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anonymous
 one year ago
Need help with differential equations :/.
Find a second solution for xy''+(x1)y'y=0, given that y1=e^x.
anonymous
 one year ago
Need help with differential equations :/. Find a second solution for xy''+(x1)y'y=0, given that y1=e^x.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you know how to use integrating factor?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh... yes I do but I thought that only worked if the equation was in standard form and a first order equation. y'+P(x)y=q(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In that case I would just multiply everything by e^(integral (p) dx). Does that work on this too?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah wait do I need to do reduction of order to get it in the right form?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait, give me a few minutes. I'm not sure if it applies here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you. I appreciate it because I am definitely lost :/.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I actually am taking Differential Equations right now for my engineering major, but we haven't covered second order differential equations with nonconstant coefficients. I would recommend however visiting this page http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx It seems to be close to what you are looking for. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay I will read over it right now. Thank you for the link :). and that is interesting. I studied molecular biology but I am switching to biosystems engineering.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If they were constants, I'd totally be able to help you using the different cases :) I can't do anything biology for the life of me, haha. I admit I find it extremely fascinating but I can never memorize all the terms and functions XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm... okay so I did find an answer key to the question but I am still not sure how the professor solved this. Apparently the answer is y2=x1 but I am still trying to figure out what she did.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I understand the basic idea a solution to the differential equation can be a scalar multiple of the given y1. but I don't understand where the integrating factor comes in... this is so frustrating :/.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand why he used integrating factor at first. It seems like a different method than reduction of order, but I actually understand what he's doing. It seems like after finding e^u, he integrating in the following form \[y_2=y_1* \int\limits \frac{ e^{\mu} }{ y_1{^2} }dx\] Then once he simplified that, he just plugged in y1 that was given from the beginning. Also note that at the end, he put an e^x instead of e^x for the last term in the parentheses. Or vice versa. Not sure which one it's suppose to be but there's an error there.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does this make sense now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I sort of understand but I don't really get why the professor is multiplying by ∫e^(−μ)/y^2)dx. Where does that come from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm assuming it's some sort of formula to follow. I wouldn't really know because I haven't learned this yet (until now though, so thank you :P). But I don't understand the logic behind it and where it was derived from. For now I would just go with it and ask your professor later for clarification.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha... okay... well, if it works on the exam, I won't question it lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the help :). Good luck with your studies as well. Hopefully we can help eachother out :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You are very welcome! And hopefully :P
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