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प्रश्न
Solve `dy/dx=x^3+y`with initial conditions y(0)=2 at x= 0.2 in step of h = 0.1 by Runge Kutta method of Fourth order.
उत्तर
Given that, ` dy/dx= x^3+y`
`f(x,y)=x^3+y, x_0=0, y_0=2 and h= 0.1`
∴ `k_1=hf(x_0,y_0)=0.1(0+2)=0.2`
`∴ K_2=hf (x_0+h/2, y_0+k_1/2)=0.1[(0.1/2)^3+2+0.2/2]=0.2100`
`∴ K_3=hf(x_0+h/2,y_0+k_2/2)=0.1[(0.1/2)^3+2+0.2100/2]=0.2105`
`∴ K_4=hf(x_0+h/2,y_0+k_3/2)=0.1[(0.1/2)^3+2+0.2100/2]=0.23105`
`∴ (k=k_1+k_2+2k_3+k_4)/6=(0.2+2(0.21)+2(0.2105)+0.23105)/6`
∴ `K=0.2120`
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Runga‐Kutta Fourth Order Formula
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