Appropriate the solution of y’ = 2x + y , y(0) = 1 using Euler’s method with step size 0.1. Approximate the value of y(0.4).

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Here, Given

f(x, y)  = 2x + y

y(0) = 1

x0 = 0 and y0 =  1

Step-size (h) = 0.1

x1 = x0 + h = 0.1

x2 = 0.2

x3 = 0.3

x4 = 0.4

We know by Euler’s method

y(xi + 1) = y(xi) + h(f(xi , yi))

Step 1:

y(x1) = y(x0) + hf(x0, y0)

y(0.1) = y(0) + hf(0.1)

= 1 + 0.1 x 1

= 1.1

(x1, y1) = (0.1, 1.1)

Step 2:

y(x2) = y(x1) + hf(x1, y1)

y(0.2) = y(o.1) + 0.1 f(0.1, 1.1)

= 1.1 + 0.1 x 1.2

= 1.1 + 1.13

= 1.23

(x2, y2) = (0.2, 1.23)

Step 3:

y(x3) = y(x2) + hf(x2, y2)

y(0.3) = y(0.2) + 0.1 f(0.2, 1.23)

= 1.23 + 0.1 x (2 x 0.2 + 1.23)

= 1.303

(x3, y3) = (0.3, 1.393)

Step 4:

y(x4) = y(x3) + hf(x3, y3)

y(0.4) = y(0.3) + 0.1 x (2 x 0.3 + 1.393)

= 1.393 + 0.1 (2 x 0.3 + 1.393)

= 1.5923

Thus y(0.4) = 1.5923

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