# How to solve this second order inhomogenous differential equation? 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.)$$\quad=\frac{1}{D^2+2D+1}e^{-t}$$

$$=\frac{1}{(D+1)^2}e^{-t}$$

$$=\frac{t^2}{2}e^{-t}$$

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

• 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? – Counter Boosting Jun 17 at 5:57
• 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 $\frac{1}{2}$. – nmasanta Jun 17 at 6:01
• 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 $C=\frac{1}{2}$. – nmasanta Jun 17 at 6:10
• Yes I understand now, thank you for the clarification! – Counter Boosting Jun 17 at 6:11
• You are most welcome. – nmasanta Jun 17 at 6:11