Let y(t) be the solution to y = 7te^{-y} satisfying

Stefan Hendricks 2021-12-31 Answered
Let \(\displaystyle{y}{\left({t}\right)}\) be the solution to \(\displaystyle{y}={7}{t}{e}^{{-{y}}}\) satisfying \(\displaystyle{y}{\left({0}\right)}={2}\).
(a) Use Euler's Method with time step \(\displaystyle{h}={0.2}\) to approximate \(\displaystyle{y}{\left({0.2}\right)},{y}{\left({0.4}\right)},\ldots,{y}{\left({1.0}\right)}\).
(b) Use separation of variables to find y(t) exactly.
(c) Compute the error in the approximations to \(\displaystyle{y}{\left({0.2}\right)},{y}{\left({0.6}\right)}\), and \(\displaystyle{y}{\left({1}\right)}\).

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Expert Answer

autormtak0w
Answered 2022-01-01 Author has 441 answers
Step 1
Let \(\displaystyle{y}={f{{\left({t},{y}\right)}}},{y}{\left({t}_{{{0}}}\right)}={y}_{{{0}}}\) be given IvP
The by Euler's method
\(\displaystyle{y}{\left({t}_{{{n}+{1}}}\right)}={y}_{{{n}+{1}}}={y}_{{{n}}}+{n}{f{{\left({t}_{{{n}}},{y}_{{{n}}}\right)}}}\), h is me step sire.
Step 2
Given \(\displaystyle{y}={7}{t}{e}^{{-{y}}}={f{{\left({t},{y}\right)}}}\) (say)
a) Then by Euler's method we have,
\(\displaystyle{y}_{{{n}+{1}}}={y}_{{{n}}}+{n}{f{{\left({t}_{{{n}}},{y}_{{{n}}}\right)}}}\), have \(\displaystyle{h}={0.2}\)
\(\displaystyle{t}_{{{0}}}={0},{y}_{{{0}}}={2}\)
\(\displaystyle\therefore{y}_{{{1}}}={y}_{{{0}}}+{0.2}{f{{\left({t}_{{{0}}},{y}_{{{0}}}\right)}}}={2}+{0.2}\cdot={2}={y}{\left({0.2}\right)}\)
\(\displaystyle\therefore{y}_{{{2}}}={y}_{{{1}}}+{0.2}{f{{\left({t}_{{{1}}},{y}_{{{1}}}\right)}}}={2}+{0.2}={y}{\left({0.189}\right)}={2.0378}={y}{\left({0.4}\right)}\)
\(\displaystyle\therefore{y}_{{{3}}}={y}_{{{2}}}+{0.2}{f{{\left({t}_{{{2}}},{y}_{{{2}}}\right)}}}={2.0378}+{0.2}{\left({0.3649}\right)}={2.11078}={y}{\left({0.6}\right)}\)
\(\displaystyle\therefore{y}_{{{4}}}={y}_{{{3}}}+{0.2}{f{{\left({t}_{{{3}}},{y}_{{{3}}}\right)}}}={2.11078}+{0.2}{\left({0.5088}\right)}={2.21254}={y}{\left({0.8}\right)}\)
\(\displaystyle\therefore{y}_{{{5}}}={y}_{{{4}}}+{0.2}{f{{\left({t}_{{{4}}},{y}_{{{4}}}\right)}}}={2.21254}+{0.2}{\left({0.6128}\right)}={2.3351}={y}{\left({1.0}\right)}\)
So we get:
\[\begin{array}{cc} k & t_{k} & y_{k} \\ 0 & 0 & 2 \\ 1 & 0.2 & 2 \\ 1 & 0.4 & 2.0378 \\ 3 & 0.6 & 2.11078 \\ 4 & 0.8 & 2.21251 \\ 5 & 1.0 & 2.3351\end{array}\]
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0
 
Ethan Sanders
Answered 2022-01-02 Author has 876 answers
Step 3
\(\displaystyle{y}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}{7}{t}{e}^{{-{y}}}\) - (1) \(\displaystyle{y}{\left({0}\right)}={2}\) -(2)
\(\displaystyle\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{e}^{{-{y}}}}}}={7}{t}{\left.{d}{t}\right.}\)
\(\displaystyle\Rightarrow\int{e}^{{{y}}}{\left.{d}{y}\right.}=\int{7}{t}{\left.{d}{t}\right.}+{c}\)
\(\displaystyle\Rightarrow{e}^{{{y}}}={\frac{{{7}{t}^{{{2}}}}}{{{2}}}}+{c}\)
Using (2) we get,
\(\displaystyle{e}^{{{2}}}={c}\)
\(\displaystyle\therefore{e}^{{{y}}}={\frac{{{7}{t}^{{{2}}}}}{{{2}}}}+{e}^{{{2}}}\)
\(\displaystyle\Rightarrow{y}{\left({t}\right)}={{\log}_{{{e}}}{\left[{\frac{{{7}{t}^{{{2}}}}}{{{2}}}}+{e}^{{{2}}}\right]}}\)
0
karton
Answered 2022-01-09 Author has 8454 answers

Step 4
\(\begin{array}{} \\\therefore y(0.2)=2.0188 \\\therefore |(y(0.7)-y_{1})=|(2.0188-2)=0.0188 \\now\ y(0.6)=2.1574 \\\therefore |(y(0.6)-y_{3})=|(2.1574-2.11078)=0.04662 \\\therefore y(1) = 2.3878 \\\therefore |(y(1)-y_{5})=|(2.3878-2.3351)=0.0527 \end{array}\)

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