planhetkk

2022-09-28

1. Write this second order ode as a first order ode.

${y}^{\u2033}(t)=y(t)+{y}^{\prime}(t)+t$

${y}^{\prime}(0)=1$

${y}^{\prime}(0)=0$

Use the Euler Method with h = 0.1. What approximation of $(y(h),{y}^{\prime}(h))$ do you get?"

I think I understand the first problem. I substitute y with ${u}_{1}$ and ${y}^{\prime}$ with ${u}_{2}$ which gives: ${y}^{\prime}={u}_{2},{y}^{\u2033}={u}_{1}+{u}_{2}+t$. Is that correct?

If so, I don't really know where to go from there.

${y}^{\u2033}(t)=y(t)+{y}^{\prime}(t)+t$

${y}^{\prime}(0)=1$

${y}^{\prime}(0)=0$

Use the Euler Method with h = 0.1. What approximation of $(y(h),{y}^{\prime}(h))$ do you get?"

I think I understand the first problem. I substitute y with ${u}_{1}$ and ${y}^{\prime}$ with ${u}_{2}$ which gives: ${y}^{\prime}={u}_{2},{y}^{\u2033}={u}_{1}+{u}_{2}+t$. Is that correct?

If so, I don't really know where to go from there.

Abigayle Lynn

Beginner2022-09-29Added 12 answers

Expanding the comment of Winther: Yes, but write ${y}^{\u2033}={u}_{2}^{\prime}$ to get the first order system:

$\begin{array}{rlr}{u}_{1}^{\prime}& ={u}_{2}& \text{and}\\ {u}_{2}^{\prime}& ={u}_{1}+{u}_{2}+t.\end{array}$ and

Now apply Euler's method one step:

$\begin{array}{rlrlr}y(h)={u}_{1}(h)& \approx {u}_{1}(0)+h\phantom{\rule{thinmathspace}{0ex}}{u}_{2}(0)& & =1+h\cdot 0& \text{and}\\ {y}^{\prime}(h)={u}_{2}(h)& \approx {u}_{2}(0)+h\phantom{\rule{thinmathspace}{0ex}}{\textstyle [}{u}_{1}(0)+{u}_{2}(0)+0{\textstyle ]}& & =0+h\cdot [1+0+0].\end{array}$ and

In the next step you would get

$\begin{array}{rlrlr}y(2h)={u}_{1}(2h)& \approx {u}_{1}(h)+h\phantom{\rule{thinmathspace}{0ex}}{u}_{2}(h)& & \approx 1+h\cdot h& \text{and}\\ {y}^{\prime}(2h)={u}_{2}(2h)& \approx {u}_{2}(h)+h\phantom{\rule{thinmathspace}{0ex}}{\textstyle [}{u}_{1}(h)+{u}_{2}(h)+h{\textstyle ]}& & \approx h+h\cdot [1+h+h].\end{array}$

and et

$\begin{array}{rlr}{u}_{1}^{\prime}& ={u}_{2}& \text{and}\\ {u}_{2}^{\prime}& ={u}_{1}+{u}_{2}+t.\end{array}$ and

Now apply Euler's method one step:

$\begin{array}{rlrlr}y(h)={u}_{1}(h)& \approx {u}_{1}(0)+h\phantom{\rule{thinmathspace}{0ex}}{u}_{2}(0)& & =1+h\cdot 0& \text{and}\\ {y}^{\prime}(h)={u}_{2}(h)& \approx {u}_{2}(0)+h\phantom{\rule{thinmathspace}{0ex}}{\textstyle [}{u}_{1}(0)+{u}_{2}(0)+0{\textstyle ]}& & =0+h\cdot [1+0+0].\end{array}$ and

In the next step you would get

$\begin{array}{rlrlr}y(2h)={u}_{1}(2h)& \approx {u}_{1}(h)+h\phantom{\rule{thinmathspace}{0ex}}{u}_{2}(h)& & \approx 1+h\cdot h& \text{and}\\ {y}^{\prime}(2h)={u}_{2}(2h)& \approx {u}_{2}(h)+h\phantom{\rule{thinmathspace}{0ex}}{\textstyle [}{u}_{1}(h)+{u}_{2}(h)+h{\textstyle ]}& & \approx h+h\cdot [1+h+h].\end{array}$

and et

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