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Determine the recursive formulas for the Taylor method of order 4 for the initial value problem y'=x2+y,y(0)=0.

Short Answer

Expert verified

yn+1=yn+h(xn2+yn)+h22(2x+x2+y)+h36(2+2x+x2+y)+h424(2+2x+x2+y)

Step by step solution

01

Find the value of f2(x,y)

Herey'=x2+y,y(0)=0

Apply the chain rule.

f2(x,y)=∂f∂x(x,y)+∂f∂y(x,y)f(x,y)

Sincef(x,y) =x2+ y

∂f∂x(x,y)=2x∂f∂y(x,y)=1

So, the equation isf2(x,y)=2x+x2+y

02

Evaluate the values of f3(x,y)  and  f4 (x,y)

Apply the same procedure as step 1

f3(x,y)=2+2x+x2+yf4(x,y)=2+2x+x+y

03

Apply the recursive formulas for order 4

The recursive formula is

xn+1=xn+hyn+1=yn+hf(xn+yn)+h22!f2(xn+yn)+.....hpp!fp(xn+yn)

xn+1=xn+hyn+1=yn+h(xn2+yn)+h22(2x+x2+y)+h36(2+2x+x2+y)+h424(2+2x+x2+y)

Where starting points are xo=0,y0=0.

Hence the solution is

role="math" localid="1664317215716" yn+1=yn+h(xn2+yn)+h22(2x+x2+y)+h36(2+2x+x2+y)+h424(2+2x+x2+y)

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