A hand-held calculator will suffice for the Problem. The problem an initial value problem and its exact solution are given. Approximate the values of x(0.2) and y(0.2)in three ways: (a) by the Euler method with two steps of size h = 0.1; (b) by the improved Euler method with a single step of size h = 0.2; and (c) by the Runge–Kutta method with a single step of size h = 0.2. Compare the approximate values with the actual values x(0.2) and y(0.2). x' =  x -2y, x(0) = 0, y' = 2x + y, y(0) = 4; x(t) = -4et sin 2t, y(t) = 4et cos 2t

Advanced Engineering Mathematics
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ISBN:9780470458365
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Chapter2: Second-order Linear Odes
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A hand-held calculator will suffice for the Problem. The problem an initial value problem and its exact solution are given. Approximate the values of x(0.2) and y(0.2)in three ways: (a) by the Euler method with two steps of size h = 0.1; (b) by the improved Euler method with a single step of size h = 0.2; and (c) by the Runge–Kutta method with a single step of size h = 0.2. Compare the approximate values with the actual values x(0.2) and y(0.2).

x' =  x -2y, x(0) = 0,

y' = 2x + y, y(0) = 4;

x(t) = -4et sin 2t, y(t) = 4et cos 2t

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Step 1

We are authorized to answer one question at a time, since you have not mentioned which question you are looking for, so we are answering the first one, please repost your question separately for the remaining question.

Step 2

Given an initial value problem are

x'=x-2y, x0=0y'=2x + y, y0 = 4

and its exact solution are given

xt=-4etsin2tyt=4etcos2t

and (steps of size) h= 0.1

Step 3

Using the  Euler method

xn+1=xn+hGtn,xn,ynyn+1=yn+hHtn,xn,yn

and tn+1=tn+h

Step 4

Now

x'=x-2yGtn,xn,yn=xn-2yny'=2x + yHtn,xn,yn=2xn+yn

Then

Iteration First: Put n=0

x1=x0+hGt0,x0,y0x1=x0+hx0-2y0

Using x0=0,y0 = 4

x1=0+0.10-2×4x1=-0.8

and 

y1=y0+hHt0,x0,y0y1=y0+h2x0+y0

Using x0=0,y0 = 4

y1=4+0.12×0+4y1=4.4

and t1=t0+h=0+0.1=0.1

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