- #1
roldy
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Homework Statement
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(x0)=y0.
Use Taylor series to determine y(x0+[itex]\Delta[/itex]x) to 4th order accuracy.
Initial condition: x0=0, y(x0)=1
step size: [itex]\Delta[/itex]x=0.1
Show 5 significant digits in the answer.
Do the calculations for only one step.
Homework Equations
[itex]\epsilon[/itex]=O([itex]\Delta[/itex]x5)
The Attempt at a Solution
dy/dx=f(x,y)
Taylor series:
y(x0+[itex]\Delta[/itex]x)=y(x0)+[itex]\Delta[/itex]xf'(x0,y(x0))+
\Delta[/itex]x21/2!f''(x0,y(x0))+[itex]\Delta[/itex]x31/3!f'''(x0,y(x0))+[itex]\Delta[/itex]x41/4!f''''(x0,y(x0))+[itex]\epsilon[/itex]+[itex] <br /> Is this correct?<br /> <br /> My solution:<br /> <br /> The derivatives:<br /> f'(x,y)=dy/dx=y+x=1+0=1<br /> f''(x,y)=d<sup>2</sup>y/dx<sup>2</sup>=dy/dx+1=1+1=2<br /> f'''(x,y)=d<sup>3</sup>y/dx<sup>3</sup>=dy<sup>2</sup>/dx<sup>2</sup>=2<br /> f''''(x,y)=d<sup>4</sup>y/dx<sup>4</sup>=d<sup>3</sup>y/dx<sup>3</sup>=2<br /> <br /> y(0+.1)=1+(.1)(1)+1/2!(.1)<sup>2</sup>(2)+1/3!(.1)<sup>3</sup>(2)+1/4!(.1)<sup>4</sup>(2)+(.1)<sup>5</sup><br /> <br /> y(.1)=1+.1+.01+.001/3+.0001/12+.00001=1.11035<br /> <br /> Did I solve this correctly? I want to be able to have something decent when I meet with the professor tomorrow.[/itex]