Toronto Math Forum

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
$-\ln|csc(2x)+cot(2x)|=\ln|tan(x)|$

There is a  mistake in step (9)
the first term should be 2y(x^2)cosy instead of 2x(y^2)cosy

Thanks and I've changed it

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

For Part(a), Note that for stationary points, we should have
\begin{equation}
    x^2+y^2-1=0
\end{equation}
And at the same time
\begin{equation}
    -2xy=0
\end{equation}
Therefore there're totally four stationary points. They are
\begin{equation}
    (x,y)=(1,0), (-1,0), (0,1) or (0,-1).
\end{equation}
\begin{equation}
   J={
\left[\begin{array}{ccc}
2x & 2y \\
-2y & -2x
\end{array}
\right ]},
\end{equation}
At point (1,0),
\begin{equation}
   J[1,0]={
\left[\begin{array}{ccc}
2 & 0 \\
0 & -2
\end{array}
\right ]},
\end{equation}
At point (-1,0),
\begin{equation}
   J[-1,0]={
\left[\begin{array}{ccc}
-2 & 0 \\
0 & 2
\end{array}
\right ]},
\end{equation}
At point (0,1),
\begin{equation}
   J[0,1]={
\left[\begin{array}{ccc}
0 & 2 \\
-2 & 0
\end{array}
\right ]},
\end{equation}
At point (0,-1),
\begin{equation}
   J[0,-1]={
\left[\begin{array}{ccc}
0 & -2 \\
2 & 0
\end{array}
\right ]},
\end{equation}

For part(b), we have
\begin{equation}
    (x^2+y^2-1)dx+2xydy=0
\end{equation}
Note that
\begin{equation}
    M_y=N_x=2y
\end{equation}
The equation is exact.
By integration
\begin{equation}
    H=\frac{1}{3}x^3+xy^2-x+h^\prime(y)
\end{equation}
\begin{equation}
    h^\prime(y)=0
\end{equation}
We choose
\begin{equation}
    h(y)=0
\end{equation}
In this way,
\begin{equation}
    H(x,y)=\frac{1}{3}x^3+xy^2-x=C
\end{equation}
I will post solution to other parts later if no one else follows.

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)

Small Error: $W_2(x)$ should be $2e^{4x}$.

For this case I don't think so since when we expand the matrix, we have to times $(-1)^{i+j}$

First we find the solution for the homogeneous system
\begin{equation}
    y^{(3)}-6y^{(2)}+11y^{(1)}-6y=0
\end{equation}
The corresponding characteristic equation is
\begin{equation}
    r^3-6r^2+11r-6=0
\end{equation}
Three roots are
\begin{equation}
    r_1=1
\end{equation}
\begin{equation}
    r_2=2
\end{equation}
\begin{equation}
    r_3=3
\end{equation}
Then the solution for the homogeneous system is

\begin{equation}
    y_c(t)=c_1e^{x}+c_2e^{2x}+c_3e^{3x}
\end{equation}
where
\begin{equation}
    y_1(t)=e^{x}
\end{equation}
\begin{equation}
    y_2(t)=e^{2x}
\end{equation}
\begin{equation}
    y_3(t)=e^{3x}
\end{equation}
Then we follow to find the required solution to the nonhomogeneous equation. We use Variation of Parameters. We have
\begin{equation}
    W[y_1,y_2,y_3]=2e^{6x}
\end{equation}
\begin{equation}
    W_1[y_1,y_2y_3]=e^{5x}
\end{equation}
\begin{equation}
    W_2[y_1,y_2y_3]=-2e^{4x}
\end{equation}
\begin{equation}
    W_3[y_1,y_2，y_3]=e^{3x}
\end{equation}
and
\begin{equation}
    g(x)=2\frac{e^{3x}}{e^{x}+1}
\end{equation}
And then we have the following integration
\begin{equation}
    \int \frac{W_1\cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{e^{2x}}{e^{x}+1}
\end{equation}
\begin{equation}
   \int\frac{e^{2x}}{e^{x}+1}=e^{x}-\ln(e^{x}+1)+c_4
\end{equation}
\begin{equation}
    \int \frac{W_2\cdot g(x)dx}{W[y_1,y_2,y_3]}=-2\int\frac{e^{x}}{e^{x}+1}
\end{equation}
\begin{equation}
    -2\int\frac{e^{x}}{e^{x}+1}=-2\ln(e^{x}+1)+c_5
\end{equation}
\begin{equation}
    \int \frac{W_3 \cdot g(x)dx}{W[y_1,y_2,y_3]}=\int\frac{1}{e^{x}+1}
\end{equation}
\begin{equation}
    \int\frac{1}{e^{x}+1}=x-\ln(e^{x}+1)+c_6
\end{equation}
And finally, the required general solution $y(t)$
\begin{equation}
    y(t)=\sum_{i=1}^3 y_i(t)\cdot\int\frac {W_i\cdot g(x)dx}{W[y_1,y_2,y_3]}
\end{equation}