MAT244--2018F > Final Exam

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Victor Ivrii:

Find the general solution $(x(t),y(t))$ of the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x' = x-2y + \sec(t)\, &&-\frac{\pi}{2}<t<\frac{\pi}{2},\\
&y' = x -\ \,y  \,.
\end{aligned}\right.
\end{equation*}
Hint: $\sec(t)=\frac{1}{\cos(t)}$.

Xu Zihan:
here is my solution.
$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ 1&-1 \end{bmatrix}+\begin{bmatrix} sec(t)\\0 \end{bmatrix}$
so $det\begin{bmatrix} 1-\lambda &-1 \\ 1& -1-\lambda \end{bmatrix}=0$
$\lambda=i/-i$ and eigenvector for i is $\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$
so $\begin{bmatrix} x'\\ y' \end{bmatrix}=C1e^{it}\begin{bmatrix} 1+i\\ 1 \end{bmatrix}$
$=C1(cost+isint)\binom{1+i}{1}$
$=C1\begin{bmatrix} cost-sint\\ cost \end{bmatrix}+C2i\begin{bmatrix} cost+sint\\ sint \end{bmatrix}$
For non homo:
$\varphi=\begin{bmatrix} cost-sint &cost+sint \\ cost& sint \end{bmatrix}$
Since $\varphi u'=g$
So $\begin{bmatrix} cost-sint &cost+sint \\ cost&sint \end{bmatrix}\begin{bmatrix} u1'\\ u2' \end{bmatrix}=\begin{bmatrix} 1/cost\\ 0 \end{bmatrix}$
so $u1'=-\frac{sint}{cost}$  which means u1=ln(cost)+C2
u2'=1 wihich means u2=t+C1
so the solution is $\begin{bmatrix} x'\\y' \end{bmatrix}=(t+C1)\binom{cost-sint}{cost}+(ln(cost)+C2)\binom{cost+sint}{sint}$

Doris Zhuomin Jia:
homo: det(A-I𝝀) = (1-𝝀)(-1-𝝀)+2 = 0 ,  𝝀= i, -i
when 𝛌 = i, A-iI = 1-i, -2 ~ 1-i , -2
1, -1-i     0     0
the eigenvector V is span{[2
1-i]}
eitV = (cos(t)+isin(t))V = [2cos(t)+ 2isin(t)
cos(t)-icos(t) + isin(t)+sin(t)]
[x,y] = c1[2cos(t)        +  c2[2sin(t)
cos(t)+sin(t)]          -cos(t)+sin(t)]

𝟇(t) = 2cos(t)         2sin(t)
cos(t)+sin(t)   -cos(t)+sin(t)
Let 𝜱(t)·u' = g(t)
2cos(t)            2sin(t)       sec(𝑡)
cos(t)+sin(t)   -cos(t)+sin(t)    0

2u'1cos(t) + 2u'2sin(t) = sec(t)
u'1cos(t)+u'1sin(t) -u'2cos(t)+u'2sin(t) = 0

u2 = ∫(cos(t)+sin(t))/2cos(t)) =1/2(t-ln|cos(t)|)+C1
u1 = ∫-1/2(cos(t) = -1/2sin(t)+C2

x  = [-1/2sin(t)+C2           ] [2cos(t)       2sin(t)]
y      [ 1/2(t-ln|cos(t)|)+C1][cost+sint   -cos(t)+sin(t)]

Xinyu Li:

--- Quote from: Xu Zihan on December 14, 2018, 11:26:25 AM ---here is my solution.
$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ 1&-1 \end{bmatrix}+\begin{bmatrix} sec(t)\\0 \end{bmatrix}$
so $det\begin{bmatrix} 1-\lambda &-1 \\ 1& -1-\lambda \end{bmatrix}=0$
$\lambda=i/-i$ and eigenvector for i is $\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$
so $\begin{bmatrix} x'\\ y' \end{bmatrix}=C1e^{it}\begin{bmatrix} 1+i\\ 1 \end{bmatrix}$
$=C1(cost+isint)\binom{1+i}{1}$
$=C1\begin{bmatrix} cost-sint\\ cost \end{bmatrix}+C2i\begin{bmatrix} cost+sint\\ sint \end{bmatrix}$
For non homo:
$\varphi=\begin{bmatrix} cost-sint &cost+sint \\ cost& sint \end{bmatrix}$
Since $\varphi u'=g$
So $\begin{bmatrix} cost-sint &cost+sint \\ cost&sint \end{bmatrix}\begin{bmatrix} u1'\\ u2' \end{bmatrix}=\begin{bmatrix} 1/cost\\ 0 \end{bmatrix}$
so $u1'=-\frac{sint}{cost}$  which means u1=ln(cost)+C2
u2'=1 wihich means u2=t+C1
so the solution is $\begin{bmatrix} x'\\y' \end{bmatrix}=(t+C1)\binom{cost-sint}{cost}+(ln(cost)+C2)\binom{cost+sint}{sint}$

--- End quote ---

Hi, is that should be 𝑑𝑒𝑡[1−𝜆 -2]=0 ???

Qinger Zhang:

--- Quote from: Xu Zihan on December 14, 2018, 11:26:25 AM ---$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ 1&-1 \end{bmatrix}+\begin{bmatrix} sec(t)\\0 \end{bmatrix}$
so $det\begin{bmatrix} 1-\lambda &-1 \\ 1& -1-\lambda \end{bmatrix}=0$
$\lambda=i/-i$ and eigenvector for i is $\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$

--- End quote ---
$\lambda=i$, eigenvector is $\begin{pmatrix} -1\\ i-1 \end{pmatrix}$