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### Messages - Alexander Jankowski

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16
##### Quiz 4 / Re: Q4--day section--problem 2
« on: March 23, 2013, 02:44:06 PM »
I updated my solution.

17
##### Quiz 4 / Q4--day section--problem 2
« on: March 23, 2013, 02:19:36 PM »
The second question was #9.2.17:

(a) Find an equation of the form $H(x,y) = c$ satisfied by the trajectories
$$\frac{dx}{dt} = 2y, \qquad \frac{dy}{dt} = 8x.$$
(b) Plot several level curves of the function $H$. These are trajectories of the given system. Indicate the direction of motion on each trajectory.

First, we determine the function $H(x,y)$:
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{8x}{2y} \Longleftrightarrow ydy = 4xdx \Longrightarrow H(x,y) = \frac{1}{2} y^2 - 2x^2 = c,$$
where $c$ is a constant of integration. For $c = -2,-1,0,1,2$, we have:

Therefore, for $c = 0$, the trajectories are two lines with slopes $2$ and $-2$ that intersect at the origin, and are separatrices. For $c \neq 0$, the trajectories are hyperbolÃ¦. In particular, for $c > 0$, the hyperbolÃ¦ lie along the ordinate; for $c < 0$, they lie along the abscissa. To determine the direction of the trajectories, we rewrite the system as the matrix equation
$$\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{cc} 0 & 2 \\ 8 & 0 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right),$$
and we plug in the vectors $(x,y)^T=(0,1)$ and $(x,y)^T=(0,-1)$. This yields, respectively,
$$\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 2 \\ 0 \end{array} \right), \qquad \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} -2 \\ 0 \end{array} \right).$$
We conclude that the hyperbolÃ¦ along the ordinate are directed counter-clockwise. For the hyperbolÃ¦ along the abscissa, we can use $(x,y)^T=(1,0)$ and $(x,y)^T=(-1,0)$ to get, respectively,
$$\frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 0 \\ 8 \end{array} \right), \qquad \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 0 \\ -8 \end{array} \right).$$
Therefore, the hyperbolÃ¦ along the abscissa for $x < 0$ are directed downwards and those for $x > 0$ are directed upwards. This can be verified with a stream plot:

18
##### Term Test 1 / Re: TT1--Problem 2
« on: February 14, 2013, 08:52:15 AM »
(a) We first re-write the differential equation as

\begin{equation*}
y'' - \frac{t\cos{t}}{\cos{t} + t\sin{t}}y' + \frac{\cos{t}}{\cos{t} + t\sin{t}} = 0.
\end{equation*}

Then, according to Abel's identity,

\begin{equation*}
W(y_1,y_2)(t) = c \exp{\left(-\int{\left(\frac{-t\cos{t}}{\cos{t} + t\sin{t}}\right)dt}\right)}.
\end{equation*}

Let $u = \cos{t} + t\sin{t}$ so that $du = t\cos{t}dt$. Then,

\begin{equation*}
W(y_1,y_2)(t) = c \exp{\int{\frac{du}{u}}} = c \exp {\ln{|t\sin{t} + \cos{t}|}} = c|t\sin{t} + \cos{t}|.
\end{equation*}

Using $W(0) = 1$ yields

\begin{equation*}
1 = C|0\cdot\sin{0} + \cos{0}| \Longleftrightarrow C = 1.
\end{equation*}

Therefore, the desired Wronskian is

\begin{equation*}
W(y_1,y_2)(t) = |t\sin{t} + \cos{t}|.
\end{equation*}

I do not yet know how to solve (b) without guessing, as applying reduction of order yields a rather unpleasant integral.

19
##### Term Test 1 / Re: TT1--Problem 1
« on: February 13, 2013, 11:08:34 PM »
I am late, but at some point, someone is going to want a coded solution that isn't rotated by 90Â° CCW.

Let $M(x,y) = x$ and let $N(x,y) = y + x^2y + y^3$. Then,

\begin{align*}
M_y(x,y) = 0, & N_x(x,y) = 2xy.
\end{align*}

So the equation is not exact, and we need an integrating factor $\mu(x,y) = \mu(y)$ to make it exact.

\begin{equation*}
\frac{d\mu(y)}{dy} = \frac{N_x - M_y}{M} \mu(y) = \frac{2xy}{x} \mu(y) = y\mu(y) \Longrightarrow \frac{d\mu(y)}{\mu(y)} = 2ydy \Longrightarrow \mu(y) = e^{y^2}
\end{equation*}

Now, we require a function $\psi(x,y)$ that satisfies $\psi_x = \mu M$ and $\psi_y = \mu N$.

\begin{equation*}
\psi_x(x,y) = xe^{y^2} \Longrightarrow \psi(x,y) = \frac{1}{2}x^2e^{y^2} + h(y)
\end{equation*}

We differentiate the result to get

\begin{equation*}
\psi_y(x,y) = x^2xe^{y^2} + h'(y) = e^{y^2}(y + x^2y + y^3) \Longleftrightarrow h'(y) = e^{y^2}(y^3 + y) \Longrightarrow h(y) = \int{(e^{y^2}(y^3 + y))}dy.
\end{equation*}

Let $u = y^2$ so that $du = 2ydy$. Then,

\begin{equation*}
h(y) = \int{(y^3e^{y^2})}dy + \int{(ye^{y^2})}dy = \frac{1}{2}\int{(ue^u)}du + \frac{1}{2}\int{e^u}du = \frac{1}{2}ue^u - \frac{1}{2}\int{e^u}du + \frac{1}{2}e^u = \frac{1}{2}ue^u - \frac{1}{2}e^u + \frac{1}{2}e^u = \frac{1}{2}y^2e^{y^2}.
\end{equation*}

Therefore, the solution is implicitly given by

\begin{equation*}
C = \frac{1}{2}e^{y^2}(x^2 + y^2).
\end{equation*}

20
##### Quiz 1 / Re: Night section, 2.2 # 26
« on: January 16, 2013, 10:29:13 PM »
The given first-order nonlinear ordinary differential equation is separable, so

$$\frac{dy}{dx} = 2(1+x)(1+y^2) \Rightarrow \frac{dy}{1+y^2} = 2(1+x)dx \Rightarrow \arctan{y} = x^2 + 2x + C \Leftrightarrow y(x) = \tan{(x^2 + 2x + C)}.$$

Using the initial condition, we find $C$:

$$0 = \tan{(0^2 + 2(0) + C)} \Leftrightarrow C = \arctan{0} = 0$$

Conclusively, the solution to the initial value problem is

$$y(x) = \tan{(x^2 + 2x)}.$$

21
##### Quiz 1 / Re: Night section: 2.1 #17
« on: January 16, 2013, 10:21:46 PM »
Let $\mu(t)$ be the integrating factor. Then,

$$y' - 2y = e^{2t} \Leftrightarrow y'\mu(t) - 2y\mu(t) = e^{2t}\mu(t).$$

We require that $\frac{d\mu(t)}{dt} = -2\mu(t)$. Thus,

$$\frac{d\mu(t)}{dt} = -2\mu(t) \Rightarrow \frac{d\mu(t)}{\mu(t)} = -2dt \Rightarrow \ln|\mu(t)| = -2t + K,$$

where $K$ is some arbitrary constant that we null to acquire the simplest integrating factor:

$$\ln|\mu(t)| = -2t \Leftrightarrow \mu(t) = e^{-2t}.$$

Now, the differential equation is

$$y'e^{-2t} - 2ye^{-2t} = e^{2t}e^{-2t} = 1.$$

Applying the product rule gives $\frac{d}{dt}(e^{-2t}y) = 1$, to which the general solution is

$$e^{-2t}y = t + C \Leftrightarrow y(t) = e^{2t}(t + C).$$

Using the initial condition, we find that $C=2$. Conclusively, the desired solution is

$$y(t) = e^{2t}(t+2).$$

Pages: 1 [2]