I can't seem to figure out where I am going wrong in my steps. I checked the answer and it is different.

The question is:

$$y'' + 2y' - 3y = 3te^t$$

The roots are: -3,1. Thus the general solution is: $$y=C_1e^{-3t} + C_2e^t$$

The particular solution i am going with is: $$y_p = Ate^t$$ $$y'_p = Ae^t + Ate^t$$ $$y''_p = 2Ae^t + Ate^t$$

Therefore: $$2Ae^t + Ate^t + 2Ae^t + 2Ate^t - 3Ate^t = 3te^t$$

$$4Ae^t = 3te^t$$

Then solving for A: $$4A=3$$ $$A=3/4$$

Thus, $y_p = \frac{3}{4}(e^t + te^t)$

Then the general solution would be: $y = C_1e^{-3t} + C_2e^t + \frac{3}{4}(e^t + te^t)$

Any guidance with my mistake would be greatly appreciated. As an aside, what does it mean when a question asks to use the stability result to determine they will have a globally stable solution of the above question. Thank you.

  • $\begingroup$ Be carefull with your variable sometimes you use t sometimes you use x... $\endgroup$ Jun 21, 2018 at 18:08
  • $\begingroup$ Thank you. I must've done it accidentally. $\endgroup$
    – fynmnx
    Jun 21, 2018 at 18:14
  • $\begingroup$ yw Safder...I know it's just a typo.. $\endgroup$ Jun 21, 2018 at 18:27

2 Answers 2


Your particular solution is not correct. Note that $4Ae^t = 3te^t$ does not imply that $A=3/4$. Moreover, what is $C$?

In the characteristic polynomial, the multiplicity of the root ${\bf 1}$ is $m=1$ . Therefore, since $f(t)=te^{{\bf1}\cdot t}$, it follows that the particular solution should have the form $$y_p=t^m(At+B)e^{{\bf1}\cdot t}=(At^2+Bt)e^t$$ where $A$ and $B$ are real constants to be found. Can you take it from here?

  • $\begingroup$ Yes I can. I had a feeling it was the particular solution. Would you happen to know how to check for global stability? Thank you. $\endgroup$
    – fynmnx
    Jun 21, 2018 at 18:16
  • $\begingroup$ What is the definition of "global stability" in your book/course? $\endgroup$
    – Robert Z
    Jun 21, 2018 at 18:18
  • $\begingroup$ This what the notes say, i'm a little confused by it. Consider a homogenous equation equation $y'' + ay' + by = f(t)$. Then the solution is in the form $y=y_h + y_s$, where $y_h, y_s$ are the homogenous and non-homogenous solutions respectively. $t \rightarrow \infty$, $ y = y_s$ if and only if a>0, b>0. Thus y_s is only globally stable if and only if a>0 and b>0 if and only if the real parts of the roots of the characteristic equation are both negative. Would you be able clarify the requirements for stability for me? $\endgroup$
    – fynmnx
    Jun 21, 2018 at 18:32
  • $\begingroup$ In your opinion, in this case, is $y$ globally stable? $\endgroup$
    – Robert Z
    Jun 21, 2018 at 18:35
  • 1
    $\begingroup$ I agree with you. $\endgroup$
    – Robert Z
    Jun 21, 2018 at 18:58

The particular solution should be: $$y_p=(At^2+Bt)e^t \implies y'_p=(At^2+(2A+B)t+B)e^t$$

Better approach

Substitute $y=ze^t$ The equation becomes $$z''+4z'=3t$$ Use the variation of constant now $(y_p=at^2+bt)$..It looks easier..

Another approach

$$y'' + 2y' - 3y = 3te^t$$ $$(y''-y')+3(y'-y)=3te^t$$ $$(y'e^{-t})'+3(ye^{-t})'=3t$$ $$(y'e^{-t})+3(ye^{-t})=\frac 32t^2+K_1$$ $$(y'e^{3t})+3(ye^{3t})=(\frac 32t^2+K_1)e^{4t}$$ $$(ye^{3t})'=(\frac 32t^2+K_1)e^{4t}$$ $$ye^{3t}=\int (\frac 32t^2+K_1)e^{4t}dt$$ $$y(t)=K_2e^{-3t}+K_1e^t+\frac 32e^{-3t}\int t^2e^{4t}dt$$ Finally $$\boxed{y(t)=K_2e^{-3t}+K_1e^t+\frac 38e^tt^2-\frac 3{16}e^tt}$$ where $$ \begin{cases} \displaystyle y_p=\frac 38e^tt(t-\frac 1{2}) \\ \displaystyle y_h=K_2e^{-3t}+K_1e^t \end{cases} $$


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