Toronto Math Forum
MAT2442018S => MAT244Tests => Final Exam => Topic started by: Victor Ivrii 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*}

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=1i
\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)=\lncot(2t)+csc(2t)+c_4
\end{equation}
Finally we get the required solution according to (6), (9), (10) and (11)

since the above solution is incomplete after μ(2)
I am attaching a copy of my solution...

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

The integration of 1/(cos t sint) is ln(sint)  ln(cost) + c
and hence you get ln(tant) + c

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
$\lncsc(2t)+cot(2t)+c$

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
$\lncsc(2x)+cot(2x)=\lntan(x)$

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