## Doubt on P.I of $e^{ax} V$

I have a doubt in finding out the Particular Integral of eaxV, where 'V' is a function of x.

I saw the book but they seem to be somewhat unclear regarding this-

Here's the derivation from the book:

If u is a function of x, then
D(eaxu)= eax Du + aeaxu= eax (D+a)u
Similarly you can do for D2,D3... and so on

and after substituting V=f(D+a)u you will get the final equation as -

$\frac{1}{f(D)}(e^{ax}V)$=$e^{ax} \frac{1}{f(D+a)}$V

Now my doubt is how do you decide where to put that f(D+a) in the right side of the above equation? I could have interchanged eax and V. This could have resulted in an entirely different answer. I couldn't get that

Thanks a lot

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 Quote by iVenky If u is a function of x, then D(eaxu)= eax Du + aeaxu= eax (D+a)u Similarly you can do for D2,D3... and so on and after substituting V=f(D+a)u you will get the final equation as - $\frac{1}{f(D)}(e^{ax}V)$=$e^{ax} \frac{1}{f(D+a)}$V Now my doubt is how do you decide where to put that f(D+a) in the right side of the above equation?
If $\frac{1}{f(D)} = Ʃb_n D^n$, $\frac{1}{f(D)}(e^{ax}V) = Ʃb_n D^n(e^{ax}V) = Ʃb_n e^{ax}(D+a)^n(V) = e^{ax} \frac{1}{f(D+a)}V$

 Quote by haruspex If $\frac{1}{f(D)} = Ʃb_n D^n$, $\frac{1}{f(D)}(e^{ax}V) = Ʃb_n D^n(e^{ax}V) = Ʃb_n e^{ax}(D+a)^n(V) = e^{ax} \frac{1}{f(D+a)}V$

I still don't get how you take that $e^ax$ out of $D^n(e^{ax}V)$ and not 'V'.

Thanks a lot :)

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## Doubt on P.I of $e^{ax} V$

 Quote by iVenky I still don't get how you take that $e^ax$ out of $D^n(e^{ax}V)$ and not 'V'.
Just by applying D(eaxu)= eax Du + aeaxu= eax (D+a)u
That took the eax out but not the u. Same with the higher powers.

 Quote by haruspex Just by applying D(eaxu)= eax Du + aeaxu= eax (D+a)u That took the eax out but not the u. Same with the higher powers.
Sorry. I overlooked it. I asked a stupid question.

Thanks a lot :)

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