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I was wondering how I should go about solving this differential equation. So far I have found that the characteristic equation has a double repeated root of $-1$, so the form will be $Ae^{-t} + Bte^{-t}$ . And I have set $u_p = Ct^2e^{-t}$ however I get stuck now when I'm trying to find $C$.


$$u''+2u'+u=e^{-t}\implies (D^2+2D+1)u=e^{-t}$$where $D\equiv\frac{d}{dt}$

The roots of the trial solution be $m=-1,-1$

So the complementary function (C.F.) is $\quad (A+Bt)e^{-t}\quad$ where $A,B$ are arbitrary constants.

Particular integral (P.I.)$\quad=\frac{1}{D^2+2D+1}e^{-t}$



So the general solution of the given differential equation is

$$u(t)=\text{C.F.}+\text{P.I.}=(A+Bt)e^{-t}+\frac{t^2}{2}e^{-t}$$ where $A,B$ are arbitrary constants.

Consider a differential equation of the form $f(D)y=X$

If $X=e^{ax}$, then

$1.$ P.I.$\quad = \frac{1}{f(D)}e^{ax}=\frac{e^{ax}}{f(a)}$, if $f(a)\neq 0$

$2.$ P.I.$\quad =\frac{1}{(D-a)^n}e^{ax}=\frac{x^n}{n!}e^{ax}$

  • $\begingroup$ I don't understand where you get these particular integrals from. Because I was taught that if X = e^(ax), then P.I = Ae^(ax) but if it's already present in the equation, then multiply by x until it's not present anymore, so in this case I used P.I = Ax^2e^(Ax). Is this incorrect? $\endgroup$ – Counter Boosting Jun 17 at 5:57
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    $\begingroup$ Yes my way is absolutely correct. you can check the answer by differentiating the final $u(t)$. In your described formula, value of your $A$ is nothing but my $\frac{1}{2}$. $\endgroup$ – nmasanta Jun 17 at 6:01
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    $\begingroup$ Since your $u_p$ is also a solution of the given differential equation, so if you put $u_p = Ct^2e^{-t}$ in $u''+2u'+u=e^{-t} \text{i.e., in }u_p''+2u_p'+u_p=e^{-t}$, you will definitely get that $C=\frac{1}{2}$. $\endgroup$ – nmasanta Jun 17 at 6:10
  • $\begingroup$ Yes I understand now, thank you for the clarification! $\endgroup$ – Counter Boosting Jun 17 at 6:11
  • $\begingroup$ You are most welcome. $\endgroup$ – nmasanta Jun 17 at 6:11

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