numerical methodsordinary differential equations
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1. Write this second order ode as a first order ode.
$y''(t)=y(t)+y'(t)+t$
$y(0) = 1$
$y'(0) = 0$
I think I understand the first problem. I substitute y with $u_1$ and $y'$ with $u_2$ which gives: $y' = u_2, y''=u_1 + u_2 + t$. Is that correct?
If so, I don't really know where to go from there.
It works for me (using this method):
This is the spreadsheet I used (energy column not shown):
The prediction is plain Euler: C2 = A2+B2*0.1, D2 = B2+0.1*(-A2).
The correction is trapezoidal: A3 = A2+0.1/2*(B2+D2), B3 = B2+0.1/2*(-A2-C2)
And so on. The way I typed the formulas in, the prediction on line 2 refers to the same time moment as row 3, which perhaps isn't logical. It could be placed one row below. But this is a matter of spreadsheet style. The point is, the method works.
The energy does grow slowly: at the end of the plot shown above, it is $12.6$ versus initial $12.5$. But this is perfectly normal for a numerical method.
$y'''(1)=y(1)+(x=1)$ To get it going.
$y(1+1h)=y(1)+y'(1)h$
$y'(1+1h)=y'(1)+y''(1)h$
$y''(1+1h)=y''(1)+y'''(1)$
$y'''(1+1h)=y(1)+(x=1)$
repeat from step 2
Best Answer
Expanding the comment of Winther: Yes, but write $y''=u_2'$ to get the first order system: \begin{align} u_1'&=u_2&\text{ and }\\ u_2'&=u_1+u_2+t. \end{align} Now apply Euler's method one step: \begin{align} y(h)=u_1(h)&\approx u_1(0)+h\,u_2(0)&&=1+h\cdot0&\text{ and }\\ y'(h)=u_2(h)&\approx u_2(0)+h\,\bigl[u_1(0)+u_2(0)+0\bigr]&&=0+hâ‹…[1+0+0]. \end{align}
In the next step you would get \begin{align} y(2h)=u_1(2h)&\approx u_1(h)+h\,u_2(h)&&≈1+h\cdot h&\text{ and }\\ y'(2h)=u_2(2h)&\approx u_2(h)+h\,\bigl[u_1(h)+u_2(h)+h\bigr]&&≈h+h⋅[1+h+h]. \end{align} et