Toronto Math Forum

Could you just post the solution? Could really learn from examples instead of words.

a clearer picture for d

Solve $1-y = 0\Rightarrow y=1$ and $x^2 - y^2 = (x+y)(x-y) = 0$, substituting $y=1$, we have 2 critical points $(-1, 1), (1,1)$

Computing the Jacobian yields
\begin{align}
J =
\begin{bmatrix}
   F_x & F_y\\
   G_x & G_y
\end{bmatrix}
=
\begin{bmatrix}
   0 & -1\\
   2x & -2y
\end{bmatrix}
\end{align}
Plugging in both critical points we have 2 linear systems, first
\begin{align}
x' =
\begin{bmatrix}
   0 & -1\\
   -2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + \sqrt{3}\, \lambda_2 = -1 - \sqrt{3}$, since $\sqrt{3} > 1$, we have $\lambda_1 > 0 > \lambda_2$, so we conclude locally $(-1, 1)$ is a saddle.
Plug in $(1,1)$
\begin{align}
x' =
\begin{bmatrix}
   0 & -1\\
   2 & -2
\end{bmatrix}
x
\end{align}
solving for eigenvalues yields $\lambda_1 = -1 + i\, \lambda_2 = -1 - i$, a complex conjugate with negative real parts, we conclude locally $(1, 1)$ is an asymptotically stable spiral.

To verify take $y_1 = e^t, y_2 = t$, then
\begin{align*}
y_1' = e^t,\ &y_2' = 1\\
y_1'' = e^t,\ &y_2'' = 0
\end{align*}

Plug in to the original equation, we check both are 0
\begin{align*}
(1-t)e^t + te^t - e^t &= 0\\
t - t &= 0
\end{align*}

Now take the Wronskian
\begin{align*}
W = e^t - te^t = (1-t)e^t
\end{align*}
Use the variation of parameters equation to find
\begin{align*}
u_1 &= -\int\frac{y_2g}{W}\\
&= -2\int te^{-2t}dt\\
&= (t+\frac{1}{2})e^{-2t}
\end{align*}
\begin{align*}
u_2 &= \int\frac{y_1g}{W}\\
&= 2\int e^{-t}dt\\
&= -2e^{-t}
\end{align*}
Putting the pieces together, we have the particular solution
\begin{align*}
Y &=  (t+\frac{1}{2})e^{-t} - 2te^{-t}\\
&= \frac{1}{2}e^{-t} - te^{-t}\\
&= e^{-t}(\frac{1}{2} - t) \\
&= -\frac{1}{2}(2t-1)e^{-t}\\
\end{align*}

a). First solve the homogenous part
\begin{align*}
y'' + 5y' + 6y &= 0\\
r^2 + 5r + 6 &=0\\
(r+2)(r+3) &=0\\
r_1 = -2,\  r_2 &= -3
\end{align*}
So solution to homogenous part is
\begin{align*}
y_c = c_1e^{-2t} + c_2e^{-3t}
\end{align*}
Next we solve \begin{align}
y'' + 5y' + 6y &= -30e^{2t}
\end{align}
Let
\begin{align*}
Y_1 &= Ae^{2t}, \text{and}\\
Y_1' &= 2Ae^{2t}\\
Y_1'' &= 4Ae^{2t}, \text{Combining with (1)}
\end{align*}
\begin{align*}
4Ae^{2t} + 10Ae^{2t} &+ 6Ae^{2t} = -30e^{2t}\\
A &= -\frac{3}{2}\\
Y_1 &= -\frac{3}{2}e^{2t}
\end{align*}

Next we solve
\begin{align}
y'' + 5y' + 6y &= 3e^{-2t}
\end{align}
Since we already have $e^{-2t}$ in our solution, let
\begin{align*}
Y_2 &= Bte^{-2t}, \text{and}\\
Y_2' &= Be^{-2t} - 2Bte^{-2t}\\
Y_2'' &= -4Be^{-2t} + 4Bte^{-2t}, \text{Combining with (2)}
\end{align*}
\begin{align*}
 (-4Be^{-2t} + 4Bte^{-2t}) + &5(Be^{-2t} - 2Bte^{-2t}) + 6(Bte^{-2t}) = 3e^{-2t}\\
B &= 3\\
Y_2 &=3te^{-2t}
\end{align*}
Combining above we have the general solution
\begin{align}
y &= c_1e^{-2t} + c_2e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
y' &= -2c_1e^{-2t} - 3c_2e^{-3t} - 3e^{2t} +3e^{-2t} - 6te^{-2t}
\end{align}
b). Plug in $y(0)=0, y'(0) = 0$ to (3) and (4)
with
\begin{align*}
c_1 + c_2  &= \frac{3}{2}\\
-2c_1 - 3c_2  &= 0\\
c_1 = \frac{9}{2}&,\ c_2 = -3
\end{align*}
And finally,
\begin{align}
y &= \frac{9}{2}e^{-2t} - 3e^{-3t} -\frac{3}{2}e^{2t} + 3te^{-2t}\\
\end{align}