# Undetermined Coefficients for solving non homogeneous equation

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.

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

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?

• 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. Jun 21, 2018 at 18:16
• What is the definition of "global stability" in your book/course? Jun 21, 2018 at 18:18
• 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? Jun 21, 2018 at 18:32
• In your opinion, in this case, is $y$ globally stable? Jun 21, 2018 at 18:35
• I agree with you. 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 $$\begin{cases} \displaystyle y_p=\frac 38e^tt(t-\frac 1{2}) \\ \displaystyle y_h=K_2e^{-3t}+K_1e^t \end{cases}$$