Question 1048644
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{dy}{dx}\ =\ \frac{a}{x^2\ +\ 1}]


If *[tex \Large \frac{dy}{dx}\ =\ 3] when *[tex \Large x\ =\ 1], then *[tex \Large a\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ dy\ =\ \frac{6}{x^2\ +\ 1}dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \int\,dy\ =\ \int\,\frac{6}{x^2\ +\ 1}dx]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 6\arctan(x)\ +\ C]


if *[tex \Large y\ =\ 3] when *[tex \Large x\ =\ 1], then *[tex \Large C\ =\ 3\ -\ 6\arctan(1)\ \approx\ -1.71239], so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ \approx\ 6\arctan(x)\ -\ 1.71239]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(2)\ \approx\ 6\arctan(2)\ -\ 1.71239]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(2)\ \approx\ -0.60524]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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