Toronto Math Forum
MAT2442018F => MAT244Tests => Final Exam => Topic started by: Victor Ivrii on December 14, 2018, 07:52:36 AM

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).

$$Homo: r^3 3r^2+4r2=0$$
$$r_1=1r_2=1+ir_3=1i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
NonHomo:
$$y^{'''}3y{''}+4y{'}2y=10e^t+10e^{t} +20cost$$
$$y^{'''}3y{''}+4y{'}2y=10e^t$$
$$y_{p1}(t)=Ate^t$$
$$y^{'}=Ate^t+ Ae^t $$
$$y^{''} = Ate^t+ 2Ae^t $$
$$y^{'}= Ate^t+ 3Ae^t $$
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
$$\therefore y_{p2}(t)=e^{t} $$
$$y^{'''}3y{''}+4y{'}2y=20cost$$
$$y_{p3}(t)=Acost+Bsint $$
$$y^{'}= Asint+Bcost$$
$$y^{''}= AcostBsint $$
$$y^{'''}= AsintBcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint10e^t+ e^{t} + 2cost+6sint $$

I think Cui calculated the homogeneous solution wrong:
$(r1)(r1)(r1) = r^3 3r^2+3r+1$
Actually:
$r^3 3r^2+4r2= (r1)(r^22r2)$ = 0
$r=1$ or $r=1i, 1+i$
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 $
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$
$Y’’’3Y’’+4Y’2Y =(A3A+4A2A)te^t+(3A6A+4A2A)e^t =Ae^t =10e^t$
$A =10, Y=10te^t$

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

Non homo part should be 10te^{t}e^{t}+2𝑐𝑜𝑠𝑡+6𝑠𝑖𝑛𝑡
How come?

I think Cui calculated the homogeneous solution wrong:
$(r1)(r1)(r1) = r^3 3r^2+3r+1$
Actually:
$r^3 3r^2+4r2= (r1)(r^22r2)$ = 0
$r=1$ or $r=1i, 1+i$
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 $
$Y’=Ate^t+Ae^t$
$Y’’=Ate^t+2Ae^t$
$Y’’’=Ate^t+3Ae^t$
$Y’’’3Y’’+4Y’2Y =(A3A+4A2A)te^t+(3A6A+4A2A)e^t =Ae^t =10e^t$
$A =10, Y=10te^t$
fixed,thx!

$$Homo: r^3 3r^2+4r2=0$$
$$r_1=1r_2=1+ir_3=1i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
NonHomo:
$$y^{'''}3y{''}+4y{'}2y=10e^t+10e^{t} +20cost$$
$$y^{'''}3y{''}+4y{'}2y=10e^t$$
$$y_{p1}(t)=Ate^t$$
$$y^{'}=Ate^t+ Ae^t $$
$$y^{''} = Ate^t+ 2Ae^t $$
$$y^{'}= Ate^t+ 3Ae^t $$
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
$$\therefore y_{p2}(t)=e^{t} $$
$$y^{'''}3y{''}+4y{'}2y=20cost$$
$$y_{p3}(t)=Acost+Bsint $$
$$y^{'}= Asint+Bcost$$
$$y^{''}= AcostBsint $$
$$y^{'''}= AsintBcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint10e^t+ e^{t} + 2cost+6sint $$
I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}e^{t}+2cost6sint$ which is same as Jingyi's

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 $2A$.

I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}e^{t}+2cost6sint$ which is same as Jingyi's
You mean $2\cos(t) + 6\sin(t)$ right?

I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}e^{t}+2cost6sint$ which is same as Jingyi's
You mean $2\cos(t) + 6\sin(t)$ right?
yes, sorry for the typo

$$Homo: r^3 3r^2+4r2=0$$
$$r_1=1r_2=1+ir_3=1i$$
$$\therefore y_c(t)=c_1e^t+c_2e^tcost+c_3e^tsint$$
NonHomo:
$$y^{'''}3y{''}+4y{'}2y=10e^t+10e^{t} +20cost$$
$$y^{'''}3y{''}+4y{'}2y=10e^t$$
$$y_{p1}(t)=Ate^t$$
$$y^{'}=Ate^t+ Ae^t $$
$$y^{''} = Ate^t+ 2Ae^t $$
$$y^{'}= Ate^t+ 3Ae^t $$
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.
$$\therefore y_{p2}(t)=e^{t} $$
$$y^{'''}3y{''}+4y{'}2y=20cost$$
$$y_{p3}(t)=Acost+Bsint $$
$$y^{'}= Asint+Bcost$$
$$y^{''}= AcostBsint $$
$$y^{'''}= AsintBcost$$
substitute above into the function:
we have A=2, B=6
$$\therefore y_{p3}(t)=2cost+6sint $$
$$\therefore y(t)= c_1e^t+c_2e^tcost+c_3e^tsint10e^t e^{t} + 2cost+6sint $$
$$ y_{p2}(t)=e^{t} $$ because W2 is negative

Writing characteristic equation: $L(k):= k^33k^2+4k2=0$. Obviously, one root $k_1=1$; then $k_2+k_3= 2$, $k_1k_2=2$ and they satisfy $k^22k+2=0\implies k_{1,2}= 1\pm i$. Then
\begin{equation}
y^*= C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)
\label{eq21}
\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 $A L'(k)_{k=1} = A(3k^26k+4)_{k=1}=10\implies A=10$,
$BL(1) =10 B=10\implies B=1$ and
$$
(C+iD)L(i)= (A+iB) (1+3i)=20\implies
C+iD= \frac{20}{1+3i}=\frac{20(13i)}{10}=
26i\implies C=2, D=6.
$$
Then
\begin{align*}
&y_{p1}= 10t e^t,\\
&y_{p2}=e^{t},\\
&y_{p3}= 2\cos(t)+6\sin(t).
\end{align*}
Finally
$$
y= \underbracket{10t e^{t}}_{y_{p1}}\underbracket{e^{t}}_{y_{p2}}+ \underbracket{2\cos(t)+6\sin(t)}_{y_{p3}} + \underbracket{C_1e^t + C_2 e^t \cos(t) + C_3 e^t\sin(t)}_{y^*}.
$$