Annihilarator technique for y'' + 4y' + 4y = 5xy^(-2x)

mammothrob

Junior Member
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Nov 12, 2005
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Im having some trouble getting the answer to this promlem...

Find the general solution to the given differetial equation. Derive your trial solution using the annihilator technique.

\(\displaystyle \[
\begin{array}{l}
y^{''} + 4y^' + 4y = 5xe^{ - 2x} \\
{\rm{so my annihilator for }}5xe^{ - 2x} {\rm{ will be,}} \\
(D^2 + 2)^2 \\
{\rm{the form my book gives me for the solution is}} \\
{\rm{cx}}^k e^{ax} \\
y_p (x) = x^m e^{ax} (A_0 + ... + A_k x^k ) \\
{\rm{where m is the multiplicity of the root = a}}{\rm{.}} \\
{\rm{I believe that my DE fits this form because }} \\
{\rm{ - 2 = a and is a repeated root}}{\rm{.}} \\
{\rm{the book claims }} \\
y(x) = e^{ - 2x} (c_1 + c_2 x + \frac{5}{6}x^3 ) \\
{\rm{I on the other hand am getting}} \\
{\rm{solutions involing trig functions}}{\rm{.}} \\
{\rm{Obviously im using a wrong form}}{\rm{.}} \\
{\rm{Can anyone walk me through this?}} \\
\end{array}\)


Any help or ideas would be great.

Thanks,

Rob


Edit:

I now know why not to use the trig form and I see why the above for fits. My concern now is why the only went to (cx) and where the heck did the (5/6) come from?

Im guessing that I have to set up some system and solve for the 5/6=A, just now sure how to do it yet.


Rob
 
mammothrob said:
Im having some trouble getting the answer to this promlem...

I now know why not to use the trig form and I see why the above for fits. My concern now is why the only went to (cx) and where the heck did the (5/6) come from?

That comes from the fact that 5/6 x^3 differentiated twice would give you '5x'

Im guessing that I have to set up some system and solve for the 5/6=A, just now sure how to do it yet.


Rob
 
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