Author Topic: FE-P4  (Read 767 times)

Victor Ivrii

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FE-P4
« on: April 11, 2018, 08:44:51 PM »
Find the general solution of the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = \hphantom{-}x +   y +\, \frac{e^{t}}{\cos(t)}\, ,\\
&y'_t = - x + y +\, \frac{e^{t}}{\sin(t)}\,.
\end{aligned}\right.
\end{equation*}

Tim Mengzhe Geng

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Re: FE-P4
« Reply #1 on: April 12, 2018, 12:16:56 AM »
First we start with the homogeneous system. The coefficient matrix is
\begin{equation}
  A={
\left[\begin{array}{ccc}
1 & 1 \\
-1 & 1
\end{array}
\right ]},
\end{equation}
The eigenvalues are
\begin{equation}
    \lambda_1=1+i
\end{equation}
\begin{equation}
    \lambda_2=1-i
\end{equation}
The eigenvector corresponding to $\lambda_1$ is
\begin{equation}
\xi^(1)=(1,i)^{T}
\end{equation}
Then we can have the solution to the homogeneous system
\begin{equation}
(x,y)^T=c_1\cdot (e^t\cos(t), -e^t\sin(t))^T+c_2\cdot(e^t\sin(t), e^t\cos(t))^T
\end{equation}
We can get a fundamental matrix from here
\begin{equation}
  \Psi={
\left[\begin{array}{ccc}
e^t\cos(t) & -e^t\sin(t) \\
e^t\sin(t) & e^t\cos(t)
\end{array}
\right ]},
\end{equation}
Note that we have
\begin{equation}
    g(t)=(\frac{e^t}{\cos(t)},\frac{e^t}{\sin(t)})^T
\end{equation}
And suppose
\begin{equation}
  \Psi \mu^\prime=g(t)
\end{equation}
We shoud have for the nonhomogeneous system
\begin{equation}
(x,y)^T=\Psi \mu
\end{equation}
This is the method of Variation of Parameters. By plugging in and integration, we have
\begin{equation}
    \mu_1(t)=c_3
\end{equation}
\begin{equation}
    \mu_2(t)=-\ln|cot(2t)+csc(2t)|+c_4
\end{equation}
Finally we get the required solution according to (6), (9), (10) and (11)

Syed Hasnain

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Re: FE-P4
« Reply #2 on: April 12, 2018, 01:37:56 PM »
since the above solution is incomplete after μ(2)
I am attaching a copy of my solution...
« Last Edit: April 13, 2018, 07:47:07 AM by Syed_Hasnain »

Tim Mengzhe Geng

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Re: FE-P4
« Reply #3 on: April 12, 2018, 03:32:57 PM »
since the above solution is incomplete after μ(2)
I am attaching a copy of my solution...
I think your integration to $\mu_2(t)$may not be correct

Syed Hasnain

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Re: FE-P4
« Reply #4 on: April 13, 2018, 02:37:46 AM »
The integration of 1/(cos t sint) is ln(sint) - ln(cost) + c
and hence you get ln(tant) + c

Tim Mengzhe Geng

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Re: FE-P4
« Reply #5 on: April 14, 2018, 02:42:24 AM »
The integration of 1/(cos t sint) is ln(sint) - ln(cost) + c
and hence you get ln(tant) + c
$1/(costsint)=2/sin(2t)=2csc(2t)$
And by the integration of 2csc(2t),
we get
$-\ln|csc(2t)+cot(2t)|+c$
« Last Edit: April 14, 2018, 02:45:58 AM by Tim Mengzhe GENG »

Tim Mengzhe Geng

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Re: FE-P4
« Reply #6 on: April 14, 2018, 03:01:14 AM »
The integration of 1/(cos t sint) is ln(sint) - ln(cost) + c
and hence you get ln(tant) + c
I'm so sorry that you're also correct  and the two result are the same since
$\tan\frac{x}{2}=\frac{\sin(x)}{1+cos(x)}$
and therefore
$-\ln|csc(2x)+cot(2x)|=\ln|tan(x)|$

Victor Ivrii

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Re: FE-P4
« Reply #7 on: April 15, 2018, 03:26:19 AM »
Solution is complete, but the answer must be written.

Also a simpler form $\ln(\tan(t))$ is preferable.

Finally \sin, \cos, and so on must be escaped by \ to provide upright letters and a proper spacing