Doubt on P.I of [itex]e^{ax} V [/itex]

[iVenky]

I have a doubt in finding out the Particular Integral of e^axV, 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(e^axu)= e^ax Du + ae^axu= e^ax (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 e^ax and V. This could have resulted in an entirely different answer. I couldn't get that

Thanks a lot

 

[haruspex]

If u is a function of x, then
D(e^axu)= e^ax Du + ae^axu= e^ax (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

 

[iVenky]

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 :)

 

[haruspex]

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(e^axu)= e^ax Du + ae^axu= e^ax (D+a)u
That took the e^ax out but not the u. Same with the higher powers.

 

[iVenky]

Just by applying D(e^axu)= e^ax Du + ae^axu= e^ax (D+a)u
That took the e^ax out but not the u. Same with the higher powers.

Sorry. I overlooked it. I asked a stupid question.


Thanks a lot :)