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**Quiz 4 / Re: Q4--day section--problem 2**

« **on:**March 23, 2013, 02:44:06 PM »

I updated my solution.

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I updated my solution.

17

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:

(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

\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.

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Yes, I prefer theory to computation. Unfortunately, I've not yet had the pleasure of taking a purely theoretical course. (*I was in engineering.*)

20

I am in the same position. I did no homework and relied purely on theory (as opposed to experience) in my solutions. Any of those questions are easily solvable in an adequate amount of time, but when time is as constrained as it was on the test, knowledge of shortcuts is the only way to score a flawless victory.

21

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*}

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*}

22

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)}.

$$

$$

\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)}.

$$

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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).

$$

$$

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).

$$

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