Initial Value Problems for Ordinary Differential

University:
California State University, Northridge
Course:
MATH 331 | Mathematical Explorations
Academic year:

2023
Views:

200

Pages:

47
Author:

Yung D

Another method 26=0, Yo= 1, h=0.1 mma = x - 20:01.00 2 h %" + 3 3! 1+ (01) (1) + (0.1) (2) + 11 21 + (0.1) 31 (2)+ ५, ५. Y(001) = + Yo+ 1.1103 y=x+y=0.14 1.1103 = 1.2103 4" 1+ 1.2103 = 2.2103 2.2103 = 16+1= = :10.2)= 1-1103+ ((1.2103) + (2.1) (2.2103)+(0.1) (2.2103) 21 3! y (0.2) = 1.24275/1 2. Sowed = 4+½², ylo)=1 by Taylor series method Find y at 2=0.1, 0.2,0.8, 0.4 Sph Taylor briel formula Y= 6+ (2-0) 4. 11 + 2 (1-0) 40+ 2! 3 3! + Guren =0, Yo=1 y = y² +x² 2 2. %= + = 1+0=! y" 2 yy' +2x y = 2 [yy" + y'y'+1] = 2% +20= 2(1)(1)+210)=2 y = 2[%%%" +42=2[02+03 =2[2+1] = 6+2 Y = Yo + 11 + 21 yo + (x-x0) 3! 111 + 1+ (2) (1)+(29) (2) + 21 (2-0) (8) + 1 + x + x²+423 3 + Y(0-1) = 1+ (0-1) + (0.1)² + 4/3 (0.1)= 1-1113 3 Y(0-2) = 1+ (0.2) + (0.2)²+4/2 (0-2)=1.25066 (0.3) 1+103)+(0.3)² +43 (0.3) 1.426 ५ (0-4) = 1+(04)+(04)²+4/3 (0.4) 1.64533 Home work using Taylor series method find y at x = oil it = x²y-1, 4(0)=1. using Taylor series method find y(1, 1) given y' = x+y, y(1)=0. 3 using Taylor teries method with the first five terms in the expansion find y(0.1) correct to three decimal places given that dy da = y = ex-y² 2 9" = ex-2yy ex-2yy" ty'yl xo = 0, y = 1 4= ex-² = 1-lco y=1+ 4.=0+ (x-xo) 40+ (x-2) = 1-2[1(1)+0]= 1-2=-1 3 21 40+ (x-x0) 40 ..... 31 (0) (0)+(2-0) 2 = 1 + (10) + (-) (1) + (-1) 3. = 1+0++(-1) :y (0.1) = 1+ (1)² -16.113 2 2 1.004833 1.0051 Employing Taylor series expansion, bind yat x = 0.1, 0.2 Cobreet to three significant digits given -2y = sex, 410)=0. Here =0, %=0, x=0.1, x2=0.2, 23=0.3 h=0.1 y = 2y+sex = 20 + 3e 1 3 = 2% +3% = 9 y" = 2y+3e 2= y(0.2) = 4,+ +y+ = 0.3486875+ (0.1) (4.012887) +0.01 (11.341286) + (0.001) (25.99808)+... 6 2 18+3 = 21 = 2y + 3ex iv % = 42+3=45 = 2y + 3ex + h4 24 + Y (0-1) = 1 = 0+ (0-1)3 + (0.01) (9) + (0.001) (21) + (0.0001) (45)+.. 2 6 = 0.3+0045 +0.0035 + 0.0001875+. = 0-3486875 = 0.349 (sures decimals) 24 4= +30= ५ 29, +3 0.3486815 x2+ Be 0.1 = 24+ 4.012887 y = 241+3 = 11.025774 ," = 2y + 30 = 25.3670608. 지 0.8110156 = 0.811 (Correct to three decimal places) The exact value of y(0.1) = 0.3486955 and y(0.2) = 0.8112658 Taylor Series method for simultaneous First order Differential equations sowa the System of equations dy 오 dz Y+x with yoo) =1, 210)=1 by taking hol to get y(0.1) and 2 (0.1). Hare yand z are dependent variables and independent. Som Yo= 1 201 = 20-x² =1-021 9" = 2-2x 4= 20-2010 = 1-210)=1 y=2"-2(1) = 20-2 = 2-2=0. Zo = Yo+to = 1+0=1 z = y'+1 z = y" 2= " =1 111 (x-x0) + (x-x0) 40" y = 0 + (x-x0) 11 ۱۱ 2! (xx) 3! + 111 ५+ Y= 1+ 12 (1)+ (2-0) (1) + (200) (0) + 2 3! y = 1+x+ ५(0-1) = 1+ (0-1) + 1.105 2 20+ 2-0) 2= 2+(ユース0) 20 12 2 31 Zo + 1+1)+(2)+(1)･･･ 31 Z(0.1) 1+ (0-1) + (0.1)² + 10 6 2(0-1) = 1.110166 HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS By Taylor's series method find y(0-1) given that y"=y+dy! Y(0)=1, y'lo)=0 Sof Here Xo=0 0=1 Given y = ytxy! y"""= y +ay" ty = 2y + xy" %"= 0 + 0% = 1 + (0) (0)=1 11 = 24% +doy!" = 2(0)+(0) (1) = = 34" +20%" = 3(1)+(0) (0) IV y = ay tay" ty" y(x)= + xy。 + + 3 2 3! 21 + 41 (1)+○+ (3)+... (3) +... =1+0+ (1) +0 + 2 y=1++ 1001)= 1+ 8 24 2 8 1.0050 2 ② Evaluate the values of ylo.1) and 4to.2) given yll-x(y)²+y²=0 910) = 1, y'(0)=0 by wing Taylor Series method. Given y" x x(y-y², xo=o o=140004 = 20 (4!)² - y=x2y' y" +(y)-2yy 2xyly" + (y) - 2yy = y' [axy" + y'-24] = (0) (0)-(1)=-1 " = 20% +(6)-2%% = 2(0) (0) (-1)+(0)-2(1)(0) = y' [2 (xy+y") + "-241) + (2xy" +4-24) y" = 2xy'y +2y'y"-2 (y)² + 2x (y")² + y'y" -2yy! = 2xy'y'" + 4y' y" + 2x(y)² - 2 (y1)²-2yy" = 200%! % +49! %" + 2xo (4)² - 2 (46) -2%%" =0+0+0-2(0)-2(1)(-1) 2 y(x) = + = 1 + (0) + 5 +x 4 41 (-1)+(2)+... + + 410-1) = 1-12+1) = 1-0.005 +0.000008 = 0.995008 =0.995 (Correct to 3 decional phe Euler and modified Enter method. Y = +ht (x0, 0) 42 = 1+h612, 41) Yn+1 Yn + ho lam, Yn), m= 0, 1, 2... @using Euler's method find y(0.2) and y(0.4), from dy = x+y, Yle)=1 with h=0.2 410.6) f(x,y) = x+y, No = 0, % = 1 24=0.2, x2 = 0.4 By Euler algorithm, Y = Yot h t (x0, yo) = 1+ (0-2) [20+] = 1+62) (0+1) 41 = 1-2 (y ( 0.2) = 1.2 2 = Yi+h21,1) = 1.2 +0.2 [21+4] = 1.2+(0-2) [0-2+1-2] 1.21 (0-2) [1.4] = 1-2+0.28 42 = 1-48 910-4) = 1.48 43 = 42 + h 612, 2) = 1.48+(0.2) [2+2] +1.48+ (0.2) [04+1.48] = 1.48 +0.376 ५(0-6)=1-256 Cising Euler's method find y(0.3) of y(2) Satisfies the initial Value problem. = ½ (2²+1) 4², y(0.2) = 1.1114. Som Criven f(x+4) = (2+1) 4, Euler, Xe = 0.2, Yo 1.1114, x1=0.3 By Y = Yoth (lo, Yo) =%+h[/½(x+1) 7 1-1114+ (01) ((6-2)²+1) (1.1114)] 2 1.1114+ [1.2846184] =1.1114 +0.0642 (0.3) 1.1756 Compute y at x = 0.25 by modified Euler mathod given y' = 2xy, ylo)01. Som Given f(x,y) = 2xy, loo, o = 1, h=0.25, 1=0.25 By modibled Euler method, Ynt1 = m+ht (2n+1½h, 4n+h6(如,㎡)) ५, 1 Yo + h (0+0.25 1+0:25 (20 %)) = 1+(0.25) [+(0-125,1+ (0.25) (0) (1) = 1+0.25 [2 (0-125) (1)) = 1+ 0.00625 = 1.00625 using modified Baler's method, Compute Yl0.1) with h=0.1 from y'= 4-2, 410)=1. Criven f(x,y) = 4-, 2000%=12,= By modified Euler method, 9₁ = + h [b[20+½, 4% + ½ (2%, 40)] =1+(0-1) [610+), 1+[%]] -