\magnification1200
\def\newp #1.{\par\noindent{\bf #1.}\ }
\def\ds{\displaystyle}
\def\frac#1#2{{{#1}\over{#2}}}
\centerline{\bf Some answers for Exam II.}
\bigskip
\newp 11.
Solve the equation 
$$t{dy\over dt} +2y=5t^3$$
with initial condition $y(1)=0$.
\bigskip
This is a problem from exam 1 since it is a 
first order linear differential equation.
First put it in standard form
$${dy\over dt} +\frac{2}{t}y=5t^2$$
so $P(t)=\frac{2}{t}$ and $Q(t)=5t^2$.

Since $\ds\int\frac{2}{t}\> dt=2\ln t+C$, we may take
$v(t)=e^{2\ln t}=e^{\ln t^2}=t^2$.
Next we have to find
$\ds\int v(t)Q(t)\>dt=\int5t^5\>dt=\frac{t^6}{6}+C$.
Since $y(t)=\frac{1}{v(t)}\cdot\ds\int v(t)Q(t)\>dt$,
$$y(t)=\frac{t^4}{6}+\frac{C}{t^2}\ .$$

\newp 12.
Find just the partial fraction decomposition 
(not the integral) of
$${2x^2-4x+1\over (x-1)^2(x-2)}$$

First we need to set up the correct equation:
$$\frac{2x^2-4x+1}{(x-1)^2(x-2)}=
\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x-2)}$$
Clearing denominators we have the more manageable 
equation
$$2x^2-4x+1=
A\cdot(x-1)(x-2)+
B\cdot(x-2)
+C\cdot(x-1)^2$$
There are two basic strategies to pursue here:
plug in judiciously chosen values to get easy
equations involving $A$, $B$ and $C$; or
equate coefficients to get three linear equations in
three unknowns and solve them.
Either works here but the easiest is plugging 
in values.

\itemitem{}Plug in $x=1$: $2-4+1=B(-1)$ so $B=1$.
\itemitem{}Plug in $x=2$: $8-8+1=C$ so $C=1$.
\itemitem{}Plug in $0$ and use the fact that $B=C=1$:
$1=A(2)+B(-2)+C$ or $1=2A-1$, so $A=1$.

Equating coefficients goes as follows:
\itemitem{}$x^2$: $2=A+C$.
\itemitem{}$x^1$: $-4=-3A+B-2C$
\itemitem{}$x^0$: $1=2A-2B+C$

There are many ways to proceed: here is one.
Eliminate $C$ using the first equation:
$C=2-A$ so
\itemitem{}$-4=-3A+B-2(2-A)$ or
$0=-A+B$.
\itemitem{} $1=2A-2B+(2-A)$ or
$-1=A-2B$.

Use the last equation to eliminate $A$:
$A=2B-1$ so
$0=-(2B-1)+B$ or
$-1=-B$, $B=1$ and hence $A=2B-1=1$ and
$C=2-A=1$.



\newp 13.
Using substitution, and then a trig substitution, find
$\ds \int_1^e \frac{dy}{y\sqrt{1+(\ln y)^2}}$

Substitute to get rid of the $\ln y$: $u=\ln y$ so
$du=\frac{dy}{y}$ and
$$\ds \int_1^e \frac{dy}{y\sqrt{1+(\ln y)^2}}=
\int_0^1\frac{du}{\sqrt{1+u^2}}$$

Now do the trig.\ substitution $u=\tan\theta$ so that
$1+u^2=1+\tan^2\theta=\sec^2\theta$.
The $\arctan$ takes values in the interval $(-\pi/2,\pi/2)$ 
and in this interval the $\cos$, and hence the $\sec$, are 
positive so
$\sqrt{1+u^2}=\sec\theta$.
When $u=0$, $\theta=0$ and when $u=1$, $\theta=\pi/4$.
Finally,
$du=\sec^2\theta\>d\theta$ so
$$\ds \int_1^e \frac{dy}{y\sqrt{1+(\ln y)^2}}=
\int_0^1\frac{du}{\sqrt{1+u^2}}=
\int_0^{\pi/4}\frac{\sec^2\theta d\theta}{\sec\theta}=
\int_0^{\pi/4}\sec\theta\>d\theta
$$
But $\int\sec\theta\>d\theta=\ln\vert\sec\theta+\tan\theta\vert+C$
so
$$\eqalign{
\int_0^{\pi/4}\sec\theta\>d\theta=&
\ln\vert\sec\theta+\tan\theta\vert\big\vert_0^{\pi/4}\cr
=&
\ln\bigl(\sec\pi/4+\tan\pi/4\bigr)-\ln\bigl(\sec0+\tan0\bigr)\cr=&
\ln(\sqrt2+1)-\ln(1+0)\cr=&
\ln(\sqrt2+1)
}$$


\newp 14.
Does $\ds \int_1^\infty \frac{(1+\sin x)}{x^{4/3}} dx$ converge
or diverge? Why?

This type of problem is not on exam 2 but you ought to be able to do it by now.
It is a first comparison test problem so first we have to {\sl guess\/}
whether it converges or diverges.
Since it looks roughly like $\frac{1}{x^{4/3}}$ and we know this improper
integral converges, we guess that our integral converges.
There is no need to justify your guess at this stage: the justification comes
when you demonstrate convergence.

Since we are trying for convergence, we need a function larger than the
one we have. 
Since $\sin x$ is between $-1$ and $1$,
$$\frac{(1+\sin x)}{x^{4/3}}\leq\frac{2}{x^{4/3}}$$
To prove our integral converges, it suffices to prove that
$$\int_1^\infty\frac{2}{x^{4/3}}\>dx$$
converges.
$$\eqalign{
\int_1^\infty\frac{2}{x^{4/3}}\>dx=&
\lim_{t\to\infty}\int_1^t\frac{2}{x^{4/3}}\>dx=
\lim_{t\to\infty}\frac{t^{-4/3+1}}{-4/3+1}-\frac{1^{-4/3+1}}{-4/3+1}\cr
=&
\lim_{t\to\infty}\frac{t^{-1/3}}{-1/3}-\frac{1^{-1/3}}{-1/3}=
\lim_{t\to\infty}-3t^{-1/3}+3=4
\cr
}$$
since $\ds\lim_{t\to\infty}t^{-1/3}=0$.
Since
$\ds\int_1^\infty\frac{2}{x^{4/3}}\>dx$ converges, and since
$\frac{(1+\sin x)}{x^{4/3}}\leq\frac{2}{x^{4/3}}$ on the
interval $[1,\infty)$, the first comparison test shows
$\ds \int_1^\infty \frac{(1+\sin x)}{x^{4/3}} dx$ converges.

\newp 15.
Determine the convergence of the sequence $\{a_n\}$ with
$\displaystyle a_n={1\over n^2}\int_1^n x\ dx$.

This will be taken up in Chapter 8.
\end