Homework Week 2 Solutions

P75: 1, 3; P76: 22.
P88: 3; P89: 16, 17.
P99: 3; P100: 12, 21; P101: 28.

Solutions posted 20110225.

page 88 number 3
Critical points are when $\frac{dy}{dt}=0$.
Thus\[
y(y-1)(y-2)=0\]
\[
y=0\mbox{ or }y-1=0\mbox{ or }y-2=0\]
\[
y=0\mbox{ or }y=1\mbox{ or }y=2\]
Between $0$ and $1$, we choose the test point $\frac{1}{2}$ to
see that $\frac{dy}{dt}>0$ for $y\in(0,1)$. Between $1$ and $2$
we choose the test point $\frac{3}{2}$ to see that $\frac{dy}{dt}<0$ for $y\in(1,2)$. And finally we choose the test point $\frac{5}{2}$ to see that $\frac{dy}{dt}>0$ for $y\in(1,\infty)$. We conclude
then that $y=0$ is an unstable equilibrium, $y=1$ is asymptotically
stable, and $y=2$ is unstable.

page 89 number 16
Critical points\[
\frac{dy}{dt}=0\]
Then\[
ry\ln(K/y)=0\]
\[
y=0\mbox{ or }\ln(K/y)=0\]
\[
y=0\mbox{ or }K/y=1\]
\[
y=0\mbox{ or }y=K\]
Now\[
\frac{dy}{dt}(K/2)=r\frac{K}{2}\ln(2)>0\]
Thus\[
\frac{dy}{dt}>0,\quad y\in(0,K)\]
And\[
\frac{dy}{dt}(2K)=r(2K)\ln(1/2)<0\] Thus\[ \frac{dy}{dt}<0,\quad y\in(K,\infty)\] We conclude $y=0$ is an unstable equilibrium solution and $y=K$ is asymptotically stable. Now for part b, we compute \begin{alignat*}{1} \frac{d}{dt}\left(ry\ln(K/y)\right) & =r\frac{dy}{dt}\ln(K/y)+ry\frac{d}{dt}\left(\ln(K)-\ln(y)\right)\\ & =r\frac{dy}{dt}\ln(K/y)-ry\frac{1}{y}\frac{dy}{dt}\\ & =r\frac{dy}{dt}(\ln(K/y)-1)=0\end{alignat*} Then\[ \ln(K/y)=1\] Then\[ K/y=e\] So\[ y=K/e\] We plug in test points to determine if $y$ is concave up or down in $(0,K/e)$ and $(K/e,K)$ For part c we want to show $ry\ln(K/y)\geq r(1-y/K)y$ for $0 < y\leq K$. Because $r$ and $y$ are positive, this is the same as showing that $\ln(K/y)\geq (1-y/K)$. So consider the function $f(y)=\ln(K/y)-1+y/K=\ln K-\ln y-1+y/K$ Then $f^{\prime}(y)=-\frac{1}{y}+\frac{1}{K}$. We solve $f^{\prime}(y)=0$. To get $y=K$. We choose a test point $y=\frac{K}{2}$ to see that $f^{\prime}(y)<0$ for $y\in(0,K)$. Since $f(K)=0$ and $f^{\prime}(y)<0$ for $y\in(0,K)$, we conclude that $f(y)\geq0$ for $y\in(0,K]$. Thus $\ln(K/y)\geq(1-y/K)$. And finally $ry\ln(K/y)\geq r(1-y/K)y$. page 89 number 17
So we make the substitution $u=\ln(y/K)=\ln y-\ln K$
So $\frac{du}{dt}=\frac{1}{y}\frac{dy}{dt}$ and the equation becomes
$\frac{du}{dt}=-ru$.
Then $\ln u=-rt+C$. Then $u=C_{1}e^{-rt}$, where $C_{1}=e^{C}$
Then $\ln(y/K)=C_{1}e^{-rt}$. So $y=Ke^{C_{1}e^{-rt}}$.
$y_{0}=y(0)=Ke^{C_{1}}$. So $C_{1}=\ln(y_{0}/K)$
We conclude the solution to the initial value problem is $y=Ke^{\ln(y_{0}/K)e^{-rt}}$

page 99 number 3
$M(x,y)=3x^{2}-2xy+2$
$N(x,y)=6y^{2}-x^{2}+3$
$M_{y}=-2x$
$N_{x}=-2x$
Then $\psi=\int Mdx=x^{3}-x^{2}y+2x+h(y)$
$N=\psi_{y}=-x^{2}+h^{\prime}(y)$
Then $h^{\prime}(y)=6y^{2}+3$. So $h(y)=2y^{3}+3y+C$
We conclude $\psi=x^{3}-x^{2}y+2x+2y^{3}+3y+C$
And so solutions are given implicitly by $x^{3}-x^{2}y+2x+2y^{3}+3y=c$.
Alternatively $\psi=\int Ndy=2y^{3}-x^{2}y+3y+g(x)$
$M=\psi_{x}=-2xy+g^{\prime}(x)$
Then $g^{\prime}(x)=3x^{2}+2$. So $g(x)=x^{3}+2x+C$

page 100 number 12
$M=\frac{x}{\left(x^{2}+y^{2}\right)^{3/2}}$
$N=\frac{y}{\left(x^{2}+y^{2}\right)^{3/2}}$
$M_{y}=-\frac{3}{2}\frac{x}{\left(x^{2}+y^{2}\right)^{5/2}}\cdot2y$
$N_{x}=-\frac{3}{2}\frac{y}{\left(x^{2}+y^{2}\right)^{5/2}}\cdot2x$
$\psi=\int Mdx=\frac{1}{2}\int\frac{dx^{2}}{\left(x^{2}+y^{2}\right)^{3/2}}=\frac{-1}{\left(x^{2}+y^{2}\right)^{1/2}}+h(y)$
$N=\psi_{y}=\frac{y}{\left(x^{2}+y^{2}\right)^{3/2}}+h^{\prime}(y)$.
So $h^{\prime}(y)=0$. So $h(y)=C$.
We conclude the solution is given implicitly by $\frac{-1}{\left(x^{2}+y^{2}\right)^{1/2}}=c$.
Of course we can make this look nicer by manipulating the equation.
We can get $\left(\frac{-1}{c}\right)^{2}=x^{2}+y^{2}$
$C=x^{2}+y^{2}$.

page 100 number 21
Prior to multiplying by the integrating factor we get
$M_{y}=1$
$N_{x}=2$
So the equation is not exact. But after multiplying the equation by
$y$ we get
$y^{2}dx+(2xy-y^{2}e^{y})dy=0$
$M=y^{2}$
$N=2xy-y^{2}e^{y}$
$M_{y}=2y$
$N_{x}=2y$
So $\psi=\int Mdx=xy^{2}+h(y)$
Then $N=\psi_{y}=2xy+h^{\prime}(y)$
Then $h^{\prime}(y)=-y^{2}e^{y}$
So $h(y)=-\int y^{2}de^{y}=-y^{2}e^{y}+\int e^{y}2ydy=-y^{2}e^{y}+2ye^{y}-\int2e^{y}dy=-y^{2}e^{y}+2ye^{y}-2e^{y}$
Then the solution is given implicitly by $xy^{2}-e^{y}(y^{2}-2y+2)=c$

page 101 number 28
$ydx+(2xy-e^{-2y})dy=0$
I try $\frac{M_{y}-N_{x}}{N}=\frac{1-2y}{2xy-e^{-2y}}$
This doesn't look like it'll end up being just a function of $x$
So I try $\frac{N_{x}-M_{y}}{M}=\frac{2y-1}{y}$. Assuming $y\neq0$.
Then $\mu_{y}=\left(\frac{2y-1}{y}\right)\mu$. So then $\int\frac{1}{\mu}d\mu=\int\left(2-\frac{1}{y}\right)dy$
$\ln\mu=2y-\ln y$. So $\mu=e^{2y-\ln y}=e^{2y}e^{-\ln y}=e^{2y}e^{\ln y^{-1}}=e^{2y}y^{-1}$.
Then we get $e^{2y}dx+\left(2xe^{2y}-y^{-1}\right)dy=0$
We should now have exact equations. I'm just gonna be careful and
check. If you have time you should too.
$M_{y}=2e^{2y}$
$N_{x}=2e^{2y}$
$\psi=\int Mdx=xe^{2y}+h(y)$
$N=\psi_{y}=2xe^{2y}+h^{\prime}(y)$
$h^{\prime}(y)=-y^{-1}$
$h(y)=-\ln y+C$
Solution given implicitly by $xe^{2y}-\ln y=c$
$y=0$ is also a solution to the equation. We assumed $y\neq0$ above,
so we consider it to be a solution and it is.