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.]


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: $$e^{\alpha t},te^{\alpha t},t^2e^{\alpha t},...,t^{n-1}e^{\alpha t}$$

  • $\begingroup$ 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

Perhaps this will grant some insight. In a way, the left hand side can be factored like a polynomial. So just as


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


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



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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