Here's the question you clicked on:
ChmE
Method of undetermined coefficients...
Must be in the form\[ay''+by'+cy=Ct ^{m}e ^{rt}\]What does C stand for? I thought it was a constant but the equation\[y''+2y'-y=t ^{-1}e ^{t}\]Does not satisfy this method. i assumed C=1 so it would, bu tit doesn't. Explanation?
Find a particular solution to the differential equation\[4y''+11y'-3y=-2te ^{-3t}\]
I believe that the method only applies for \(m\geq0\), but I cannot recall to save my life.
\[4r ^{2}+11r-3=0\rightarrow (4r-1)(r+3)\rightarrow r= \frac{ 1 }{ 4 },-3\]
That makes sense I'm going to look that up. Thanks
@across I found it you are right
since -3 does appear as a root s=1 which is the multiplicity of the root -3
Ok @UnkleRhaukus I could use a little help on my last problem of the night. I posted the question in this stream 3rd from the top.
\[Y _{p}= t(Ae ^{-3t})\]Do I have to include the other root as well?
@Directrix @Hero Can I have a little push in the right direction
@Hero do you see a mistake I have made or is anything coming to you about what to do next?
Sorry, I'm busy doing my own work at the moment.
The answer in the back of the book is \[(\frac{ t }{ 13 }-\frac{ 8 }{ 169 })te ^{-3t}\]
@UnkleRhaukus Just in time I was about to give up. Do you mind helping me with this last question?
where are you up to
I feel like I need to plug in \[Y _{p},Y'_{p}, Y''_{p}\]into the original to solve for A. I am up to writing an equation for Yp
I feel like I have my roots and s correct
\[4y''+11y'-3y=-2te ^{-3t}\] \[4r^2+11r-3=0\]\[(4r-1)(r+3)=0\]\[r=1/4,-3\] \[y_c=ae^{t/4}+be^{-3t}\]
Oh ya I forgot, when you have two diff roots you need to use two different unknowns. What about the t^s? My teacher threatened us if we forgot that on the test
As i understand it t^s, where s is the multiplicity of your roots needs to be out in front.
\[y_p=t\left(c\cdot e^{-3t}+d\cdot t\cdot e^{-3t}\right)\]
Why do you have two terms in the ( ) if m=1 shouldn't there be 1. And I don't mean to come across like I'm questioning you, just trying to undersstand
its a second order equation there will be two solutions
So I see your Yp matches the books except I need to now solve for c and d by plugging Yp into the original equation right?.........Ok
take the first and second derivative of y_p
Right then plug it in. I can do that on my own later. Thanks for your help again. I'm calling it a night