(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I posted this already but decided to revive this thread since I re-worked the problem.

Consider dy/dx=x+y, a function of both x and y subject to initial condition, y(x_{0})=y_{0}.

Use Taylor series to determine y(x_{0}+[itex]\Delta[/itex]x) to 4th order accuracy.

Initial condition: x_{0}=0, y(x_{0})=1

step size: [itex]\Delta[/itex]x=0.1

Show 5 significant digits in the answer.

Do the calculations for only one step.

2. Relevant equations

[itex]\epsilon[/itex]=O([itex]\Delta[/itex]x^{5})

3. The attempt at a solution

dy/dx=f(x,y)

Taylor series:

y(x_{0}+[itex]\Delta[/itex]x)=y(x_{0})+[itex]\Delta[/itex]xf'(x_{0},y(x_{0}))+

\Delta[/itex]x^{2}1/2!f''(x_{0},y(x_{0}))+[itex]\Delta[/itex]x^{3}1/3!f'''(x_{0},y(x_{0}))+[itex]\Delta[/itex]x^{4}1/4!f''''(x_{0},y(x_{0}))+[itex]\epsilon[/itex]+[itex]

Is this correct?

My solution:

The derivatives:

f'(x,y)=dy/dx=y+x=1+0=1

f''(x,y)=d^{2}y/dx^{2}=dy/dx+1=1+1=2

f'''(x,y)=d^{3}y/dx^{3}=dy^{2}/dx^{2}=2

f''''(x,y)=d^{4}y/dx^{4}=d^{3}y/dx^{3}=2

y(0+.1)=1+(.1)(1)+1/2!(.1)^{2}(2)+1/3!(.1)^{3}(2)+1/4!(.1)^{4}(2)+(.1)^{5}

y(.1)=1+.1+.01+.001/3+.0001/12+.00001=1.11035

Did I solve this correctly? I want to be able to have something decent when I meet with the professor tomorrow.

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# Homework Help: Taylor series homework check

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