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 April 3rd, 2018, 07:05 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 Which series can solve ? How to get the solution... $\displaystyle xy'=e^x$
 April 3rd, 2018, 09:51 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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. Thanks from topsquark
 April 3rd, 2018, 01:57 PM #3 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,855 Thanks: 750 Math Focus: Wibbly wobbly timey-wimey stuff. 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.) -Dan
 April 3rd, 2018, 04:13 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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. Thanks from topsquark
 April 3rd, 2018, 04:40 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra 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. Thanks from topsquark

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