To determine: a) The origin [left(0, 0right)] is a critical point of the systems. [frac{dx}{dt}=y + x left(x^{2} + y^{2}right), frac {dx}{dt}= -x + y left(x^{2} + y^{2}right)] and [frac{dx}{dt}=y - x left(x^{2} + y^{2}right), frac {dx}{dt}= -x - y left(x^{2} + y^{2}right)]. Futhermore, it is a center of the corresponding linear system. b) The systems [frac{dx}{dt}=y + x left(x^{2} + y^{2}right), frac {dx}{dt}= -x + y left(x^{2} + y^{2}right)] and [frac{dx}{dt}= y - y + x left(x^{2} + y^{2}right), frac {dx}{dt}= -x - y left(x^{2} + y^{2}right)] are almost linear. c) To prove: [frac{dr}{dt} < 0] and [r rightarrow 0 as t rightarrow infty], hence the critical point for the system [frac{dx}{dt}= y - x left(x^{2} + y^{2}right), frac {dx}{dt}= -x - y left(x^{2} + y^{2}right)] is asymptoticall

To determine: a) The origin [left(0, 0right)] is a critical point of the systems. [frac{dx}{dt}=y + x left(x^{2} + y^{2}right), frac {dx}{dt}= -x + y left(x^{2} + y^{2}right)] and [frac{dx}{dt}=y - x left(x^{2} + y^{2}right), frac {dx}{dt}= -x - y left(x^{2} + y^{2}right)]. Futhermore, it is a center of the corresponding linear system. b) The systems [frac{dx}{dt}=y + x left(x^{2} + y^{2}right), frac {dx}{dt}= -x + y left(x^{2} + y^{2}right)] and [frac{dx}{dt}= y - y + x left(x^{2} + y^{2}right), frac {dx}{dt}= -x - y left(x^{2} + y^{2}right)] are almost linear. c) To prove: [frac{dr}{dt} < 0] and [r rightarrow 0 as t rightarrow infty], hence the critical point for the system [frac{dx}{dt}= y - x left(x^{2} + y^{2}right), frac {dx}{dt}= -x - y left(x^{2} + y^{2}right)] is asymptoticall

Question
Alternate coordinate systems
asked 2020-11-08
To determine:
a) The origin \(\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}\) is a critical point of the systems.
\(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\). Futhermore, it is a center of the corresponding linear system.
b) The systems \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\) are almost linear.
c) To prove: \(\displaystyle{\left[{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}\ {<}\ {0}\right]}{\quad\text{and}\quad}{\left[{r}\rightarrow\ {0}\ {a}{s}\ {t}\rightarrow\ \infty\right]},\)</span> hence the critical point for the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\) is asymptotically stable and the solution of the initial value problem for \(\displaystyle{\left[{r}\ {w}{i}{t}{h}\ {r}={r}_{{{0}}}\ {a}{t}\ {t}={0}\right]}\) becomes unbounded as \(\displaystyle{\left[{t}\rightarrow{\frac{{{1}}}{{{2}}}}\ {r}{\frac{{{2}}}{{{0}}}}\right]}\), hence the critical point for the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\) is unstable.

Answers (1)

2020-11-09

a)
The systems of equations are \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\ {\quad\text{and}\quad}\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}.\)
Formula used: The points, if any, where \(\displaystyle{\left[{f}{\left({x}\right)}={0}\right]}\) are called critical pointsof the autonomous system \(\displaystyle{\left[{x}^{{{p}{r}{i}{m}{e}}}={f}{\left({x}\right)}\right]}.\)
Proof: The critical points of the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\) are found by solving the equations \(\displaystyle{\left[{y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}={0}\ {\quad\text{and}\quad}\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}={0}\right]}.\)
From the equation \(\displaystyle{\left[{x}^{{{2}}}\ +\ {y}^{{{2}}}=\ -{\frac{{{y}}}{{{x}}}}\right]}\) and from the second equation \(\displaystyle{\left[{x}^{{{2}}}\ +\ {y}^{{{2}}}={\frac{{{x}}}{{{y}}}}\right]},\)
\(\displaystyle\Rightarrow\ -{\frac{{{y}}}{{{x}}}}={\frac{{{x}}}{{{y}}}}\)
\(\displaystyle\Rightarrow\ {x}^{{{2}}}\ +\ {y}^{{{2}}}={0}\)
Thus, the only critical point is \(\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}.\)
Now,
The critical points of the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\) are found by solving the equations \(\displaystyle{\left[{y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}={0}\ {\quad\text{and}\quad}\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}={0}\right]}.\)
From the first equation \(\displaystyle{\left[{x}^{{{2}}}\ +\ {y}^{{{2}}}={\frac{{{y}}}{{{x}}}}\right]}\) and from the second equation \(\displaystyle{\left[{x}^{{{2}}}\ +\ {y}^{{{2}}}=\ -{\frac{{{x}}}{{{y}}}}\right]},\)
\(\displaystyle\Rightarrow\ {\frac{{{y}}}{{{x}}}}=\ -{\frac{{{x}}}{{{y}}}}\)
\(\displaystyle\Rightarrow\ {x}^{{{2}}}\ +\ {y}^{{{2}}}={0}\)
Thus, the only critical point is \(\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}\).
For both the systems, the linearized system is \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y},\ {\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\right]}\).
Since \(\displaystyle{\left[{F}\ {\left({x},\ {y}\right)}={y}\ {\quad\text{and}\quad}\ {G}\ {\left({x},\ {y}\right)}=\ -{x}\right]}\), the Jacobian matrix is given by,
\([J \left(x,\ y \right)=\left(\begin{array}{c}F_{x} & F_{y}\\ G_{x} & G_{y}\end{array}\right)=\left(\begin{array}{c}0 & 1\\ -1 & 0\end{array}\right)]\)
Near the critical point \(\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}\), the Jacobian matrix becomes,
\([J\ \left(0,\ 0\right)=\left(\begin{array}{c}F_{x}\ \left(0,\ 0\right) & F_{y}\ \left(0,\ 0\right)\\ G_{x}\ \left(0,\ 0\right) & G_{y}\ \left(0,\ 0\right)\end{array}\right)=\left(\begin{array}{c}0 & 1\\ -1 & 0\end{array}\right)]\) Hence, the corresponding linear system near the critical point \(\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}\) is \([\frac{dx}{dt}\ \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0 & 1\\ -1 & 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)]\)
Let \([A=\left(\begin{array}{c}0 & 1\\ -1 & 0\end{array}\right)]\ and\ the eigenvalues [\lambda_{1},\ \lambda_{2}]\) and the eigenvalues \(\displaystyle{\left[\lambda_{{{1}}},\ \lambda_{{{2}}}\right]}.\)
The characteristics equation is \(\displaystyle{\left[{\left|{A}\ -\lambda{I}\right|}={0}\right]},\)
\(\Rightarrow \begin{vmatrix}- \lambda & 1 \\ -1 & - \lambda \end{vmatrix}=0\)
\(\displaystyle\Rightarrow\lambda^{{{2}}}\ +\ {1}={0}\)
\(\displaystyle\Rightarrow\lambda^{{{2}}}=\ -{1}\)
\(\displaystyle\Rightarrow\lambda=\sqrt{{-{1}}}\)
\(\displaystyle\Rightarrow\lambda=\pm\ {i}\)
Since \(\displaystyle{\left[\lambda=\mu\ +\ {i}\nu\ {\quad\text{and}\quad}\ \mu={0}\right]},\) it can be concluded the critical point \(\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}\) is a center point of the corresponding linear system.
b)
Formula used: Conditions for the almost linear system: \(\displaystyle{\left[{\frac{{{g}_{{{1}}}\ {\left({x},\ {y}\right)}}}{{{r}}}}\rightarrow{0},{\frac{{{g}_{{{2}}}\ {\left({x},\ {y}\right)}}}{{{r}}}}\rightarrow{0}\ {a}{s}\ {r}\rightarrow{0}\right]}.\)
Propf:
Consider the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]},\)
The equivalent system is,
\([\frac{d}{dt}\ \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0 & 1\\ -1 & 0\end{array}\right)\ \left(\begin{array}{c}x\\ y\end{array}\right)\ +\ \left(\begin{array}{c}x\left(x^{2}\ +\ y^{2}\right)\\ y\left(x^{2}\ +\ y^{2}\right)\end{array}\right)]\)
Comparing this system with \(\displaystyle{\left[{x}^{{{p}{r}{i}{m}{e}}}={A}{x}\ +\ {g{{\left({x}\right)}}}\right]}\) gives,
\([g(x)=\left(\begin{array}{c}g_{1}\ \left(x,\ y\right)\\ g_{2}\ \left(x,\ y\right)\end{array}\right)= \left(\begin{array}{c}x\left(x^{2}\ +\ y^{2}\right)\\ y\left(x^{2}\ +\ y^{2}\right)\end{array}\right)]\)
By using polar coordinates \(\displaystyle{\left[{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},\right]}\)
\(\displaystyle{\frac{{{{g}_{{{1}}}{\left({x},\ {y}\right)}}}}{{{r}}}}={\frac{{{x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}}}{{{r}}}}={\frac{{{r}{\cos{\theta}}{\left({\left({r}{\cos{\theta}}\right)}^{{{2}}}\ +\ {\left({r}{\sin{\theta}}\right)}^{{{2}}}\right)}}}{{{r}}}}\)
\(\displaystyle={\cos{\theta}}{\left({r}^{{{2}}}{{\cos}^{{{2}}}\theta}\ +\ {r}^{{{2}}}{{\sin}^{{{2}}}\theta}\right)}\)
\(\displaystyle={r}^{{{2}}}{\cos{\theta}}{\left({{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}\right)}\)
As \(\displaystyle{{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}={1},\)
\(\displaystyle={r}^{{{2}}}{\cos{\theta}}\rightarrow\ {0}\ {a}{s}\ {r}\rightarrow\ {0}\)
Similarly,
\(\displaystyle{\frac{{{{g}_{{{2}}}{\left({x},\ {y}\right)}}}}{{{r}}}}={\frac{{{y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}}}{{{r}}}}={\frac{{{r}{\cos{\theta}}{\left({\left({r}{\cos{\theta}}\right)}^{{{2}}}\ +\ {\left({r}{\sin{\theta}}\right)}^{{{2}}}\right)}}}{{{r}}}}\)
\(\displaystyle={\sin{\theta}}{\left({r}^{{{2}}}{{\cos}^{{{2}}}\theta}\ +\ {r}^{{{2}}}{{\sin}^{{{2}}}\theta}{r}{i}{g}{h}{t}\right.}\)
\(\displaystyle{\left[={r}^{{{2}}}{\sin{\theta}}{\left({{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}\right)}\right]}\)
\(\displaystyle{\left[{A}{s}\ {{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}={1},\right]}\)
\(\displaystyle{\left[={r}^{{{2}}}{\sin{\theta}}\rightarrow\ {0}\ {a}{s}\ {r}\rightarrow\ {0}\right]}\)
Hence, the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\) is almost a linear system.
Now, consider the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]},\)
The equivalent system is,
\([\frac{d}{dt}\ \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0 & 1\\ -1 & 0\end{array}\right)\ \left(\begin{array}{c}x\\ y\end{array}\right)\ +\ \left(\begin{array}{c}-x\ (x^{2}\ +\ y^{2})\\ -y\ (x^{2}\ +\ y^{2})\end{array}\right)]\)
Comparing this system with \(\displaystyle{\left[{x}^{{{p}{r}{i}{m}{e}}}={A}{x}\ +\ {g{{\left({x}\right)}}}\right]}\) gives,
\(\displaystyle{\left[{g}\ {\left({x}\right)}={\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{g}_{{{1}}}\ {\left({x},\ {y}\right)}\backslash{g}_{{{2}}}\ {\left({x},\ {y}\right)}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}={\left({b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}-{x}\ {\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\backslash-{y}\ {\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\right)}\right]}\)
By using polar coordinates \(\displaystyle{\left[{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},\right]}\)
\(\displaystyle{\frac{{{{g}_{{{1}}}{\left({x},\ {y}\right)}}}}{{{r}}}}={\frac{{-{x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}}}{{{r}}}}={\frac{{-{r}{\cos{\theta}}{\left({\left({r}{\cos{\theta}}\right)}^{{{2}}}\ +\ {\left({r}{\sin{\theta}}\right)}^{{{2}}}\right)}}}{{{r}}}}\)
\(\displaystyle=\ -{\cos{\theta}}{\left({r}^{{{2}}}{{\cos}^{{{2}}}\theta}\ +\ {r}^{{{2}}}{{\sin}^{{{2}}}\theta}\right)}\)
\(\displaystyle=\ -{r}^{{{2}}}{\cos{\theta}}{\left({{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}\right)}\)
As \(\displaystyle{{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}={1},\)
\(\displaystyle=\ -{r}^{{{2}}}{\cos{\theta}}\rightarrow\ {0}\ {a}{s}\ {r}\rightarrow\ {0}\)
Similarly,
\(\displaystyle{\frac{{{{g}_{{{2}}}{\left({x},\ {y}\right)}}}}{{{r}}}}={\frac{{-{y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}}}{{{r}}}}={\frac{{-{r}{\cos{\theta}}{\left({\left({r}{\cos{\theta}}\right)}^{{{2}}}\ +\ {\left({r}{\sin{\theta}}\right)}^{{{2}}}\right)}}}{{{r}}}}\)
\(\displaystyle=\ -{\sin{\theta}}{\left({r}^{{{2}}}{{\cos}^{{{2}}}\theta}\ +\ {r}^{{{2}}}{{\sin}^{{{2}}}\theta}\right)}\)
\(\displaystyle=\ -{r}^{{{2}}}{\sin{\theta}}{\left({{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}\right)}\)
As \(\displaystyle{{\cos}^{{{2}}}\theta}\ +{{\sin}^{{{2}}}\theta}={1},\)
\(\displaystyle=\ -{r}^{{{2}}}{\sin{\theta}}\rightarrow\ {0}\ {a}{s}\ {r}\rightarrow\ {0}\)
Hence, the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}\ {\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}\) is almost a linear system.
c)
Formula used:
\(\displaystyle{\left[{r}^{{{2}}}={x}^{{{2}}}\ +\ {y}^{{{2}}}\right]}\)
Proof:
For the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}\ {\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]},\)
By using \(\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},\)
\(\displaystyle\Rightarrow\ {r}^{{{2}}}={x}^{{{2}}}\ +\ {y}^{{{2}}}\)
Differentiating both sides with respect to t,
\(\displaystyle\Rightarrow\ {2}{r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={2}{x}\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\ +\ {2}{y}\ {\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\)
\(\displaystyle\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={x}\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\ +\ {y}\ {\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\)
\(\displaystyle\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={x}{\left({y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right)}\ +\ {y}\ {\left(-{x}\ -\ {y}\ {\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right)}\)
\(\displaystyle{\left[\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={x}{y}\ -\ {x}^{{{2}}}\ {\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\ -\ {x}{y}\ -\ {y}^{{{2}}}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right.}\)
\(\displaystyle\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={x}{y}\ -\ {x}^{{{4}}}\ -\ {x}^{{{2}}}\ {y}^{{{2}}}\ -\ {x}{y}\ -\ {x}^{{{2}}}\ {y}^{{{2}}}\ -{y}^{{{4}}}\)
\(\displaystyle\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}=-{x}^{{{4}}}\ -\ {2}{x}^{{{2}}}{y}^{{{2}}}\ -\ {y}^{{{4}}}\)
\(\displaystyle\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}=\ -{\left({x}^{{{4}}}\ +\ {2}{x}^{{{2}}}{y}^{{{2}}}\ +\ {y}^{{{4}}}\right)}\)
\(\displaystyle\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}=\ -{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}^{{{2}}}\)
As \(\displaystyle{x}^{{{2}}}\ +\ {y}^{{{2}}}={r}^{{{2}}}\)
\(\displaystyle\Rightarrow\ {r}\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}=\ -{r}^{{{4}}}\)
\(\displaystyle\Rightarrow\ {\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}=\ -{r}^{{{3}}}\ {<}\ {0}\)
\(\displaystyle\Rightarrow\ {r}^{{-{3}}}\ {d}{r}=\ -{\left.{d}{t}\right.}\)
Integrating on both sides, \(\displaystyle\Rightarrow\ {\frac{{{r}^{{-{2}}}}}{{-{2}}}}=\ -{t}\ +\ {c}\)
\(\displaystyle\Rightarrow\ {\frac{{{1}}}{{-{2}{r}^{{{2}}}}}}=\ -{t}\ +\ {c}\)
\(\displaystyle\Rightarrow\ {\frac{{{1}}}{{{r}^{{{2}}}}}}={2}\ {t}\ -\ {2}\ {c}\)
\(\displaystyle\Rightarrow\ {\frac{{{1}}}{{{r}^{{{2}}}}}}={2}\ {t}\ +\ {C}\ {w}{h}{e}{r}{e}\ {C}=\ -{2}{c}\)
\(\displaystyle\Rightarrow\ {r}^{{{2}}}={\frac{{{1}}}{{{2}{t}\ +\ {C}}}}\)
Hence, the origin is an asymptotically stable equilibrium point.
Now,
For the system \(\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}\ {\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]},\)
By using \(\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},\)

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we have to sketch the graph f(y) versus y, determine the critical points, and classify each one as asymptotically stable or unstable.Thing is, how do you get the critical points?
asked 2021-03-10

A system of linear equations is given below.
\(2x+4y=10\)
\(\displaystyle-{\frac{{{1}}}{{{2}}}}{x}+{3}={y}\)
Find the solution to the system of equations.
A. (0, -3)
B. (-6, 0)
C. There are infinite solutions.
D. There are no solutions.

asked 2021-05-12
4.7 A multiprocessor with eight processors has 20attached tape drives. There is a large number of jobs submitted tothe system that each require a maximum of four tape drives tocomplete execution. Assume that each job starts running with onlythree tape drives for a long period before requiring the fourthtape drive for a short period toward the end of its operation. Alsoassume an endless supply of such jobs.
a) Assume the scheduler in the OS will not start a job unlessthere are four tape drives available. When a job is started, fourdrives are assigned immediately and are not released until the jobfinishes. What is the maximum number of jobs that can be inprogress at once? What is the maximum and minimum number of tapedrives that may be left idle as a result of this policy?
b) Suggest an alternative policy to improve tape driveutilization and at the same time avoid system deadlock. What is themaximum number of jobs that can be in progress at once? What arethe bounds on the number of idling tape drives?
asked 2021-03-22
A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant but starts at .100 at P and increases linerly with distance past P, reaching a value of .600 at 12.5 m past point P. (a) Use the work energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid iff the friciton coefficient didn't increase, but instead had the constant value of .1?
asked 2021-05-16
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
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