MAT244--2018F > Final Exam

FE-P2

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Qi Cui:

--- Quote from: Zhiya Lou on December 14, 2018, 09:28:44 AM ---I think Cui calculated the homogeneous solution wrong:
$(r-1)(r-1)(r-1) = r^3 -3r^2+3r+1$

Actually:
$r^3 -3r^2+4r-2= (r-1)(r^2-2r-2)$ = 0
$r=1$ or $r=1-i, 1+i$


So, Homogeneous solution is
$y= c_1e^t + c_2e^t\cos(t) + c_3e^t\sin(t)$

Non homogenous part for $10e^t$
We should assume $Y= Ate^t $
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$

$Y’’’-3Y’’+4Y’-2Y =(A-3A+4A-2A)te^t+(3A-6A+4A-2A)e^t =-Ae^t =10e^t$
$A =-10, Y=-10te^t$

--- End quote ---
fixed,thx!

Xu Zihan:

--- Quote from: Qi Cui on December 14, 2018, 08:52:58 AM ---$$Homo: r^3 -3r^2+4r-2=0$$
$$r_1=1r_2=1+ir_3=1-i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
Non-Homo:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t+10e^{-t} +20cost$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$y_{p1}(t)=Ate^t$$
$$y^{'}=Ate^t+ Ae^t $$
$$y^{''} = Ate^t+ 2Ae^t $$
$$y^{'}= Ate^t+ 3Ae^t  $$
substitute above into the function:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$we have: A=-10$$
$$\therefore y_{p1}(t)=-10e^t $$
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
$$y^{'''}-3y{''}+4y{'}-2y=10e^{-t}$$
$$y_{p2}(t)=Ae^{-t} $$
$$y^{'}= -Ae^{-t}$$
$$y^{''}= Ae^{-t} $$
$$y^{'''}= -Ae^{-t} $$
substitute above into the function:
$$y^{'''}-3y{''}+4y{'}-2y=10e^t$$
we have A=1
$$\therefore y_{p2}(t)=e^{-t}  $$
$$y^{'''}-3y{''}+4y{'}-2y=20cost$$
$$y_{p3}(t)=Acost+Bsint $$
$$y^{'}= -Asint+Bcost$$
$$y^{''}= -Acost-Bsint $$
$$y^{'''}= Asint-Bcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint-10e^t+ e^{-t}  + 2cost+6sint $$

--- End quote ---

I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}-e^{-t}+2cost-6sint$ which is same as Jingyi's

Tzu-Ching Yen:
I also think one of the particular solution should be $10te^{t}$ $L'(1) = 1$, by $AL'(1) = 10$ $A = 10$. In your solution the $e^{t}$ part had an extra $-2A$.

Tzu-Ching Yen:

--- Quote from: Xu Zihan on December 14, 2018, 10:34:57 AM ---I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}-e^{-t}+2cost-6sint$ which is same as Jingyi's

--- End quote ---
You mean $2\cos(t) + 6\sin(t)$ right?

Xu Zihan:

--- Quote from: Tzu-Ching Yen on December 16, 2018, 11:07:35 AM ---
--- Quote from: Xu Zihan on December 14, 2018, 10:34:57 AM ---I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}-e^{-t}+2cost-6sint$ which is same as Jingyi's

--- End quote ---
You mean $2\cos(t) + 6\sin(t)$ right?

--- End quote ---
yes, sorry for the typo

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