MAT244--2018F > Final Exam

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Victor Ivrii:
Typed solutions only. No uploads

Find the general solution by method of the undetermined coefficients:
\begin{equation*}
y'''-3y''+4y'- 2y= 20\cosh(t)+20\cos(t);
\end{equation*}
Hint: All roots are integers (or complex integers).

Qi Cui:
$$Homo: r^3 -3r^2+4r-2=0$$
$$r_1=1r_2=1+ir_3=1-i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
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
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint-10e^t+ e^{-t}  + 2cost+6sint $$

Zhiya Lou:
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 $
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$

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

Jingyi Wang:
Non homo part should be 10tet-e-t+2𝑐𝑜𝑠𝑡+6𝑠𝑖𝑛𝑡

Qi Cui:

--- Quote from: Jingyi Wang on December 14, 2018, 10:06:29 AM ---Non homo part should be 10tet-e-t+2𝑐𝑜𝑠𝑡+6𝑠𝑖𝑛𝑡

--- End quote ---
How come?

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