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

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RolyPolyBest ResponseYou've already chosen the best response.0
I 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..)
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
For the left, it looks like CauchyEuler equation...
 one year ago

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

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

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

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

nubeerBest ResponseYou've already chosen the best response.0
hmm i dont know.. but why doesn't look right to you?
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
http://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.
 one year ago

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

RolyPolyBest ResponseYou've already chosen the best response.0
And wolf gives the same answer :(
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
letting \[y=v(\frac{1}{x})\]\[y'=\frac{v}{x^2}+\frac{v'}{x}\]\[x\left(v/x^2 + v'/x\right)+v/x=1/x\]\[v'=\frac{1}{x}\]\[v=\ln x\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
\[y=v\frac{1}{x}=\frac{\ln x}{x}\] the second term of the solution given by wolf @RolyPoly
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
How do you come up with y = v/x ???
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
variation of parameter uses \(y=v\cdot u_1\) where \(u_1\) is a solution to the homogeneous DE. i used \[u_1 = \frac{1}{x}\]
 one year ago

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

RolyPolyBest 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?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
the complementary solution is \[y_c=c_1(\frac{1}{x})\]
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
In 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 \(u_2\) ?!
 one year ago
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