I have a doubt in finding out the Particular Integral of e(adsbygoogle = window.adsbygoogle || []).push({}); ^{ax}V, 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^{ax}u)= e^{ax}Du + ae^{ax}u= e^{ax}(D+a)u

Similarly you can do for D^{2},D^{3}... and so on

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

[itex]\frac{1}{f(D)}(e^{ax}V)[/itex]=[itex]e^{ax} \frac{1}{f(D+a)}[/itex]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

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

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