Problem 2 (morning)

[Victor Ivrii]

Consider equation
\begin{equation}
y'''-2y''+4y'-8y=15\cos (t).
\label{2-1}
\end{equation}
(a) Write a differential equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

(b) Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

(c) Find the general solution of (\ref{2-1}).

[Yiheng Bian]

No double-dipping
(a):
$$
W=ce^{-\int{p(t)}}dt\\
W=ce^{-\int{-2}dt}\\
W=ce^{2t}
$$



(b):
We can get:
$$
r^3-2r^2+4r-8=0\\
(r-2)(r^2+4)=0\\
r_1=2,r_2=2i,r_3=-2i
$$
Therefore:
$$
y=c_1e^2t+c_2sin2t+c_3cos2t
$$
$$
\begin{vmatrix}
e^2t & sin2t & cos2t \\
2e^2t & 2cos2t & -2sin2t \\
4e^2t & -4sin2t & -4cos2t
\end{vmatrix}=e^{2t}({-8(cos2t)^2}-8(sin2t)^2)-sin2t(-8e^{2t}cos2t+8e^{2t}sin2t)+cos2t(-8e^{2t}sin2t-8e^{2t}cos2t)=-8e^{2t}-8e^{2t}=-16e^{2t}
$$
compare with (a) we get
$$
c=-16
$$



(c):
Let
$$
y=Acost+Bsint
$$
Therefore:
$$
y'=-Asint+Bcost\\
y''=-Acost-Bsint\\
y'''=Asint-Bcost
$$
Next We take these into equation and get:
$$
(-3A-6B)sint+(3B-6A)cost=15cost\\
-3A-6B=0:3B-6A=15\\
A=-2,B=1
$$
Finally:
$$
y=c_1e^2t+c_2sin2t+c_3cos2t-2cost+sint
$$

[Lan Cheng]

a) We can see the coefficient of y” is -2. Therefore, $W=Ce^{2t}.$

b) First, we should solve for $y'''-2y"+4y'-8y=0.$

Let $r^{3}-2r^{2}+4r-8=0.$

$r_{1}=2,r_{2}=2i,r_{3}=-2i.$

thus, $y_{c}(t)=C_{1}e^{2t}+cos(2t)+sin(2t).$

$W=det\begin{bmatrix}e^{2t} & cos(2t) & sin(2t)\\
2e^{2t} & -2sin(2t) & 2cos(2t)\\
4e^{2t} & -4cos(2t) & -4sin(2t)
\end{bmatrix}=16e^{2t}.$

Therefore, the answer in b) respond that the differential equation in a) is correct.

c) Let $y_{p}(t)=Acos(t)+Bsin(t).$

$y'=-Asin(t)+Bcos(t).y"=-Acos(t)-Bsin(t).y'''=Asin(t)-Bcos(t).$

Plug in, we get $\begin{cases}
-3A-6B=0 & -6A+3B=15.\end{cases}\begin{cases}
A=-2 & B=1.\end{cases}$

Therefore, $y(t)=C_{1}e^{2t}+cos(2t)+sin(2t)-2cos(t)+sin(t).$ Two constants missing

[Ruojing Chen]

$$y'''-2y''+4y'-8y=15cost$$
a) $$p(t)=-2$$
$$w=ce^{-\int p(t)}=ce^{\int2dt}=ce^{2t}$$
b)$$y'''-2y''+4y'-8y=0$$
$$r^3-2r^2+4r-8=0$$
$$(r-2)(r^2+4)=0$$
$$\therefore r=2,\pm2i$$
$$y_c(t)=c_1e^{2t}+c_2cos2t+c_3sin2t$$
$$w(y_1,y_2,y_3)(t)=\left[\begin{matrix}
e^{2t} & cos2t & sin2t \\
2e^{2t} & -2sin2t & 2cos2t \\
4e^{2t} & -4cos2t & -4sin2t
\end{matrix}\right]=16e^{2t}$$
$$\therefore w=ce^{2t}=16e^{2t}, c=16$$
c)Assume $$y_p(t)=Acost+Bsint$$
$$y'=-Asint+Bcost$$$$y''=-Acost-Bsint$$$$y'''=Asint-Bcost$$
By plugging in y,y',y'',y''' into the function
$$Asint-Bcost+2Acost+2Bsint-4Asint+4Bcost-8Acost-8Bsint=15cost$$
$$\begin{equation}
\left\{
             \begin{array}{lr}
             A+2B-4A-8B=0, &  \\
            -B+2A+4B-8A=15
           
             \end{array}
\right.
\end{equation} $$
$$\Rightarrow

\begin{equation}
\left\{
             \begin{array}{lr}
             A+2B=0, &  \\
            B-2A=5
           
             \end{array}
\right.
\end{equation}$$
$$\therefore \begin{equation}
\left\{
             \begin{array}{lr}
             A=-2, & \\
             B=1, &
             \end{array}
\right.
\end{equation}$$

$$y(t)=c_1e^{2t}+c_2cos2t+c_3sin2t-2cost+sint$$

OK, except LaTeX sucks:
 
2)  "operators" should be escaped: \cos, \sin, \tan, \ln

$$
\boxed{y= -2\cos(t)+\sin(t)   + C_1e^{2t} +C_2\cos(2t) +C_3\sin(2t).}
$$

[Yiran Wang]

This is the solution

[Yiran Wang]

 

[Yiran Wang]

 

[Mingdi Xie]

I think Lan Cheng miss out $c_2$ and $c_3$ term in front of $\cos2t$ and $\sin2t$