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using the taylor method perform two steps when $y' = -2t - y$ when $y(0) = -1$ and $h = 0.1$

what is $\frac{df}{dt},\quad \frac{d^2f}{dt^2}$ ...

I found $\frac{df}{dt} = -2 - y'$ and I do not think my $\frac{d^2f}{dt^2}$ is correct please help. I need steps.

But then d3f/dt3 = -2 - y' the same as df/dt. That's why I thought I was wrong. Do you get the same d3f/dt3?

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2 Answers 2

Assuming $f$ is your solution:

In that $y(t)=f(t)$ and in that case you are asked to find the numerical value of its derivatives. So $f'=y'=-2t-y$; which at start when $t=0$ and $y=-1$ has the value of $+1$.

$y'=-2t-y$

$y''=(-2t-y)'= -2-y'= -2-(-2t-y)=-2+2t+y$

$y'''=2+y'=2-2t-y$

$y''''=-2-y'=-2-(-2t-y)=-2+2t+y$

$y^{v}=2+y'=2-2t-y$

Note $y^{(2)}=y^{(4)}=y^{(6)}=\cdots$, and

$y^{(3)}=y^{(5)}=y^{(7)}=\cdots$ and these are negative of above number.

Assuming $f$ is the right hand side of your equation

$f=-2t-y$

$f'=-2-y'=-2-(-2t-y)=-2+2t+y$

$f''=2+ y'=2-2t-y$

$f'''=-2-y'=-2+2t+y$ same as $f'$ all odd order will be same

$f''''=2+y'=2-2t-y$ same as $f''$ all even order will be same, and negative of above

The general solution, obtained by integrating factor is $y=-2(t-1)+Ce^{-t}$ and for $y(0)=-1$ we get $C=-3$.

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What do you consider to be your $f$? is it same as $y$ or is it the right hand side of your original equation. –  Maesumi Feb 22 '13 at 21:02
    
If it is your rght hand side then your derivative list will be off from mine by a shift of 1. –  Maesumi Feb 22 '13 at 21:03
    
I consider f to be the right hand side of my y' = -2t - y. –  user63574 Feb 22 '13 at 21:04
    
both versions are included. The reason you see the same thing show up every other turn is that solution is mainly built by $e^{-t}$ which alternates when you differentiate it. –  Maesumi Feb 22 '13 at 21:14

You are given that $y'=-2t-y$ You can then take more derivatives: $y''=-2-y',\ \ y'''=-y''$ and then substitute in if you want: $y''=-2+2t+y,\ \ y'''=2-2t-y$

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