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
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.

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 $$\displaystyle{x}={r}{\cos{\theta}},\ {y}={r}{\sin{\theta}},$$

Relevant Questions

The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius $$\displaystyle{R}={7.4}\times{10}^{{-{15}}}$$ m. In nuclear fission, this can divide into twosmaller nuclei, each of 1/2 the charge and 1/2 the voume of theoriginal uranium-236 nucleus. This is one of the reactionsthat occurred n the nuclear weapon that exploded over Hiroshima, Japan in August 1945.
A. Find the radii of the two "daughter" nuclei of charge+46e.
B. In a simple model for the fission process, immediatelyafter the uranium-236 nucleus has undergone fission the "daughter"nuclei are at rest and just touching. Calculate the kineticenergy that each of the "daughter" nuclei will have when they arevery far apart.
C. In this model the sum of the kinetic energies of the two"daughter" nuclei is the energy released by the fission of oneuranium-236 nucleus. Calculate the energy released by thefission of 10.0 kg of uranium-236. The atomic mass ofuranium-236 is 236 u, where 1 u = 1 atomic mass unit $$\displaystyle={1.66}\times{10}^{{-{27}}}$$ kg. Express your answer both in joules and in kilotonsof TNT (1 kiloton of TNT releases 4.18 x 10^12 J when itexplodes).
Assume that a ball of charged particles has a uniformly distributednegative charge density except for a narrow radial tunnel throughits center, from the surface on one side to the surface on the opposite side. Also assume that we can position a proton any where along the tunnel or outside the ball. Let $$\displaystyle{F}_{{R}}$$ be the magnitude of the electrostatic force on the proton when it islocated at the ball's surface, at radius R. As a multiple ofR, how far from the surface is there a point where the forcemagnitude is 0.44FR if we move the proton(a) away from the ball and (b) into the tunnel?
The student engineer of a campus radio station wishes to verify the effectivencess of the lightning rod on the antenna mast. The unknown resistance $$\displaystyle{R}_{{x}}$$ is between points C and E. Point E is a "true ground", but is inaccessible for direct measurement because the stratum in which it is located is several meters below Earth's surface. Two identical rods are driven into the ground at A and B, introducing an unknown resistance $$\displaystyle{R}_{{y}}$$. The procedure for finding the unknown resistance $$\displaystyle{R}_{{x}}$$ is as follows. Measure resistance $$\displaystyle{R}_{{1}}$$ between points A and B. Then connect A and B with a heavy conducting wire and measure resistance $$\displaystyle{R}_{{2}}$$ between points A and C.Derive a formula for $$\displaystyle{R}_{{x}}$$ in terms of the observable resistances $$\displaystyle{R}_{{1}}$$ and $$\displaystyle{R}_{{2}}$$. A satisfactory ground resistance would be $$\displaystyle{R}_{{x}}{<}{2.0}$$ Ohms. Is the grounding of the station adequate if measurments give $$\displaystyle{R}_{{1}}={13}{O}{h}{m}{s}$$ and R_2=6.0 Ohms?
The graph of y = f(x) contains the point (0,2), $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{-{x}}}{{{y}{e}^{{{x}^{{2}}}}}}}$$, and f(x) is greater than 0 for all x, then f(x)=
A) $$\displaystyle{3}+{e}^{{-{x}^{{2}}}}$$
B) $$\displaystyle\sqrt{{{3}}}+{e}^{{-{x}}}$$
C) $$\displaystyle{1}+{e}^{{-{x}}}$$
D) $$\displaystyle\sqrt{{{3}+{e}^{{-{x}^{{2}}}}}}$$
E) $$\displaystyle\sqrt{{{3}+{e}^{{{x}^{{2}}}}}}$$
Car 1 has a mass of m1 = 65 ❝ 103 kg and moves at a velocity of v01 = +0.81 m/s. Car 2, with a mass of m2 = 92 ❝ 103 kg and a velocity of v02 = +1.2 m/s, overtakes car 1 and couples to it. Neglect the effects of friction in your answer.
(a) Determine the velocity of their center of mass before the collision m/s
(b) Determine the velocity of their center of mass after the collision m/s
(c) Should your answer in part (b) be less than, greater than, or equal to the common velocity vf of the two coupled cars after the collision? less than greater than equal to
The problem question is:
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={a}{y}+{b}{y}^{{2}}$$
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?

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.

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?