1)Verify that the given function pair is a solution to the first-order system. \(x=-e^t\), \(y=e^t\) \(\frac{dx}{dt}=-y\), \(\frac{dy}{dt}=-x\)
\(\frac{dx}{dt}=-e^t\) \(\frac{dx}{dt}=-y\) \(y=e^t\)
\(\frac{dy}{dt}=e^t\) \(\frac{dy}{dt}=-x\) \(x=-e^t\)
\(x=-e^t\), \(y=e^t\) is a solution to the fist order system
\(\frac{dx}{dt}=-y,\frac{dy}{dt}=-x\)
First derivative of the function: \(x=\frac{dx}{dt}\), \(y=\frac{dy}{dt}\)
Rest point of the autonomous system \(\frac{dx}{dt}=0\), \(\frac{dy}{dt}=0\)
\(-(y-1)=0\)
\(y=1\)
\(x-2=0\)
\(x=2\) The rest point of the autonomous system is (2,1)
Show that the two trajectories leading to \((m/n, a/b)\) shown in Figure 12.8 are unique.
\[\frac{dy}{dx}=\frac{(m-nx)y}{(a-by)x}\]
\(\frac{dt}{dx}=(\frac{dx}{dt})^-1=\frac{1}{(a-by)x}\)
\(\frac{dy}{dx}=\frac{dy}{dt}\)x\(\frac{dt}{dx}=\frac{(m-nx)y}{(a-by)x}\)
Separate variables, integrate, and exponentiate to obtain \[y^ae^{-by}=Kx^me^-nx\]
\(\frac{(a-by)}{y}dy=\frac{(m-nx)}{x}dx\)
\(\int \frac{(a-by)}{y}dy=\int \frac{m-mx}{x}dx\)
\(\int (\frac{(a)}{y}-b)dy=\int(\frac{m}{x}-n)dx\)
\(a\)ln\(|y|-by=m\)ln\(|x|-nx+k_1\)
ln\(\frac{|y|^a}{|x|^b}=by-nx+k_1\)
\(\frac{y^a}{x^b}=e^{by-mx+k_1}\)
\(y^ae^{-by}=x^me^{-nx}e^{k_1}\)
\(y^ae{-by}=Kx^me^{-nx}\)
Let \(f(y)=y^a/e^{by}\) and \(g(x)=x^m/e^{nx}\). Show that \(f(y)\) has a unique maximum of \(M_y=(a/eb)^a\) when \(y=a/b\) as shown in Figure 12.12. Similarly, show that \(g(x)\) has a unique maximum \(M_x=(x/en)^m\) when \(x=m/n\)
\(f^1 = 0\) to find the maxima
\(\frac{df(y)}{dy}=\frac{d\frac{y^a}{e^{by}}}{dy}\)
\(\frac{e^{by}ay^{a-1}-y^abe^{by}}{e^{2by}}=0\)
\(e^{by}y^{a-1}(a-yb)=0\)
\(y=\frac{a}{b}\) or \(y=0\)
\(\frac{d^2f(y)}{dy^2}=\frac{d\frac{ay^{a-1}-y^ab}{e^{by}}}{dy}\)
\(\frac{e^{by}{a(a-1)y^{a-2}-ay^{a-1}}-(ay^{a-1}-y^q)be^{by}}{e^{2by}}\)
\(\frac{d^2f(y)}{dy^2}=\frac{e^a(0-0)-(0-0)be^a}{e^{2a}}=0\)
\(\frac{d^2f(y)}{dy^2}=\frac{(\frac{a}{b})^{a-2}x-a}{e^a}\)
Since a and b are positive constants, \(f^{11}\le0\)
The calue of maximum is: \(M_y=(\frac{a}{eb})^a\)
\[\frac{y^a}{e^{by}}=(\frac{M_y}{M_x}) (\frac{x^m}{e^{nx}})\]
\(lim_{y\rightarrow \frac{a}b}[y^ae^{-by}]=lim_{x\rightarrow \frac{m}{n}}K[x^me{-nx}]\)
\(\frac{lim_{y\rightarrow\frac{a}b}[y^ae^{-by}]}{lim_x\rightarrow[x^me{-nx}]}=K\)
\(lim_{y\rightarrow \frac{a}b}[y^ae^{-by}]\{lim_{x\rightarrow \frac{m}n}[x^me^{-nx}]\}^{-1}=K\)
\(lim_{y\rightarrow \frac{a}b}[y^ae^{-by}]lim_{x\rightarrow \frac{m}n[x^{-m}e^{nx}]=K}\)
\(lim_{y\rightarrow \frac{m}n}lim_{y\rightarrow \frac{a}b}([y^ae^{-by}][x^{-m}e^{nx}])=K\)
\(lim_{y\rightarrow \frac{a}b\ _x\rightarrow \frac{m}n}([y^ae^{-by}][x^{-m}e^{nx}])=K\)
\((\frac{a}{eb})^a(\frac{en}{x})^m=K\), \(\frac{M_y}{m_x}=K\)
\((\frac{y^a}e^{by})=(\frac{M_y}M_x)(\frac{x^m}{e^{nx}})\)
The funcation is decreaasing in the range
\((\frac{a}{b},\infty),f(y_0) < M_y\)
This implies: \(\frac{M_y}{M_x}(\frac{x^m}{e^{nx}})=\frac{y_0^a}{e^{by_0}} < M_y\)
\(\frac{M_y}{M_x}\frac{x^m}{e^{nx}} \lt M_y\)
Impiles \(\frac{x^m}{e^{nx}} \lt M\)
Apply the first and second derivative tests to the function \(f(y)=y^a/e^{by}\) to show that \(y=a/b\) is a unique critical point that yields the relative maximum \(f(a/b)\). Show also that \(f(y)\) approaches zero as \(y\) tends to infinity.
Set \(f^1=0\) to find the maximum \(e^{by}y^{a-1}(a-yb)=0\)
\(y=\frac{a}b\space or\space y=0\)
Second derivative test \(f^{11}=\frac{e^{by}\left\{a(a-1)y^{a-2}-ay^{a-1}\right\}-(ay^{a-1}-y^a)be^{by}}{e^{2by}}\)
\(\frac{e^a\left\{0-0\right\}-(0-0)be^{by}}{e^{2a}}=0\)
\(\frac{(\frac{a}{b})^{a-2}x-a}{e^a}\)
Since \(a\) and \(b\) are positive, \(f^{11}\lt0\), \(y=\frac{a}b\) is a unique critical point and relative maximum = \(M_y=(\frac{a}{eb})^a\)
Limit The limit \(lim_{y\rightarrow\infty\frac{y^a}{e^{by}}}\) approaches zero does ot satisfy a useful limit.
Apply the LaHopitalas rule \(lim_{y\rightarrow\infty}\frac{ay^{a-1}}{be^{by}}\)
\(lim_{y\rightarrow\infty}\frac{a!}{b^{a+1}e^{b(\infty)}}\)
\(\frac{a!}{b^{a+1}(\infty)}=0\)
Proves that as \(y\rightarrow \infty \space f(y)\rightarrow0\)