Question

Question 2: [25 pts] Consider the initial value problem y' = y/x, y(1) = 4. a) Approximate the value of the solution at x = 1.4 with step size h = 0.2. b) Approximate the value of the solution at x = 1.4 with step size h = 0.1. c) Does the error change at x = 1.4 with step size h = 0.2, if the initial data is changed as y(0.6) = 4. Explain.

Answer 1

+ 24 Given y's Y ; 4(1) - 4. a At X=1,4 and h=0.2 method, using Euler's Yuri Ynth. f(xn, Yn) where, Xnt) = Xn th. h= 0.2 - Xo=1, Yo = 4 We have op (x,y)= x Step 1: ohh X = 10 + h = 1+ ) 5 9, = +5 + x ) Y(x,) = Y( 5 ) 0) = 4+ ļxf(1,4) 5 4+ (4) 4.8 Step-2; X2 = x, th = ult Y(X2) = 7/7/5) = 72 = 4, th.f(x,yy.) 4.8+ **,4-5) Ś *(4) li 4.8+ = 5.6 Y(14) = 5.6 b) Now, Xal4 and h = o.1 Euler's method, By using he 0.1 = here, ) 10 Xo = 1, Y=4 , f(x,y)=% Step-s; X = X, th = 0 +0. 1 = 0 4(x1)=Y (1.1) = 4, = Yth. f(xo, yo) = 4 + (0.1) F(1,4) 4 + (0,1)x 4 - 4.4

Step 2: X2= x, th="!+ 0.1 = 1.2 Y(X2)= y(1:2) = Y, = 4, thi f(x, y,) 4.4 + (0.1), f(11,4.4) 4.4 + (011)x 4 4.8 Step-3: X3 = x2 th - 12 + Ol- 13 Y(X3) - Y(1-3): 42 = y2 + h. f(x2,42) = 4.8 +(0.1). +(1.2, 4.8) Il 4.8 + 0,4 = 5.2 Step-4; X 4 = X3th 1.3 tool =1.4 -; y(x4)= Y(1-4) = Y4 = 43+ hof (x3 -43) 5.2 + (0.1)*P(1.3,5.2) = = 5.2 + (0-1)*4 5.6 -17 (1.4): 5,6 Here, Y (o. 6 )= 4, 6 – 02 C using Eulers method, 0.6+ 0.2=0.8 Step) : X,= Xo tha Y(x)= y(0.8) = 4, = Yo th. f(xo, yo) 4 + (0:2)x f(0,6,4) 5.3 - 1. step-2 X2= x,th = 0.8+ 0.2 Y(X2) = 4(1) = Y = y, th. f(x,,4,) = 5,3 + (0.2), f(0.8,5-3) = 5:3+ (0.2) (6.6667) 6.6667

Step 3: X 3 = x 2 th=1+0.2 = 1.2 y (1.2) Y3 Yth. f (x2, Y, ) = 6.6667 + (0.2). f (', 6.6667) Step 4: X4 - = 4+ Ex 1.2 +0.2 = 1.4 Y (X4) = 4(1.4) = 44 = Yzt hit (x3, 43) = 8 + (0.2). F(1:2,8) = 8+ (0.2) * ( 6.666667) = 9.3333 - (0.184): EEEE'b = 9.3333 there Hence, we can conclude that with b=0.2 and you)=4 when X = 1.4 a change when and x=1-4, h = 0.2 and y(0.6)= 4. Because, y(1-4) far y(0.6)= 4 کے 7 9.3333 but, is 5.6 y(1.4) fox y(= 4

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