Toronto Math Forum

Solving for eigenvalues:
\begin{gather*}
(4-r)(-6-r)+24 = 0\\
\implies - 24 - 4r + 6r + r^2 = 0\\
\implies r^2 + 2r = 0 \implies r(r+2) = 0
\end{gather*}
eigenvalues are $0$ and $-2$.

When $r = 0$,  Nullspace of $\begin{pmatrix}
    4 & -3 \\
    8 & -6
  \end{pmatrix}$ equals to  Nullspace of $\begin{pmatrix}
    4 & -3 \\
    0 & 0
  \end{pmatrix}$, which is equal to the span of  $\begin{pmatrix}
    3 \\
    4
  \end{pmatrix}$.
When $r = -2$,  Nullspace of $\begin{pmatrix}
    6 & -3 \\
    8 & -4
  \end{pmatrix}$, which is equal to Nullspace of $\begin{pmatrix}
    6 & -3 \\
    0 & 0
  \end{pmatrix}$, which is a span of $\begin{pmatrix}
    1 \\
    2
  \end{pmatrix}$.

Then
$$\mathbf{x} = c_1\begin{pmatrix}3\\4\end{pmatrix} + c_2e^{-2t}\begin{pmatrix}1\\2\end{pmatrix}$$

As $t$ goes to infinity, the solutions tend to $c_1\begin{pmatrix}3\\4\end{pmatrix}$. A sketch of the phase portrait:

$M_y = 8xy$
$N_x = 4xy $
$\implies$ The equation is not exact

$\frac{M_y-N_x}{N} = \frac{2}{x}$

$\frac{d\mu}{dx} = \frac{2}{x}\mu$

$\mu = x^2$

The integrating factor is $\mu = x^2$.

New equation: $4x^3y^2 + 3x^2 lnx + x^2 + 2x^4yy'=0$
$M_y = 8x^3y = N_x$ (The equation is exact)
$\phi_x = M$
$\phi = x^4y^2 + x^3\ln x + h(y)$
$\phi_y = 2x^4y + h'(y) = N$
$h'(y)=0$
$h(y)=C$

Thus, $\phi= x^4y^2 + x^3\ln x  = C$

For particular solution passing (1,1):
$1^41^2 + 1^2\ln 1  = C$
$C=1$
Thus, $\phi= x^4y^2 + x^2\ln x  = 1$

*typed solutions coming soon* (I have class until 9)



*typed solutions coming soon*

*I will type up the solutions soon*



*Typed solutions to come soon*