Consider that math modeling following initial valu problem displaystylefrac{{left.{d}{y}right.}}{{left.{d}{t}right.}}={3}-{2}{t}-{0.5}{y},{y}{left({0}right)}={1} We would liketo find an approximation solution with the step siza h = 0.05 What is the approximation of y(0.1)?

Consider that math modeling following initial valu problem displaystylefrac{{left.{d}{y}right.}}{{left.{d}{t}right.}}={3}-{2}{t}-{0.5}{y},{y}{left({0}right)}={1} We would liketo find an approximation solution with the step siza h = 0.05 What is the approximation of y(0.1)?

Question
Modeling
asked 2021-01-16
Consider that math modeling following initial valu problem
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}={3}-{2}{t}-{0.5}{y},{y}{\left({0}\right)}={1}\)
We would liketo find an approximation solution with the step siza \(h = 0.05\)
What is the approximation of \(y(0.1)?\)

Answers (1)

2021-01-17

Given Initial Value Problem is:
\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}={3}-{2}{t}-{0},{5}{y},{y}{\left({0}\right)}={1},{t}_{{0}}={0}\)
and step size \(h = 0,05\)
Here \(\displaystyle f{{\left({t},{y}\right)}}={3}-{2}{t}-{0},{5}{y}\)
\(\displaystyle f{{\left({t}_{{{n}-{1}}},{y}_{{{n}-{1}}}\right)}}={3}-{2}{t}_{{{n}-{1}}}-{0},{5}{y}_{{{n}-{1}}}\)
\(\displaystyle={3}-{2}{\left\lbrace{t}_{{0}}+{\left({n}-{1}\right)}{h}\right\rbrace}-{0},{5}{y}_{{{n}-{1}}}\)
Now we have \(\displaystyle{t}_{{0}}={0}{\quad\text{and}\quad}{y}_{{0}}={1}\)
for \(n = 1\) \(\displaystyle\therefore f{{\left({t}_{{0}},{y}_{{0}}\right)}}={3}-{2}{\left\lbrace{0}+{\left({1}-{1}\right)}{h}\right\rbrace}-{0},{5}\)
\(\displaystyle={3}-{2}\times{0}-{0},{5}\times{1}\)
\(= 3 - 0,5\)
\(= 2,5\)
\(\displaystyle\therefore{y}_{{1}}={y}_{{0}}+{h} f{{\left({t}_{{0}},{y}_{{0}}\right)}}\)
\(\displaystyle={1}+{0},{05}\times{2},{5}\)
\(= 1 + 0,125\)
\(= 1,125\)
Now, \(\displaystyle{t}_{{1}}={t}_{{0}}+{\left({n}-{1}\right)}{h}={0}+{\left({2}-{1}\right)}\times{0},{05}={0},{05},{{f}_{{{0}{|}><}}{n}}={2}{{f}_{{{0}{|}><}}}\)
\(n = 2,\)
\(\displaystyle f{{\left({t}_{{1}},{y}_{{1}}\right)}}={3}-{2}{\left\lbrace{t}_{{0}}+{\left({n}-{1}\right)}{h}\right\rbrace}-{0},{5}{y}_{{1}}\)
\(\displaystyle={3}-{2}{\left\lbrace{0}+{\left({2}-{1}\right)}{0},{05}\right\rbrace}-{0},{5}\times{1},{125}\)
\(\displaystyle={3}-{2}\times{0},{05}-{0},{5625}\)
\(\displaystyle={3}-{0},{1}-{0},{05625}\)
\(= 2,3375\)
\(\displaystyle\therefore{y}_{{2}}={y}_{{1}}={h} f{{\left({t}_{{1}},{y}_{{1}}\right)}}\)
\(\displaystyle={1},{125}+{0},{05}\times{2},{3375}\)
\(\displaystyle={1},{125}+{0},{1169}{\left[{f}{o}{u}{r}\ {d}{e}{c}{i}{m}{a}{l}\ {p}{l}{a}{c}{e}{s}\right]}\)
\(= 1,2419\)
Again, \(\displaystyle{t}_{{2}}={t}_{{0}}+{\left({3}-{1}\right)}{h}\)
\(\displaystyle={0}+{2}\times{0},{05},{n}={3}\)
\(= 0,1\)
\(\displaystyle\therefore f{{\left({t}_{{2}},{y}_{{2}}\right)}}={3}-{2}{t}_{{2}}-{0},{5}{y}_{{2}}\)
\(\displaystyle={3}-{2}\times{0},{1}-{0},{5}\times{1},{2419}\)
\(\displaystyle={3}-{0},{2}-{0},{6210}{\left[{f}{o}{u}{r}\ {d}{e}{c}{i}{m}{a}{l}\ {s}{p}{l}{a}{c}{e}{s}\right]}\)
\(= 2,1790\)
Hence
\(\displaystyle{y}_{{3}}={y}_{{2}}+{h} f{{\left({t}_{{2}},{y}_{{2}}\right)}}\)
\(\displaystyle={1},{2419}+{0},{05}\times{2},{1790}\)
\(= 1,2419\ +\ 0,1090\)
\(= 1,3509\)
Here \(t_{2} = 0,1\ \text{and at}\ t_{2} = 0.1,\ y_{3} = 1.3509\)
\(\displaystyle\therefore{y}_{{3}}={y}{\left({0.1}\right)}={1.3509}\)

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