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anonymous
 3 years ago
Solve by undetermined coefficients/variation of parameters:
xy' + y = 1/x (x>0)
anonymous
 3 years ago
Solve by undetermined coefficients/variation of parameters: xy' + y = 1/x (x>0)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know using integrating factor is a better way (at least for me), but still I would like to do it using these two methods (or at least one of them..)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the left, it looks like CauchyEuler equation...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[xy_c' + y_c=0\]\[y_c = c_1x^{1}\] But what about the particular solution?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0variation of parameter. the other method would not work as we dont have the general for for 1/x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are the two equations:\[u_1'(x^{1}) = 0\]\[u_1'(\frac{1}{x^2})=\frac{1}{x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm... \[u_1'(\frac{1}{x^2})=\frac{1}{x}\]\[u_1' = x\]\[u_1=\frac{x^2}{2}\]? But that doesn't seem right..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm i dont know.. but why doesn't look right to you?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=xy%27+%2B+y+%3D+1%2Fx ok try to look on this.. as it's not opening up for me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The answer to the question is \[y=\frac{c_1}{x}+\frac{logx}{x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And wolf gives the same answer :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0letting \[y=v(\frac{1}{x})\]\[y'=\frac{v}{x^2}+\frac{v'}{x}\]\[x\left(v/x^2 + v'/x\right)+v/x=1/x\]\[v'=\frac{1}{x}\]\[v=\ln x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y=v\frac{1}{x}=\frac{\ln x}{x}\] the second term of the solution given by wolf @RolyPoly

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How do you come up with y = v/x ???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0variation of parameter uses \(y=v\cdot u_1\) where \(u_1\) is a solution to the homogeneous DE. i used \[u_1 = \frac{1}{x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I... just realized that I didn't (and don't) know variation of parameters works!!! :'(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y_p=u_1y_1+u_2y_2\]In this case, only one complementary solution, so \[y_p = u_1y_1\]where \(y_1=x^{1}\) \[y_p'=\frac{u_1'}{x}\frac{u_1}{x^2}\] Sub these into the DE and solve \(u_1\) Is this the idea?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the complementary solution is \[y_c=c_1(\frac{1}{x})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0In case there are two, I can use the equations \[u_1'y_1+u_2'y_2 =0\]\[u_1'y_1'+u_2'y_2' =f\]to get \(u_1\) and \(u_2\) ?!
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