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April 3rd, 2018, 07:05 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 536 Thanks: 81  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: 3,264 Thanks: 902 
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 $\displaystyle \frac{e^x}{x}= \frac{1}{x}+ 1+ \frac{1}{2}x+ \frac{1}{6}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^{n1}+ \cdot\cdot\cdot$. Find y by integrating that term by term. 
April 3rd, 2018, 01:57 PM  #3 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,201 Thanks: 899 Math Focus: Wibbly wobbly timeywimey 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: 3,264 Thanks: 902  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.

April 3rd, 2018, 04:40 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 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.


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