# 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
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 $$y(0.1)?$$

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 $$t_{2} = 0,1\ \text{and at}\ t_{2} = 0.1,\ y_{3} = 1.3509$$
$$\displaystyle\therefore{y}_{{3}}={y}{\left({0.1}\right)}={1.3509}$$

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
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At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
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