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Messages - Jared Jubas-Malz

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1
Quiz-7 / Re: Q7-0901
« on: March 31, 2018, 10:52:58 PM »
I attached another phase portrait at the end of this post with the saddle point emphasized.
For the bonus, this is what I did. I'm not entirely sure if it's correct.
Begin by setting $M=x+\cos(y)-1$ and $N=(1+x)\sin(y)$. Since it's not exact, we look for an integrating factor $\mu=\mu(x)$:
\begin{align}\frac{d\mu}{dx}=\frac{M_{y}-N_{x}}{N}\mu\end{align}
\begin{align}\frac{d\mu}{dx}=\frac{M_{y}-N_{x}}{N}=\frac{-\sin(y)-\sin(y)}{(1+x)\sin(y)}=\frac{-2}{1+x}\end{align}
Therefore $(8 )$ becomes:
\begin{align}\frac{d\mu}{dx}=\frac{-2}{1+x}\mu\end{align}
Separating and integrating gives:
\begin{align}\mu=\frac{1}{(1+x)^2}\end{align}
Plugging the integrating factor back into the original equation:
\begin{align}\frac{x+\cos(y)-1}{(1+x)^2}dx+\frac{\sin(y)}{1+x}dy\end{align}
which is exact. Then,
\begin{align}H_{x}=\frac{x+\cos(y)-1}{(1+x)^2}\end{align}
\begin{align}H_{y}=\frac{\sin(y)}{1+x}\end{align}
Integrating $(13)$ with respect to x:
\begin{align}H(x,y)=\ln(x+1)-\frac{\cos(y)-2}{x+1}+h(y)\end{align}
Plugging into $(14)$:
\begin{align}H_{y}=\frac{\partial}{\partial(y)}(\ln(x+1)-\frac{\cos(y)-2}{x+1}+h(y))=\frac{\sin(y)(x+1)}{(x+1)^2}+h'(y)=\frac{\sin(y)}{1+x}\end{align}
so $h'(y)=0\implies h(y)=c$.
Therefore this implies that
\begin{align}H(x,y)=\ln(x+1)-\frac{\cos(y)-2}{x+1}=C\end{align}

2
Quiz-7 / Re: Q7-T0101
« on: March 30, 2018, 01:29:23 PM »
The critical points would be when $x'=0$ and $y'=0$:
\begin{align}0=x^2+xy-x=x(x+y-1)\end{align}
\begin{align}0=2y^2+xy-3y=y(2y+x-3)\end{align}
From $(1)$ and $(2)$ the critical points would be $(0,0)$, $(0,\frac{3}{2})$, $(1,0)$ and $(-1,2)$.
The Jacobian matrix would be:
\begin{align}J=\begin{pmatrix}F_{x}&F_{y}\\G_{x}&G_{y}\end{pmatrix}=\begin{pmatrix}-2x-y+1&-x\\-y&-4y-x+3\end{pmatrix}\end{align}
At $(0,0)$:
\begin{align}J=\begin{pmatrix}1&0\\0&3\end{pmatrix}\end{align}
Since it is a triangular matrix, the eigenvalues would just be the diagonal entries so $r_{1}=1$ and $r_{2}=3$. This means that $0<r_{1}<r_{2}$ so the nonlinear system would be an unstable node.
At $(0,\frac{3}{2})$:
\begin{align}J=\begin{pmatrix}\frac{-1}{2}&0\\\frac{-3}{2}&-3\end{pmatrix}\end{align}
Since it is a triangular matrix, the eigenvalues would just be the diagonal entries so $r_{1}=\frac{-1}{2}$ and $r_{2}=-3$. This means that $0>r_{1}>r_{2}$ so the nonlinear system would be an asymptotically stable node.
At $(1,0)$:
\begin{align}J=\begin{pmatrix}-1&-1\\0&2\end{pmatrix}\end{align}
Since it is a triangular matrix, the eigenvalues would just be the diagonal entries so $r_{1}=-1$ and $r_{2}=2$. This means that $r_{1}<0<r_{2}$ so the nonlinear system would be an unstable saddle point.
At $(-1,2)$:
\begin{align}J=\begin{pmatrix}1&1\\-2&-4\end{pmatrix}\end{align}
Solving gives eigenvalues $r_{1}=\frac{-3+\sqrt17}{2}$ and $r_{2}=\frac{-3-\sqrt17}{2}$. Since $r_{2}<0<r_{1}$ the nonlinear system would be an unstable saddle point.

3
Quiz-7 / Re: Q7-T0201
« on: March 30, 2018, 01:28:49 PM »
The critical points would be when $x'=0$ and $y'=0$:
\begin{align}0=x+x^2+y^2\end{align}
\begin{align}0=y-xy\end{align}
From $(1)$ and $(2)$ the critical points would be $(0,0)$ and $(-1,0)$.
The Jacobian matrix would be:
\begin{align}J=\begin{pmatrix}F_{x}&F_{y}\\G_{x}&G_{y}\end{pmatrix}=\begin{pmatrix}2x+1&2y\\-y&1-x\end{pmatrix}\end{align}
At $(0,0)$:
\begin{align}J=\begin{pmatrix}1&0\\0&1\end{pmatrix}\end{align}
Since it is a triangular matrix, the eigenvalues would just be the diagonal entries so $r_{1}=r_{2}=1$. This means that $r_{1}=r_{2}>0$ so the nonlinear system would be an unstable proper node or spiral point.
At $(-1,0)$:
\begin{align}J=\begin{pmatrix}-1&0\\0&2\end{pmatrix}\end{align}
Since it is a triangular matrix, the eigenvalues would just be the diagonal entries so $r_{1}=-1$ and $r_{2}=2$. Since $r_{1}<0<r_{2}$ the nonlinear system would be an unstable saddle point.

4
Quiz-7 / Re: Q7-0901
« on: March 30, 2018, 01:21:33 PM »
The critical points would be when $x'=0$ and $y'=0$:
\begin{align}0=(1+x)\sin(y)\end{align}
\begin{align}0=1-x-\cos(y)\end{align}
From $(1)$, $\sin(y)$ would be $0$ when $y=n\pi$ where $n=0, 1, 2, 3,...$. Since there is a $\cos(y)$ in $(2)$, the previous equation can be split into two cases: $y=2n\pi$ where $n=0, 1, 2, 3,....$ and $y=n\pi$ where $n=1, 3, 5,....$. Then in case 1 $\cos(y)=1$ and in case 2 $cos(y)=-1$. Plugging both of these into $(2)$ gives the critical points:
\begin{align}(0, 2n\pi) \quad where \quad n=0, 1, 2, 3,...\end{align}
\begin{align}(2, n\pi) \quad where \quad n=1, 3, 5,...\end{align}
The Jacobian matrix would be:
\begin{align}J=\begin{pmatrix}F_{x}&F_{y}\\G_{x}&G_{y}\end{pmatrix}=\begin{pmatrix}\sin(y)&-\cos(y)(1+x)\\-1&\sin(y)\end{pmatrix}\end{align}
At $(0, 2n\pi)$:
\begin{align}J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\end{align}
This gives eigenvalues $r_{1}=i$ and $r_{2}=-i$. Since $r_{1}, r_{2}=\lambda\pm i\mu$ where $\lambda=0$ the nonlinear system would be an indeterminate center or spiral point.
At $(2, n\pi)$:
\begin{align}J=\begin{pmatrix}0&-3\\-1&0\end{pmatrix}\end{align}
The eigenvalues would be $r_{1}=\sqrt{3}$ and $r_{2}=-\sqrt{3}$. Since $r_{2}<0<r_{1}$ the nonlinear system would be an unstable saddle point.

5
Term Test 2 / Re: TT2--P3M
« on: March 28, 2018, 10:43:24 AM »
I found my mistake. I think you're correct.

6
Term Test 2 / Re: TT2--P3M
« on: March 28, 2018, 01:26:54 AM »
I got something different for part (b). I got that $u_{2}'$ should be:
\begin{align}u_{2}'=\frac{-6e^{3t}}{e^t+1}\end{align}
Therefore $u_{2}$ would be:
\begin{align}u_{2}(t) = -3e^{2t}+6e^t-6\ln(e^t+1)+c_{2}\end{align}
Then the general solution would be:
\begin{align}\textbf{x}(t)=(18\ln(e^t+1)+c_{1})e^{-t}\begin{pmatrix}1\\0\end{pmatrix}+(-3e^{2t}+6e^t-6\ln(e^t+1)+c_{2})e^{-3t}\begin{pmatrix}1\\-2\end{pmatrix}\end{align}
Subbing in the initial condition $\textbf{x}(0)=\begin{pmatrix}1-3\ln2\\-3-3\ln2\end{pmatrix}$:
\begin{align}1-3\ln2=c_{1}+3+c_{2}\end{align}
\begin{align}-3-3\ln2=-6-2c_{2}\end{align}
Solving these equations for the constants gives:
\begin{align}c_{1}=\frac{-19-9\ln2}{2}\end{align}
\begin{align}c_{2}=\frac{-3+3\ln2}{2}\end{align}
Therefore the general solution for the IVP is:
\begin{align}\textbf{x}(t)=\frac{-19-9\ln2}{2}e^{-t}\begin{pmatrix}1\\0\end{pmatrix}+\frac{-3+3\ln2}{2}e^{-3t}\begin{pmatrix}1\\-2\end{pmatrix}+18\ln(e^t+1)e^{-t}\begin{pmatrix}1\\0\end{pmatrix}-3e^{-t}\begin{pmatrix}1\\-2\end{pmatrix}+6e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}-6\ln(e^t+1)e^{-3t}\begin{pmatrix}1\\-2\end{pmatrix} \end{align}

7
MAT244--Misc / Re: What s/w have you used for plots?
« on: March 24, 2018, 05:06:39 PM »

8
Term Test 2 / Re: TT2--P4D
« on: March 22, 2018, 05:04:05 PM »
Setting $\begin{pmatrix}-3&-2\\2&-3\end{pmatrix} = A$, the eigenvalues can be found by taking $det(A-\lambda I)$:
\begin{align}det(A-\lambda I) = det\begin{pmatrix}-3-\lambda&-2\\2&-3-\lambda\end{pmatrix}=\lambda^2+6\lambda+13=0\end{align}
The eigenvalues would be $\lambda_{1}=-3+2i \quad and \quad \lambda_{2}=-3-2i$. Using $\lambda_{1}=-3+2i\quad$gives:
\begin{align}\begin{pmatrix}-2i&-2\\2&-2i\end{pmatrix}\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \implies ix_{1}=-x_{2}\end{align}
Therefore, the corresponding eigenvector would be: $$\textbf{v}=\begin{pmatrix}i\\1\end{pmatrix}$$
We want to find the real/imaginary solutions of $e^{(-3+2i)t}\begin{pmatrix}i\\1\end{pmatrix}$. Separating the exponential then applying Euler's formula:
\begin{align}e^{-3t}\times e^{2it}\begin{pmatrix}i\\1\end{pmatrix}=e^{-3t}(cos(2t)+isin(2t))\begin{pmatrix}i\\1\end{pmatrix}=e^{-3t}\begin{pmatrix}icos(2t)&-sin(2t)\\cos(2t)&isin(2t)\end{pmatrix}\end{align}
Separating the real and imaginary parts of (3):
\begin{align}e^{-3t}\begin{pmatrix}-sin(2t)\\cos(2t)\end{pmatrix}+ie^{-3t}\begin{pmatrix}cos(2t)\\sin(2t)\end{pmatrix}\end{align}
Therefore the general real solution will be:
\begin{align}\textbf{x}(t)=c_{1}e^{-3t}\begin{pmatrix}-sin(2t)\\cos(2t)\end{pmatrix}+c_{2}ie^{-3t}\begin{pmatrix}cos(2t)\\sin(2t)\end{pmatrix}\end{align}

9
Term Test 2 / Re: TT2--P3
« on: March 22, 2018, 01:18:14 AM »
I got a different answer for part (b). Aside from a missing constant, I got everything the same up until $(6)$ so I will start from there. Rewriting $(6)$ I got:
$$\textbf{x}(t)=(ln(e^t+1)+c_{1})e^t\begin{pmatrix}1\\1\end{pmatrix}+c_{2}e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}=ln(e^t+1)e^t\begin{pmatrix}1\\1\end{pmatrix}+c_{1}e^t\begin{pmatrix}1\\1\end{pmatrix}+c_{2}e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}$$
Plugging in the initial condition:
$$\textbf{x}(0)=\begin{pmatrix}3\\0\end{pmatrix}$$
This gives:
$$3=ln(2)+c_{1}+c_{2} \implies c_{1}=3-ln(2)-c_{2}\qquad(7)$$
and:
$$0=ln(2)+c_{1}-2c_{2}\qquad(8 )$$
Plugging $(7)$ into $(8 )$ gives $c_{2}=1$. Plugging this back into (7) gives $c_{1}=2-ln(2)$. Therefore the constants are:
$$c_{1}=2-ln(2)\quad and \quad c_{2}=1$$
Therefore the general solution would be:
$$\textbf{x}(t)=ln(e^t+1)e^t\begin{pmatrix}1\\1\end{pmatrix}+(2-ln(2))e^t\begin{pmatrix}1\\1\end{pmatrix}+e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}$$

10
Term Test 2 / Re: TT2--P2
« on: March 21, 2018, 09:08:55 PM »
Part (a)$\\$
The equation for the Wronskian would be:
$$W = c\times exp[-\int p_{1}(t)dt]$$
Since there is no $y''$ term, $p_{1}(t)$ would be $0$:
$$W = c\times exp[-\int 0 dt]=c\times exp(0)=c$$
Therefore, the Wronskian would be a constant.

Part (b)$\\$
Consider the homogeneous equation:
$$y''' - 3y' + 2y = 0$$
The characteristic equation would be:
$$r^3-3r+2=0$$
Solving this gives:
$$(r-1)^2(r+2)\rightarrow r_{1}=r_{2}=1, r_{3}=-2$$
Therefore, the homogeneous solution would be:
$$y_{c}(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{-2t}\qquad(2)$$
Computing the Wronskian:
$$W=\begin{array}{|c c c|}e^t &te^t &e^{-2t}\\e^t&(t+1)e^t&-2e^{-2t}\\e^t&(t+2)e^t&4e^{-2t}\end{array}=4(t+1)+2(t+2)-t(4+2)+(t+2)-(t+1)=9$$
Therefore, the Wronskian is a constant just as expected based on part (a).

Part (c)$\\$
The particular solution should be of the form:
$$Y(t)=Ae^{-2t}$$
Since $e^{-2t}$ is part of the homogeneous solution, we look for solutions of the form:
$$Y(t)=Ate^{-2t}\qquad(3)$$
Differentiating this:
$$Y'(t)=Ae^{-2t}-2Ate^{-2t} \qquad(4)$$
Differentiating again:
$$Y''(t)=-4Ae^{-2t}+4Ate^{-2t}$$
Differentiating once more:
$$Y'''(t)=12Ae^{-2t}-8Ate^{-2t} \qquad(5)$$
Plugging (3), (4) and (5) into (1):
$$12Ae^{-2t}-8Ate^{-2t}-3Ae^{-2t}+6Ate^{-2t}+2Ate^{-2t}=18e^{-2t}$$
Simplifying gives:
$$9Ae^{-2t}=18e^{-2t}$$
Therefore, $A=2$. Subbing this value of A into (3) and combining it with (2) gives the general solution:
$$y(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{2t}+2te^{-2t}$$

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