5+ h 2 15 +5 10 10 15 1 5 = 5+ 10 2 5+ + 5 100+ 3 + V10 5.25219748894888

Step 2. 22 = xi +h 1 10 + 1 10 5 Ÿ2 = y; +h. f(21,yı) = 5.25219748894888 + hf Go 5.25219748894888 10 = (5.25219748894888) + 1 10 . (2.81698707760627) = 5.53389619670951 h Y2 = 3 + ((21,91) + f (72;ğ2)) = 5.25219748894888 h 1 f 2 10 45.25219748894888) + s ( *(5,5.53389619670951 :)) 10 (2.81698707760627 + (3.4592027124244)) 5.25219748894888 + 5.56600697845042 2

Step 3. 1 x3 = 22 +h + 1 3 10 10 5 İs = y2 +h. f (72, y) = 5.50600697845042 +hıs (5) (5,5.56600697845042 (5.56600697845042) + 1 10 (3.47244003950092) = 5.91325098240051 h 93 = y2 + (f(x2, 92) + f (03,Ī3)) = 5.56600697845042 h f ,5.56600697845042) +f 2 + 3 10 5.91325098240051 = 5.56600697845042 + = 5.94991363026332 (3.47244003950092 + (4.20569299675723)) 2

Step 4. 24 = x3 +h= 3 1 10 + 2 5 10 VA = y3 +h. f (x3,43) = 5.94991363026332+ hºf st ,5.94991363026332 1 = (5.94991363026332) + (4.22421856847489) = 6.37233548711081 10 h YA = y3 + (f(x3, 93) + f (x4,ĩa)) 5.94991363026332 +(5.1400136302632) + (6, 6.37233548711052)) = To (4.22421856847489 +(5.07328272211048)) 5.94991363026332 + 6.41478869479259 2

Step 5. 25 = X4+h - 2 5 + 1 10 2 js = ya th• f (, vy) = 6.41478869479259 +hıs (6,6. +(6, 6-4147896279259) = (6.41478869479259) + 1 10 (5.09865879195275) = 6.92465457398787 43 = + (f(xa» ) + f (x5, ģ3) = 6.41478869479259 h f Core ,6.41478869479259) +1 (7,6:124654 26.92465457398787 (5.09865879195275 + (6.09380112763683)) 6.41478869479259 + = 6.97441169077207 2 Hence, the required value is y(0.5)=6.9744

For h=0.05 : Step 1. 1 1 x = xo + h = 0+ 20 20 Ti yo+h. f(x0,yo) = 5+h: f(0,5) = 1 = (5) + (15) 5 20 + 5 20 h y = yo +5fCE (f(x0, y) + f (21,71)) = = 5+ +($10,5+5 (+ +5)) - = 5+ 5+ 15 400 + + 15 20 + 5 = 5.11881469779236 2

Step 2. 22 = x1 + h = 1 20 + 1 20 1 10 Z2 = y +h. f(21, yı) = 5.11881469779236 + hf G ,5.11881469779236 20 (5.11881469779236) + 1 20 (2.51842050226708) = 5.24473572290571 h Y2 = y + z(f(x1,91) + f (x2, Ķ2)) 5.11881469779236 h 1 f ,5.11881469779236) + f 2 20 + ,5.24473572290571 1)) = 5.11881469779236 + -(2.51842050226708+ (2.81461237221203)) (2.5 5.25214051965434

Step 3. 1 x3 = 22 +h 1 10 + 3 20 20 Y3 = y2 +h. f (x2, y2) = 5.25214051965434 +hf , 5.25214051965434 10 1 (5.25214051965434) + (2.81696895152479) = 5.39298896723058 20 h 43 = y2 + 3 (f(x2, y2) + f (x3,ỹz)) 5.25214051965434 h 1 f ,5.25214051965434) + f 2 10 + 3 ,5.39298896723058 20 8)) 5.25214051965434 + (2.81696895152479+ (3.13122932877941)) 2 5.40084547666194

Step 4. 3 1 1 24 = x3 +h + 20 20 94 = y; +h. f(x3, y3) = 5.40084547666194 + hf 3 20 ,5.40084547666194 1 (5.40084547666194) + (3.13409873971645) = 5.55755041364776 20 h YA = y3 +3 (f(x3; Y3) + f (x4,ỹa)) = 5.40084547666194 h 3 + f 5.40084547666194) + f 2 20 :)) 20 2 = 5.40084547666194 + = 5.56592184070508 (3.13409873971645 + (3.4689558220092))

Step 5. 1 5 = 2; +h 1 5 + 1 20 is = Y4+h. f (x4,44) = 5.56592184070508 +hf ,5.56592184070508 (5.56592184070508) + 1 20 (3.47240496840397) = 5.73954208912528 h Ys = y2 + (f(x4, y4) + f (x5; Ï3)) = 5.56592184070508 h + f 15.56592184070508) +- ( ,5.73954208912528 2 (3.47240496840397 + (3.83061966636436)) 5.56592184070508 + = 5.74849745657429 2

Step 6. 1 3 26 = x +h 1 4 20 10 T6 = ys+h. f (x5, Ys) = 5.74849745657429+hf (1,5 5.74849745657429 1 (5.74849745657429) + (3.83472680336787) = 5.94023379674268 20 h 46 = 45 + (f(x5,4s) + f (x6, Ķc)) 5.74849745657429 + (+(7,5.748497456574 1.74849745657429) + s (0,5.94023379674268 (3.83472680336787 + (4.21932962392438) = 5.74849745657429 + 5.9498488672566 2

Step 7. 3 27 = 26 +h + 1 7 20 20 10 öz = 40 + h. f (165 %) = 5.9498488672566+ h-5 (10 „5.9498488672566 (5.9498488672566) + 1 20 (4.22418586431827) = 6.16105816047251 h 97 = 46 + 5 (F(16, 96) + f (27,ỹx)) - 5.9498488672566 h 3 f ,5.9498488672566) + f 2 10 ( ,6.16105816047251 :)) 20 (4.22418586431827+ (4.63851824871177)) 5.9498488672566 + 6.17141647008235 2

Step 8. 28 27 +h= 7 20 + 1 20 2 5 Ģs = y; +h. f (x7, y) = 6.17141647008235 + hf -16 7 20 ,6.17141647008235 1 (6.17141647008235) + (4.64422934253278) = 6.40362793720899 20 h Y8 = yz+ż (f(x7,97) + f (xs, įs)) = 6.17141647008235 h 2 2 + )) (+(30 36.17141647008235) + s ( ,6.40362793720899 20 = 6.17141647008235 + - 6.41482195953269 -(4.64422934253278 + (5.09199023548096)) 2

Step 9. 1 29 = x's +h 2 5 + 9 20 20 19 = ++. $ (38, ) = 6.41482195953269 + hºf (6,6.4148219595 (6.41482195953269) + 1 20 (5.09867866477896) = 6.66975589277164 h Yg = yg + ž (f(x8,48) + f (x9,ỹg)) = 6.41482195953269 h 2 f ,6.41482195953269) + f 9 20 6.66975589277164 :)) (5.09867866477896+ (5.58397720599568)) = 6.41482195953269 + 6.68188835630206 2

Step 10. 9 210 = X9 + h = 1 1 20 20 V10 = yg + h. f (xg, yy) = 6.68188835630206 +hf 9 20 ,6.68188835630206 = (6.68188835630206) + 1 20 (5.59178464468061) = 6.96147758853609 Y10 910 = y2 + (f(x9 , yg) + f (110, İno)) = 6.68188835630206 h 9 , 6.68188835630206 ) + f 20 + (: (20,6.6818825630206 - (1,6.961477 6.96147758853609 (5.59178464468061 + (6.11920001030592)) = 6.68188835630206 + = 6.97466297267672 2

Hence, the required value is y(0.5) = 6.9747