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I am trying to get the particular solution to the equation -

$y''' + 4y'' + 5y' + 2y = e^{-t} $

We are taught the method of undetermined coefficients to solve such equations. However, one of the solutions of the homogenous equation is of the form of the particular equation (so when I substitute it, I get LHS$ = 0$, while RHS is not $0$ ).

Please give me a hint on how should I proceed to get the particular solution.

[This question is a part of the tricky questions set given to us. The actual differential equation is a bit more complicated, but I have reduced it to the point I have got stuck in.]

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Since $P(D)y=(D+1)^3(D+2)y=0$ so your particular solution is $$y_p=At^2e^{-t}$$ where $A$ is an unknown constant. Note that the differential operator $D^n$ annihilates each of the functions: $$1,t,t^2,...,t^{n-1}$$ and differential operator $(D-\alpha)^n$ annihilates each of the following functions: $$e^{\alpha t},te^{\alpha t},t^2e^{\alpha t},...,t^{n-1}e^{\alpha t}$$

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  • $\begingroup$ Thanks for help. I think , the equ is $P(D)y=(D+1)^2(D+2)y=0$. Will the solution still be one you gave above? $\endgroup$ Sep 3 '12 at 9:12
  • $\begingroup$ In fact, $P(D)y=(D+1)^2(D+2)y=e^{-t}$, but if you take $(D+1)$ one more time of both sides; you get mine. :-) $\endgroup$
    – Mikasa
    Sep 3 '12 at 9:16
  • $\begingroup$ Ok. Thanks. Didn't think about that! :) $\endgroup$ Sep 3 '12 at 9:19
  • $\begingroup$ Always a good teacher! +1 $\endgroup$
    – amWhy
    Mar 14 '13 at 2:05
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Perhaps this will grant some insight. In a way, the left hand side can be factored like a polynomial. So just as

$s^3+4s^2+5s+2=(s+1)(s^2+3s+2)$

if you let $u=y''+3y'+2y$, this equation can be rewritten as

$y'''+4y''+5y'+2y=(y''+3y'+2y)'+y''+3y'+2y=u'+u=e^{-t}$

So let's solve this for $u$ using integrating factors.

$u'e^t+ue^t=(ue^t)'=1$

$ue^t=t+C,u=y''+3y'+2y=te^{-t}+Ce^{-t}$

If, as you put it, this solution of the homogenous equation had not been of the form of the particular equation, the right side before integration would have been of the form $e^{nt}$. Instead, we've had a $te^{-t}$ term introduced.

Something similar will occur when we make the substitution $v=y'+2y$, where we will again have $e^t$ as an integrating factor.

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