P8: 22
P24: 1, 5; P25: 7, 9, 15, 17.
P39: 13, 17; P40: 30
P47: 1, 3; P48: 7, 8; P50: 31, 36
P77: 31.
Select solutions have been posted 2/24/2011. Problems with solutions have hyperlinks to them. If you want to see a solution which is not posted, please let me know. You may also request solutions to problems outside the homework set as well.
page 8 number 22
Let $V$ be the volume of the spherical raindrop, $S$ be its surface
area, and $K$ a positive constant. Then the first sentence tells
us that
\[
\frac{dV}{dt}=-K\cdot S.\]
But we want a differential equation for the volume. We know that $V=\frac{4}{3}\pi r^{3}$
and $S=4\pi r^{2}$. So we can solve for $r$ in terms of $V$ and
use that in the formula for $S$.\[
r=\left(\frac{3}{4\pi}V\right)^{1/3}\]
\[
S=4\pi\left(\frac{3}{4\pi}V\right)^{2/3}\]
\[
\frac{dV}{dt}=-CV^{2/3}\]
where $C$ is a positive constant.
page 25 number 7
\[
\frac{d}{dt}\cosh t=\sinh t\]
\[
\frac{d}{dt}\sinh t=\cosh t\]
\[
\frac{d}{dt}\cos t=-\sin t\]
\[
\frac{d}{dt}\sin t=\cos t\]
page 25 number 15
\[
y^{\prime}+2y=0\]
\[
re^{rt}+2e^{rt}=0\]
\[
r+2=0\]
\[
r=-2\]
page 25 number 17
\[
y^{\prime\prime}+y^{\prime}-6y=0\]
\[
r^{2}e^{rt}+re^{rt}-6e^{rt}=0\]
\[
r^{2}+r-6=0\]
\[
\left(r-2\right)\left(r+3\right)=0\]
\[
r-2=0\mbox{ or }r+3=0\]
\[
r=2\mbox{ or }r=-3\]
page 39 number 13
\[
y^{\prime}-y=2te^{2t},\quad y(0)=1\]
\[
\mu=e^{\int(-1)dt}=e^{-t}\]
\[
\left(\mu y\right)^{\prime}=\mu(2te^{2t})=2te^{t}\]
\[
\mu y=\int2te^{t}dt\]
Recall that the integral $\int te^{t}dt$ can be found by integration
by parts: \[
\int udv=uv-\int vdu.\]
I would write\[
\int te^{t}dt=\int tde^{t}=te^{t}-\int e^{t}dt.\]
Thus\[
\int te^{t}dt=te^{t}-e^{t}+C=(t-1)e^{t}+C.\]
In any case,\[
\mu y=2(t-1)e^{t}+C.\]
Multiplying both sides by $\frac{1}{\mu}$ we get
\[
y=2(t-1)e^{2t}+Ce^{t}.\]
Now we solve the initial value problem by plugging in $t=0$ and obtaining
\[
1=y(0)=-2+C\]
Thus, $C=3$. And so the solution is\[
y=2(t-1)e^{2t}+3e^{t}.\]
page 39 number 17
\[
y^{\prime}-2y=e^{2t},\quad y(0)=2\]
\[
\mu=e^{-2t}\]
\[
\left(\mu y\right)^{\prime}=1\]
\[
\mu y=t+C\]
\[
y=(t+C)e^{2t}\]
\[
2=y(0)=C\]
\[
y=(t+2)e^{2t}\]
page 40 number 30
\[
y^{\prime}-y=1+3\sin t,\quad y(0)=y_{0}\]
\[
\mu=e^{-t}\]
\[
\left(\mu y\right)^{\prime}=e^{^{-t}}\left(1+3\sin t\right)\]
\[
\mu y=\int e^{-t}(1+3\sin t)dt\]
In this situation, it's easier to do the integral in two pieces. So
I would write\[
\int e^{-t}(1+3\sin t)dt=\int e^{-t}dt+3\int e^{-t}\sin tdt\]
The second term we integrate by parts twice.
\begin{alignat*}{1}
\int e^{-t}\sin tdt & =\int\sin td\left(-e^{-t}\right)\\
& =\sin t\cdot(-e^{-t})-\int(-e^{-t})d\left(\sin t\right)\\
& =-e^{-t}\sin t+\int e^{-t}\cos tdt\\
& =-e^{-t}\sin t+\int\cos td(-e^{-t})\\
& =-e^{-t}\sin t+\cos t\cdot(-e^{-t})-\int(-e^{-t})d\left(\cos t\right)\\
& =-e^{-t}\sin t-e^{-t}\cos t+\int e^{-t}\cdot\left(-\sin t\right)dt\end{alignat*}
\[
2\int e^{-t}\sin tdt=-e^{-t}(\sin t+\cos t)\]
\[
\int e^{-t}\sin tdt=-\frac{1}{2}e^{-t}(\sin t+\cos t)\]
Thus,\[
\mu y=-e^{-t}-\frac{3}{2}e^{-t}(\sin t+\cos t)+C\]
\[
y=-(1+\frac{3}{2}\sin t+\frac{3}{2}\cos t)+Ce^{t}\]
We see that the only way for the solution to remain finite as $t\to\infty$
is if $C=0$
Thus we require that \[
y_{0}=y(0)=-(1+\frac{3}{2})=-\frac{5}{2}\]
page 47 number 1
\[
y^{\prime}=x^{2}/y\]
\[
\int ydy=\int x^{2}dx\]
\[
y^{2}=\frac{x^{3}}{3}+C\]
\[
y=\sqrt{\frac{x^{3}}{3}+C}\]
page 47 number 3
\[
y^{\prime}+y^{2}\sin x=0\]
\[
y^{\prime}=-y^{2}\sin x\]
If $y\neq0$ then we can divide both sides by $y$. This is the point
at which we would separately consider $y=0$ as a solution. Which
it is.
\[
\int\frac{1}{y^{2}}dx=\int(-\sin x)dx\]
\[
-\frac{1}{y}=\cos x+C\]
Note: I tend to check my antiderivatives when I can. I can also add
any constants I need as well. For example, I might think the antiderivative
of $\frac{1}{y^{2}}$ is $K\cdot\frac{1}{y^{3}}$ for some constant
$K$. But I check $(K\cdot y^{-3})^{\prime}=-3K\cdot y^{-4}\neq y^{-2}$.
On the other hand, I might correctly think the antiderivative of $\frac{1}{y^{2}}$
is $K\cdot\frac{1}{y}$. But I still have to figure out the constant.
So I check $(K\cdot y^{-1})^{\prime}=-K\cdot y^{-2}$. I realize that
the constant I should choose is $K=-1$. Of course I do this in my
head, but whether you do it in your head or on paper, checking, in
this situation, is valuable.
page 50 number 31
\begin{alignat*}{1}
\frac{dy}{dx} & =\frac{x^{2}+xy+y^{2}}{x^{2}}\\
& =\frac{x^{2}+xy+y^{2}}{x^{2}}\cdot\frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}\\
& =1+\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^{2}\end{alignat*}
Now $v=\frac{y}{x}$. So $y=vx$ and so $\frac{dy}{dx}=\frac{dv}{dx}\cdot x+v$.
Alternatively $\frac{dv}{dx}=\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^{2}}=\frac{1}{x}\frac{dy}{dx}-\frac{v}{x}$
and hence $x\frac{dv}{dx}=\frac{dy}{dx}-v$. This second way takes
much longer to solve for $\frac{dy}{dx}$, but I'm just illustrating
the possibility. In any case, making these substitutions we get
\[
x\cdot\frac{dv}{dx}+v=1+v+v^{2}\]
\[
x\cdot\frac{dv}{dx}=1+v^{2}\]
\[
\int\frac{1}{1+v^{2}}dv=\int\frac{1}{x}dx\]
Recall (and memorize!)
\[
\frac{d}{dt}\arctan t=\frac{1}{1+t^{2}}\]
Thus\[
\arctan v=\ln|x|+C\]
\[
v=\tan\left(\ln|x|+C\right)\]
\[
\frac{y}{x}=\tan(\ln|x|+C)\]
\[
y=x\cdot\tan\left(\ln|x|+C\right)\]
Note: The book's solution is $\arctan\frac{y}{x}-\ln|x|+C$
page 50 number 36
\[
-x^{2}\frac{dy}{dx}=-(x^{2}+3xy+y^{2})\]
\[
\frac{dy}{dx}=1+3\left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)^{2}\]
Using the same substitutions worked out in the previous problem:
\[
x\cdot\frac{dv}{dx}+v=1+3v+v^{2}\]
\[
x\cdot\frac{dv}{dx}=1+2v+v^{2}=(1+v)^{2}\]
\[
\int\frac{1}{(1+v)^{2}}dv=\int\frac{1}{x}dx\]
\[
\frac{-1}{1+v}=\ln|x|+C\]
\[
\frac{-1}{1+\frac{y}{x}}=\ln|x|+C\]
Multiplying the term on the right hand side by $\frac{x}{x}$ and
both sides of the equation by a minus sign, and replacing the constant
$-C$ by a new constant $c$ we obtain the book's answer:
\[
\frac{x}{x+y}+\ln|x|=c\]
page 77 number 31
\[
\frac{dy}{dt}=(\Gamma\cos t+T)y-y^{3}\]
Assuming $y\neq0$ (at which point we check to see that $y=0$ is
a solution to the differential equation), then $v=y^{1-3}=y^{-2}$.
$\frac{dv}{dt}=-2y^{-3}\frac{dy}{dt}$.
\[
y^{-3}\frac{dy}{dt}=(\Gamma\cos t+T)y^{-2}-1\]
\[
\frac{1}{-2}\frac{dv}{dt}=(\Gamma\cos t+T)v-1\]
Multiplying both sides by $-2$ and moving one of the terms from the
right to the left side of the equation, I obtain:
\[
\frac{dv}{dt}+2(\Gamma\cos t+T)v=2.\]
\[
\mu=e^{\int2(\Gamma\cos t+T)dt}=e^{2\Gamma\sin t+2Tt}\]
\[
\left(\mu v\right)^{\prime}=2\mu\]
\[
\mu v=\int2\mu dt=2\int_{t_{0}}^{t}\mu(s)ds+C\]
Note: $2\int_{t_{0}}^{t}\mu(s)ds$ is an antiderivative of $2\mu$.
To check this we take it's derivative with respect to $t$. By the
fundamental theorem of calculus $\frac{d}{dt}\left(2\int_{t_{0}}^{t}\mu(s)ds\right)=2\mu(t)$.
We choose this antiderivative, because there is no easy way to integrate
$2\mu$.
\[
v=\frac{1}{\mu}\left(2\int_{t_{0}}^{t}\mu(s)ds+C\right)\]
\[
\frac{1}{y^{2}}=\frac{1}{\mu}\left(2\int_{t_{0}}^{t}\mu(s)ds+C\right)\]
\[
y=\pm\sqrt{\mu(t)/\left(2\int_{t_{0}}^{t}\mu(s)ds+C\right)}\]
Note: $s$ is a dummy-variable. Thus $\int_{t_{0}}^{t}\mu(s)ds=\int_{t_{0}}^{t}\mu(r)dr=\dots$.
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