If, \(\frac{{dy}}{{dx}} = x + y,y\left( 0 \right) = 1\) using Runge’s method the value of y at x = 0.2, when h = 0.2 is

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  1. 1.2
  2. 1.4
  3. 1
  4. 1.48

Answer (Detailed Solution Below)

Option 1 : 1.2

Detailed Solution

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Concept:

Range-Kutta method(R-K method):

To solve \(\frac{{dy}}{{dx}}\) = f(x, y) with condition y(x0) = y0

  • Let 'h' denotes the interval between equidistant values of x 
  • If the initial values are (x0 , y0)
  • Then the first increment in y is computed from the formula given by,

k1 = h f(x0 , y0)

k2 = h f(x0 + \(\frac{{h}}{{2}}\), y0 + \(\frac{{k_1}}{{2}}\))

k3 = h f(x0 + \(\frac{{h}}{{2}}\), y0 + \(\frac{{k_2}}{{2}}\))

k4 = h f(x0 + h, y0 + k3)

Δy = \(\frac{{1}}{{6}}\)(k1 + 2k2 + 2k3 + k4)

y1 = y0 + Δy

where,

Δy is the change in y

y0 is the intial value of y

y1 is the changed value of y

Calculation:

Given:

x0 = 0; y0 = 1; h = 0.2;

f(x, y) = x + y;

k1 = h f(x0 , y0) = 0.2 (0 + 1) = 0.2;

k2 = h f(x0 + \(\frac{{h}}{{2}}\), y0 + \(\frac{{k_1}}{{2}}\)) = 0.2 × f(0 + 0.1, 1 + 0.1) = 0.2 × f(0.1, 1.1)

⇒ k2 = 0.2 × (0.1 + 1.1) = 0.24;

k3 = h f(x0 + \(\frac{{h}}{{2}}\), y0 + \(\frac{{k_2}}{{2}}\)) = 0.2 × f(0 + 0.1, 1 + 0.12) = 0.2 × f(0.1, 1.12)

⇒ k3 = 0.244;

k4 = h f(x0 + h, y0 + k3) = 0.2 × f(0 + 0.2, 1 + 0.244) = 0.2 × f(0.2,1.244)

⇒ k4 = 0.2888;

Δy = \(\frac{{1}}{{6}}\)(k1 + 2k2 + 2k3 + k4)

⇒ Δy = 1/6 (0.2 + (2 × 0.24) + (2 × 0.244) + 0.2888) = 0.2428;

Now 

y1 = y0 + Δy

⇒ y1 = 1 + 0.2428 = 1.24

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