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I was reading my notes and at some point I write that in solving a first order linear DE using the integrating factor the final solution should have only one arvitrary constant arising from the integration of $Q(x)e^{\int P(x)dx}$ and so $e^{\int P(x)dx}$ does not contain an arbitrary constant.

Essentially what this means is that if $$\frac{dy}{dx}+P(x)\cdot y=Q(x) \Rightarrow y=e^{-\int P(x)dx}\int Q(x)e^{\int P(x)dx}dx,$$the only constant that would determine a particular solution would be the one arising from the second integral and not the one in the exponent. Why is this?

I understand that since this is a firstorder DE is should only involve one constant, but what about what I ask above?


If I define $\int P(x)dx=f(x) + C$ and substitute above, this would yield $$y=e^{-f(x)+c}\int Q(x)e^{-f(x)+c}dx=e^{-f(x)}e^{2c}\int Q(x)e^{-f(x)}dx$$and now also define $\int Q(x)e^{-f(x)}dx=g(x)+k$ we have$$y=e^{-f(x)}e^{2c}\left(g(x)+k\right)=e^{2c}\cdot e^{-f(x)}g(x)+ke^{2c}\cdot e^{-f(x)}$$ so wouldn't the constant of integration $c$ also affect the final solution?

Thanks in advance!

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Study the derivation of solving it once again. Or you are just done with the function you gave,it satisfy the lenier DE. –  Mr.ØØ7 Mar 31 '13 at 13:57
    
I understand how to solve a linear DE using the IF. I just don't get why there is only one constant involved. –  Ryuky Mar 31 '13 at 14:07
    
Ok , do you have a reason why $C$ is not added in the Integration by parts[$\int fx.gx$] second integral (while we integrate g): $-\int( f'x\int (gx) .dx)dx$ .No, These are definations got by derivating a particular function to get the function whose integral is needed. In linear DE we multiply with $ e^{-P(x).dx}$ To get a function which is anti derivative of required one.It's a way , a defination to solve it.If we use a $C$ we would end up getting a wrong function. –  Mr.ØØ7 Mar 31 '13 at 14:13
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It's just typos. You should get $$y=e^{-(f(x)+C)}\int{Q(x)e^{f(x)+C} dx}=e^{-f(x)}\int{Q(x)e^{f(x)} dx}$$ –  Ivan Loh Mar 31 '13 at 14:36
    
@IvanLoh Yes, you are right. Thanks! –  Ryuky Mar 31 '13 at 14:44
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