In each of Problems l-3, verify that \(x(t),y(t)\) is a solution of the given system of equations, and find its orbit. \begin{align*} \dot{x} & =1\\ \dot{y} & =2\left ( 1-x\right ) \sin \left ( \left ( 1-x\right ) ^{2}\right ) \\ x\left ( t\right ) & =1+t\\ y\left ( t\right ) & =\cos \left ( t^{2}\right ) \end{align*}
solution
Since \(x\left ( t\right ) =1+t\) then \(\dot{x}=1\). Verified OK. And since \(y\left ( t\right ) =\cos \left ( t^{2}\right ) \) then \(\dot{y}=-2t\sin \left ( t^{2}\right ) \). But \(t=x-1\), hence \(\dot{y}=-2\left ( x-1\right ) \sin \left ( \left ( 1-x\right ) ^{2}\right ) \) or \[ \dot{y}=2\left ( 1-x\right ) \sin \left ( \left ( 1-x\right ) ^{2}\right ) \] Verified OK. Both solutions verified. Now we need to find system orbit. The Orbit is given by the equation \[ \frac{dy}{dx}=\frac{g\left ( x,y\right ) }{f\left ( x,y\right ) }\] When we write the given system in the following form\begin{align*} \dot{x} & =f\left ( x,y\right ) \\ \dot{y} & =g\left ( x,y\right ) \end{align*}
We see now that \(f\left ( x,y\right ) =1\) and \(g\left ( x,y\right ) =2\left ( 1-x\right ) \sin \left ( \left ( 1-x\right ) ^{2}\right ) \). Therefore\begin{align*} \frac{dy}{dx} & =\frac{2\left ( 1-x\right ) \sin \left ( \left ( 1-x\right ) ^{2}\right ) }{1}\\ & =2\left ( 1-x\right ) \sin \left ( \left ( 1-x\right ) ^{2}\right ) \end{align*}
This is first order ODE. Since separable, we can solve it by integration\[ y\left ( x\right ) =\int 2\left ( 1-x\right ) \sin \left ( \left ( 1-x\right ) ^{2}\right ) dx \] Let \(u=\left ( 1-x\right ) ^{2}\), then \(\frac{du}{dx}=2\left ( 1-x\right ) \left ( -1\right ) =-2\sqrt{u}\). Substituting in the above gives\begin{align*} y\left ( x\right ) & =\int 2\sqrt{u}\sin \left ( u\right ) \frac{du}{-2\sqrt{u}}\\ & =-\int \sin \left ( u\right ) du\\ & =-\left ( -\cos \left ( u\right ) \right ) +C\\ & =\cos \left ( u\right ) +C\\ & =\cos \left ( \left ( 1-x\right ) ^{2}\right ) +C \end{align*}
Therefore the equation of the orbit is \[ y\left ( x\right ) =\cos \left ( \left ( 1-x\right ) ^{2}\right ) +C \] For different values of \(C\), different orbit results.
In each of Problems l-3, verify that \(x(t),y(t)\) is a solution of the given system of equations, and find its orbit. \begin{align*} \dot{x} & =e^{-x}\\ \dot{y} & =e^{e^{x}-1}\\ x\left ( t\right ) & =\ln \left ( 1+t\right ) \\ y\left ( t\right ) & =e^{t} \end{align*}
Solution
\begin{align*} \frac{dx}{dt} & =\frac{d}{dt}\ln \left ( 1+t\right ) \\ \dot{x} & =\frac{1}{1+t} \end{align*}
But \(e^{-x}=e^{-\ln \left ( 1+t\right ) }=\frac{1}{1+t}\). Verified OK. And \begin{align*} \frac{dy}{dt} & =\frac{d}{dt}e^{t}\\ \dot{y} & =e^{t} \end{align*}
But \(x-1=\ln \left ( 1+t\right ) -1\). Hence \(\ln \left ( 1+t\right ) =x\). Therefore \(1+t=e^{x}\) or \(t=e^{x}-1\). Therefore \(\dot{y}=e^{t}=e^{e^{x}-1}\). Verified OK.
Now we need to find system orbit. The Orbit is given by the equation \[ \frac{dy}{dx}=\frac{g\left ( x,y\right ) }{f\left ( x,y\right ) }\] When we write the given system in the following form\begin{align*} \dot{x} & =f\left ( x,y\right ) \\ \dot{y} & =g\left ( x,y\right ) \end{align*}
We see now that \(f\left ( x,y\right ) =e^{-x}\) and \(g\left ( x,y\right ) =e^{e^{x}-1}\). Therefore\[ \frac{dy}{dx}=\frac{e^{e^{x}-1}}{e^{-x}}\] Integrating\[ \int dy=\int \frac{e^{e^{x}-1}}{e^{-x}}dx \] Let \(e^{x}=u,du=e^{x}dx\). Hence the RHS \(\int \frac{e^{e^{x}-1}}{e^{-x}}dx=\int \frac{e^{u-1}}{\frac{1}{u}}\frac{du}{u}=\int e^{u-1}du=e^{u-1}=e^{e^{x}-1}\). The above becomes\[ y=e^{e^{x}-1}+C \] The orbits are given by the above equation for different \(C\)
In each of Problems l-3, verify that \(x(t),y(t)\) is a solution of the given system of equations, and find its orbit. \begin{align*} \dot{x} & =1+x^{2}\\ \dot{y} & =\left ( 1+x^{2}\right ) \sec ^{2}x\\ x\left ( t\right ) & =\tan t\\ y\left ( t\right ) & =\tan \left ( \tan t\right ) \end{align*}
solution
Orbits given by\begin{align*} \frac{dy}{dx} & =\frac{\left ( 1+x^{2}\right ) \sec ^{2}x}{1+x^{2}}\\ & =\sec ^{2}x \end{align*}
Hence \[ \int dy=\int \sec ^{2}xdx \] But \(\sec ^{2}x=\frac{1}{\cos ^{2}x}\). And \(\frac{d}{dx}\frac{\sin x}{\cos x}=\frac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=\frac{1}{\cos ^{2}x}\). Hence \(\int \sec ^{2}xdx=\tan x\). Therefore the above gives\[ y=\tan x+C \] The orbits are given by the above equation for different \(C\). (do not know why book gives only \(y=\tan x\))
Find the orbits of each of the following systems\begin{align*} \dot{x} & =y+x^{2}y\\ \dot{y} & =3x+xy^{2} \end{align*}
Solution
The Orbit is given by the equation \[ \frac{dy}{dx}=\frac{g\left ( x,y\right ) }{f\left ( x,y\right ) }\] When we write the given system in the following form\begin{align*} \dot{x} & =f\left ( x,y\right ) \\ \dot{y} & =g\left ( x,y\right ) \end{align*}
We see now that \(f\left ( x,y\right ) =y+x^{2}y\) and \(g\left ( x,y\right ) =3x+xy^{2}\). Therefore\begin{align*} \frac{dy}{dx} & =\frac{3x+xy^{2}}{y+x^{2}y}\\ & =\frac{x\left ( 3+y^{2}\right ) }{y\left ( 1+x^{2}\right ) }\\ & =\frac{x}{\left ( 1+x^{2}\right ) }\frac{\left ( 3+y^{2}\right ) }{y} \end{align*}
Hence it is separable. \begin{align*} \int \frac{y}{3+y^{2}}dy & =\int \frac{x}{1+x^{2}}dx\\ \frac{1}{2}\ln \left ( 3+y^{2}\right ) & =\frac{1}{2}\ln \left ( 1+x^{2}\right ) +C_{2}\\ \ln \left ( 3+y^{2}\right ) & =\ln \left ( 1+x^{2}\right ) +C_{1} \end{align*}
Therefore\begin{align*} 3+y^{2} & =e^{\ln \left ( 1+x^{2}\right ) +C_{1}}\\ & =e^{C_{1}}e^{\ln \left ( 1+x^{2}\right ) }\\ & =C\left ( 1+x^{2}\right ) \end{align*}
Hence\begin{align*} y^{2} & =C\left ( 1+x^{2}\right ) -3\\ y\left ( x\right ) & =\pm \sqrt{C\left ( 1+x^{2}\right ) -3} \end{align*}
The above gives the equations for the orbit. For each \(C\) value, there is a different orbit curve. Now we need to find equilibrium points, since these are orbits also. We need to solve\begin{align*} 0 & =y+x^{2}y\\ 0 & =3x+xy^{2} \end{align*}
Or\begin{align*} 0 & =y\left ( 1+x^{2}\right ) \\ 0 & =x\left ( 3+y^{2}\right ) \end{align*}
First equation gives \(y=0\) as only real solution. When \(y=0\) then second equation gives \(x=0\). Hence \(\left ( 0,0\right ) \) is also an orbit. So the orbits are\begin{align*} y^{2} & =C\left ( 1+x^{2}\right ) -3\qquad C\neq 3\\ \left ( x,y\right ) & =\left ( 0,0\right ) \end{align*}
And when \(C=3\) we obtain orbits \(y^{2}=3\left ( 1+x^{2}\right ) -3=3x^{2}\), with additional orbits (notice that we have to exclude \(x=0\) from each one below, since \(x=0\) is allready included in \(\left ( x,y\right ) =\left ( 0,0\right ) \))
\begin{align*} y & =\sqrt{3}x\qquad x>0\\ y & =\sqrt{3}x\qquad x<0\\ y & =-\sqrt{3}x\qquad x>0\\ y & =-\sqrt{3}x\qquad x<0 \end{align*}
Hence there are 6 possible orbits in total.
Draw the phase portraits of each of the following systems of differential equations\[ \mathbf{\dot{x}=}\begin{pmatrix} 4 & -1\\ -2 & 5 \end{pmatrix} \mathbf{x}\] solution\begin{align*} \det \left ( A-\lambda I\right ) & =0\\\begin{vmatrix} 4-\lambda & -1\\ -2 & 5-\lambda \end{vmatrix} & =0\\ \left ( 4-\lambda \right ) \left ( 5-\lambda \right ) -2 & =0\\ \lambda ^{2}-9\lambda +18 & =0 \end{align*}
Hence \begin{align*} \lambda _{1} & =6\\ \lambda _{2} & =3 \end{align*}
Case \(\lambda _{1}=6\)\begin{align*} \begin{pmatrix} 4-\lambda & -1\\ -2 & 5-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 4-6 & -1\\ -2 & 5-6 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} -2 & -1\\ -2 & -1 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From first row \(-2v_{1}-v_{2}=0\). Hence \(v_{2}=-2v_{1}\). Therefore the first eigenvector is \(\mathbf{v}^{1}=\begin{pmatrix} v_{1}\\ -2v_{1}\end{pmatrix} =v_{1}\begin{pmatrix} 1\\ -2 \end{pmatrix} =\begin{pmatrix} 1\\ -2 \end{pmatrix} \) by setting \(v_{1}=1\)
Case \(\lambda _{1}=3\)
\begin{align*} \begin{pmatrix} 4-\lambda & -1\\ -2 & 5-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 4-3 & -1\\ -2 & 5-3 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 1 & -1\\ -2 & 2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From first row \(v_{1}-v_{2}=0\). Hence \(v_{2}=v_{1}\). Therefore the second eigenvector is \(\mathbf{v}^{2}=\begin{pmatrix} v_{1}\\ v_{1}\end{pmatrix} =v_{1}\begin{pmatrix} 1\\ 1 \end{pmatrix} =\begin{pmatrix} 1\\ 1 \end{pmatrix} \) by setting \(v_{1}=1\)
Since eigenvalues are both real and both are positive, then \(\left ( 0,0\right ) \) is unstable node. Here is a the Phase portrait. The lines marked red and blue are the two eigenvectors found above. The arrows are all leaving \(\left ( 0,0\right ) \) which means this is unstable equilibrium point.
Draw the phase portraits of each of the following systems of differential equations\[ \mathbf{\dot{x}=}\begin{pmatrix} 3 & -1\\ 5 & -3 \end{pmatrix} \mathbf{x}\] Solution
\begin{align*} \det \left ( A-\lambda I\right ) & =0\\\begin{vmatrix} 3-\lambda & -1\\ 5 & -3-\lambda \end{vmatrix} & =0\\ \left ( 3-\lambda \right ) \left ( -3-\lambda \right ) +5 & =0\\ \lambda ^{2}-4 & =0 \end{align*}
Hence \begin{align*} \lambda _{1} & =2\\ \lambda _{2} & =-2 \end{align*}
We see that one eigenvalue is stable and one is not stable.
Case \(\lambda _{1}=2\)\begin{align*} \begin{pmatrix} 3-\lambda & -1\\ 5 & -3-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 3-2 & -1\\ 5 & -3-2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 1 & -1\\ 5 & -5 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From first row \(v_{1}-v_{2}=0\). Hence \(v_{2}=v_{1}\). Therefore the first eigenvector is \(\mathbf{v}^{1}=\begin{pmatrix} v_{1}\\ v_{1}\end{pmatrix} =v_{1}\begin{pmatrix} 1\\ 1 \end{pmatrix} =\begin{pmatrix} 1\\ 1 \end{pmatrix} \) by setting \(v_{1}=1\)
Case \(\lambda _{1}=-2\)\begin{align*} \begin{pmatrix} 3-\lambda & -1\\ 5 & -3-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 3+2 & -1\\ 5 & -3+2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 5 & -1\\ 5 & -1 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From first row \(5v_{1}-v_{2}=0\). Hence \(v_{2}=5v_{1}\). Therefore the first eigenvector is \(\mathbf{v}^{1}=\begin{pmatrix} v_{1}\\ 5v_{1}\end{pmatrix} =v_{1}\begin{pmatrix} 1\\ 5 \end{pmatrix} =\begin{pmatrix} 1\\ 5 \end{pmatrix} \) by setting \(v_{1}=1\).
Since one eigenvalue is stable and one is not, then \(\left ( 0,0\right ) \) is unstable saddle point. Here is a the Phase portrait. The lines marked red and blue are the two eigenvectors found above.
Draw the phase portraits of each of the following systems of differential equations\[ \mathbf{\dot{x}=}\begin{pmatrix} 2 & 1\\ -5 & -2 \end{pmatrix} \mathbf{x}\] solution
\begin{align*} \det \left ( A-\lambda I\right ) & =0\\\begin{vmatrix} 2-\lambda & -1\\ -5 & -2-\lambda \end{vmatrix} & =0\\ \left ( 2-\lambda \right ) \left ( -2-\lambda \right ) +5 & =0\\ \lambda ^{2}+1 & =0 \end{align*}
Hence \begin{align*} \lambda _{1} & =i\\ \lambda _{2} & =-i \end{align*}
The real part is zero. Hence \(\left ( 0,0\right ) \) equilibrium point is called CENTER. it is stable, but not asymptotically stable.
Case \(\lambda _{1}=i\)\begin{align*} \begin{pmatrix} 2-\lambda & 1\\ -5 & -2-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 2-i & 1\\ -5 & -2-i \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From second row \(-5v_{1}-\left ( 2+i\right ) v_{2}=0\). Hence \(v_{2}=-\frac{5}{\left ( 2+i\right ) }v_{1}\). Therefore the first eigenvector is \(\mathbf{v}^{1}=\begin{pmatrix} v_{1}\\ -\frac{5}{\left ( 2+i\right ) }v_{1}\end{pmatrix} =v_{1}\begin{pmatrix} 1\\ -\frac{5}{\left ( 2+i\right ) }\end{pmatrix} =\) \(\begin{pmatrix} -\left ( 2+i\right ) \\ 5 \end{pmatrix} \) by setting \(v_{1}=1\)
Case \(\lambda _{1}=-i\)\begin{align*} \begin{pmatrix} 2-\lambda & 1\\ -5 & -2-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 2+i & 1\\ -5 & -2+i \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From second row \(-5v_{1}+\left ( -2+i\right ) v_{2}=0\). Hence \(v_{2}=-\frac{5}{\left ( -2+i\right ) }v_{1}\). Therefore the first eigenvector is \(\mathbf{v}^{1}=\begin{pmatrix} v_{1}\\ -\frac{5}{\left ( -2+i\right ) }v_{1}\end{pmatrix} =v_{1}\begin{pmatrix} 1\\ -\frac{5}{\left ( -2+i\right ) }\end{pmatrix} =\) \(\begin{pmatrix} -2+i\\ 5 \end{pmatrix} \) by setting \(v_{1}=1\)
\(\left ( 0,0\right ) \) equilibrium point is called CENTER with curves making closed circles around \(\left ( 0,0\right ) \) as shown below
