FE-P2

[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 10te^t-e^-t+2𝑐𝑜𝑠𝑡+6𝑠𝑖𝑛𝑡

[Qi Cui]

Non homo part should be 10te^t-e^-t+2𝑐𝑜𝑠𝑡+6𝑠𝑖𝑛𝑡

How come?

[Qi Cui]

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 $
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$

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

fixed,thx!

[Xu Zihan]

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

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]

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

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

[Xu Zihan]

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

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

yes, sorry for the typo

[Qinger Zhang]

$$Homo: r^3 -3r^2+4r-2=0$$
$$r_1=1r_2=1+ir_3=1-i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
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:
$$y^{'''}-3y{''}+4y{'}-2y=10e^{-t}$$
 I think A should be -1 here.
$$\therefore y_{p2}(t)=-e^{-t}  $$
$$y^{'''}-3y{''}+4y{'}-2y=20cost$$
$$y_{p3}(t)=Acost+Bsint $$
$$y^{'}= -Asint+Bcost$$
$$y^{''}= -Acost-Bsint $$
$$y^{'''}= Asint-Bcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint-10e^t- e^{-t}  + 2cost+6sint $$

$$ y_{p2}(t)=-e^{-t}  $$ because W2 is negative

[Victor Ivrii]

Writing characteristic equation: $L(k):= k^3-3k^2+4k-2=0$. Obviously, one root $k_1=1$; then $k_2+k_3= 2$, $k_1k_2=2$ and they satisfy $k^2-2k+2=0\implies k_{1,2}= 1\pm i$. Then
\begin{equation}
y^*= C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)
\label{eq-2-1}
\end{equation}
is a general solution to the homogeneous equation.

Solving inhomogeneous equations with RHE $f_1=10e^{t}$, $f_2=10 e^{-t}$, $f_3=\cos(t)$:
\begin{align*}
&y_{p1}= At e^t,\\
&y_{p2}=Be^{-t},\\
&y_{p3}= C\cos(t)+D\sin(t).
\end{align*}
Here $A L'(k)|_{k=1} = A(3k^2-6k+4)|_{k=1}=10\implies A=10$,
$BL(-1) =-10 B=10\implies B=-1$ and
$$
(C+iD)L(i)= (A+iB) (1+3i)=20\implies
C+iD= \frac{20}{1+3i}=\frac{20(1-3i)}{10}=
2-6i\implies C=2, D=6.
$$
Then
\begin{align*}
&y_{p1}= 10t e^t,\\
&y_{p2}=-e^{-t},\\
&y_{p3}= 2\cos(t)+6\sin(t).
\end{align*}
Finally
$$
y= \underbracket{10t e^{t}}_{y_{p1}}\underbracket{-e^{-t}}_{y_{p2}}+ \underbracket{2\cos(t)+6\sin(t)}_{y_{p3}} + \underbracket{C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)}_{y^*}.
$$