Author Topic: FE-P4  (Read 3904 times)

Victor Ivrii

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FE-P4
« on: December 14, 2018, 07:55:46 AM »

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

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Re: FE-P4
« Reply #1 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 $\begin{bmatrix} x'\\y' \end{bmatrix}=(t+C1)\binom{cost-sint}{cost}+(ln(cost)+C2)\binom{cost+sint}{sint}$

« Last Edit: December 14, 2018, 12:15:23 PM by Xu Zihan »

Doris Zhuomin Jia

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Re: FE-P4
« Reply #2 on: December 14, 2018, 11:30:19 AM »
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)]
« Last Edit: December 14, 2018, 11:54:23 AM by Doris Zhuomin Jia »

Xinyu Li

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Re: FE-P4
« Reply #3 on: December 17, 2018, 05:52:13 PM »
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 $\begin{bmatrix} x'\\y' \end{bmatrix}=(t+C1)\binom{cost-sint}{cost}+(ln(cost)+C2)\binom{cost+sint}{sint}$

Hi, is that should be 𝑑𝑒𝑡[1−𝜆 -2]=0

Qinger Zhang

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Re: FE-P4
« Reply #4 on: December 17, 2018, 08:26:55 PM »
$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ 1&-1 \end{bmatrix}+\begin{bmatrix} sec(t)\\0 \end{bmatrix}$
so $det\begin{bmatrix} 1-\lambda &-1 \\ 1& -1-\lambda \end{bmatrix}=0$
$\lambda=i/-i$ and eigenvector for i is $\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$
$\lambda=i$, eigenvector is $\begin{pmatrix} -1\\ i-1 \end{pmatrix}$
« Last Edit: December 17, 2018, 08:29:48 PM by Qinger Zhang »

Victor Ivrii

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FE-P4 official
« Reply #5 on: December 20, 2018, 02:01:42 AM »
$\renewcommand{\Re}{\operatorname{Re}}$
Characteristic equation:
$$\left|\begin{matrix} 1-k & -2\\ 1 & -1-k \end{matrix}\right|= k^2+1=0\implies k_{1,2}=\pm i.$$
Finding eigenvector corresponding to $k_1=i$
$$\begin{pmatrix} 1-i & -2\\ 1 & -1-i \end{pmatrix}\begin{pmatrix} \alpha\\ \beta \end{pmatrix}=0\implies \alpha =(1+i)\beta\implies \mathbf{e}_1 =\begin{pmatrix} 1+i\\ 1 \end{pmatrix}$$
and $\mathbf{e}_2$ is complex conjugate. Then the general solution to the homogeneous equation is
\begin{align}
\begin{pmatrix}x\\y\end{pmatrix}=&\Re \Bigl[(C_1+iC_2) (\cos (t)+i\sin(t))\begin{pmatrix} 1+i\\ 1 \end{pmatrix}\Bigr]\notag\implies \\
&x=  C_1 (\cos(t)-\sin(t))+ C_2( -\cos(t)-\sin(t) )\\
&y=  C_1\cos(t)    -C_2\sin(t)  \,.
\label{eq-4-1}
\end{align}
We are looking for solution to inhomogeneous equation in the same form albeit with variable $C_1,C_2$. Then
\begin{align*}
&\left\{\begin{aligned}
&C'_1 (\cos(t)-\sin(t))+ C'_2( -\cos(t)-\sin(t) )=\sec(t),\\
\end{aligned}\right.\implies\\[4pt]
&\left\{\begin{aligned}
&C'_1\sin(t)+ C'_2\cos(t)=-\sec(t),\\
&C'_1\cos(t)      -\ C'_2\sin(t) =0
\end{aligned}\right.\implies\\
&\left\{\begin{aligned}
&C'_1= -\sec(t)\sin(t)=-\tan(t)\implies C_1=\ln (\cos(t))+c_1\\
&C'_2=-\sec(t)\cos(t)=-1\implies C_2=-t +c_2\,.
\end{aligned}\right.
\end{align*}
Finally
\begin{align*}
&x=  [\ln (\cos(t))+c_1] (\cos(t)-\sin(t))+ [t -c_2]( \cos(t)+\sin(t) )\,\\
&y=[\ln (\cos(t))+c_1]\cos(t)   +[t-c_2]\sin(t) \,.
\end{align*}