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November 30th, 2017, 03:39 AM   #1
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Domain of a particular solution

The function defined by $\displaystyle y=e^{1/2-y/x}$ ($\displaystyle y$ is a function of $\displaystyle x$) is a particular solution to the DE $\displaystyle y'=\frac{y^2}{x^2+xy}$. Its graph passes through $\displaystyle (2, 1)$. Can I find the domain of the function? I have studied calculus III. Thanks.
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December 1st, 2017, 04:56 AM   #2
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That function is $\displaystyle y= e^{1/2}e^{-y/x}$. Multiply on both sides by $\displaystyle e^{y/x}$ to get $\displaystyle ye^{y/x}= e^{1/2}$. Let u= y/x. Then y= xu so the equation becomes $\displaystyle xue^u= e^{1/2}$ and then $\displaystyle ue^u= \frac{e^{1/2}}{x}$. From that $\displaystyle u= \frac{y}{x}= W\left(\frac{e^{1/2}}{x}\right)$, where "W" is "Lambert's W function", the inverse function to $\displaystyle f(x)= xe^x$. Finally, $\displaystyle y= xW\left(\frac{e^{1/2}}{x}\right)$. The domain of the W function is $\displaystyle x\ge -\frac{1}{e}$ so this function has domain $\displaystyle x\ge -e^{3/2}$.

See A Brief Look into the Lambert W Function

Last edited by Country Boy; December 1st, 2017 at 05:00 AM.
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December 1st, 2017, 02:36 PM   #3
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Originally Posted by Country Boy View Post
. . . so this function has domain $\displaystyle x\ge -e^{3/2}$.
How did you get that?
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