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April 3rd, 2018, 08:05 AM   #1
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Which series can solve ?

How to get the solution...
$\displaystyle xy'=e^x$
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April 3rd, 2018, 10:51 AM   #2
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That equation can be written as $\displaystyle y'= \frac{e^x}{x}$.

The MacLaurin series for $\displaystyle e^x= 1+ x+ \frac{1}{2}x^2+ \frac{1}{6}x^3+ \cdot\cdot\cdot+ \frac{1}{n!}x^n+ \cdot\cdot\cdot$ and so $\displaystyle \frac{e^x}{x}= \frac{1}{x}+ 1+ \frac{1}{2}x+ \frac{1}{6}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^{n-1}+ \cdot\cdot\cdot$.

Find y by integrating that term by term.
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April 3rd, 2018, 02:57 PM   #3
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Technical point: Since we have to approximate using a power series shouldn't we be using a Taylor expansion instead of a Maclaurin? (Just for the sake of convergence issues. It's practically the same process.)

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April 3rd, 2018, 05:13 PM   #4
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If we were given an initial condition of the form y(a)= Y with a not 0, then we would prefer expand $\displaystyle e^x$ in a Taylor's series about a.
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April 3rd, 2018, 05:40 PM   #5
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The series given converges everywhere (that it is defined), so there is no need to search for other (more complicated) series. Although, given an initial condition not at zero might make a different series easier to work with.
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