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Ordinary differential equations - Example

Example [1].

Calculate by the Runge-Kutta’s method integral of the differential equation at the initial condition on the segment [0, 0.5] with the step of integration

S o l u t i o n.  Let’s calculate . For this purpose at the first we’ll consistently calculate :

Now we receive:

and, consequently,

The subsequent approaches are similarly calculated. Results of calculations are tabulated:


Results of numerical integration of the differential
equation (1) by the fourth order Runge-Kutta’s method

i

x

y

k = 0.1 ( x + y )

Δ y

0

1

1

1

0.1

0.05

1.05

1.1

0.22

0.05

1.055

1.105

0.221

0.1

1.1105

1.210

0.1210

1/6 0.6620= 0.1103

1

0.1

1.1103

1.210

0.1210

0.15

1.1708

1.321

0.2642

0.15

1.1763

1.326

0.2652

0.2

1.2429

1.443

0.1443

1/6 0.7947= 0.1324

2

0.2

1.2427

1.443

0.1443

0.25

1.3149

1.565

0.3130

0.25

1.3209

1.571

0.3142

0.3

1.3998

1.700

0.1700

1/6 0.9415= 0.1569

3

0.3

1.3996

1.700

0.1700

0.35

1.4846

1.835

0.3670

0.35

1.4904

1.840

0.3680

0.4

1.5836

1.984

0.1984

1/6 1.1034= 0.1840

4

0.4

1.5836

1.984

0.1984

0.45

1.6828

2.133

0.4266

0.45

1.6902

2.140

0.4280

0.5

1.7976

2.298

0.2298

1/6 1.2828= 0. 2138

5

0.5

1.7974

So , у (0.5) =1.7974.

For comparison the exact decision of the differential equation (1) is:

whence

Thus, exact and numerical solutions of the equation (1) have coincided up to the fourth decimal place.

The fourth order Runge-Kutta’s method also is widely applied to the numerical solution of systems of ordinary differential equations.


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