Third Examination Solutions. As given Thursday, November 19, 1998.

\begin{tabular} {\vert c\vert c\vert c\vert c\vert c\vert c\vert c\vert c\vert p... ... \hline Points\rule[-4pt]{0pt}{18pt} & & & & & & & & \\ \hline \end{tabular}
Problem1. % latex2html id marker 381 \( (\alph{enumi}) \) Find the solution of the differential equation \( {\displaystyle \frac{dy}{dx}=xy^{2}} \) which satisfies the initial condition \( y(0)=2 \).

Solution Use the method of separation of variables. First put the equation in differential form with all \( y \)-variables on the left-hand side and all \( x \)-variables on the right-hand side:  \begin{equation} \frac{dy}{y^{2}}=x\,dx\mbox{\,.} \end{equation}

(1)


Next integrate using the power rule on each side:  \begin{equation} \int\frac{dy}{y^{2}}=-\frac{1}{y}=\int x\,dx=\frac{x^{2}}{2}+C\mbox{\,.} \end{equation}

(2)


Now solve for \( C \) by setting \( x=0 \) and \( y=2 \) to get \( {\displaystyle -\frac{1}{2}=\frac{0}{2}+C} \), which means that \( C=-1/2 \), so that equation (2) becomes  \begin{equation} -\frac{1}{y}=\frac{x^{2}}{2}-\frac{1}{2}=\frac{x^{2}-1}{2}\mbox{\,.} \end{equation}

(3)


Finally, solve equation (3) for \( y \) by multiplying each side by \( y \) and then multiplying each side by \( {\displaystyle \frac{2}{x^{2}-1}} \) to get:  \begin{equation} y=-\frac{2}{x^{2}-1}\mbox{\,.} \end{equation}

(4)


Problem2. % latex2html id marker 394 \( (\alph{enumi}) \) The differential equation \( {\displaystyle \frac{dP}{dt}=k(1-P)} \), where \( k \) is a positive constant, models performance \( P(t) \) of a trainee learning a skill. Performance is measured on a scale of 0 (no skill) to 1 (perfect performance), and \( t \) is elapsed time in hours since the beginning of training. Use the method of separation of variables to solve this differential equation for \( P(t) \), assuming the initial condition \( P(0)=1/3 \).

Solution In this problem performance \( P \) is the dependent variable; it is a function of the independent variable \( t \). In other words, \( P=P(t) \). Solve the differential equation by separating the variables in differential form: \( P \) and \( dP \) on the left; \( t \) and \( dt \) on the right:  \begin{equation} \frac{dP}{1-P}=k\,dt\mbox{\,.} \end{equation}

(5)


Next integrate each side:  \begin{equation} \int\frac{P}{1-P}=-\ln(1-P)=\int k\,dt=kt+C\mbox{\,.} \end{equation}

(6)


Now evaluate the constant \( C \) using the initial condition \( P(0)=1/3 \): \( -\ln(1-1/3)=k\cdot 0+C \Rightarrow C=\ln(2/3)\). Substitute this value in equation (6) and multiply both sides by \( -1 \) to obtain  \begin{equation} \ln(1-P)=-kt-\ln\left(\frac{2}{3}\right)= -kt+\ln\left(\frac{3}{2}\right)\mbox{\,,} \end{equation}

(7)


where we have used the fact that \( {\displaystyle -\ln\left(\frac{a}{b}\right)=\ln\left(\frac{b}{a}\right)} \).

Finally, solve for \( P \) by applying the exponential function to each side of (7) to obtain:  \begin{eqnarray} 1-P & = & e^{-kt+\ln(3/2)} \nonumber \\ P & = & 1-e^{-kt+\ln(3/2)} \nonumber \\ P & = & 1-\frac{3}{2}e^{-kt} \end{eqnarray}



(8)






In the final equation of (8) we have used the fact that \( {\displaystyle e^{\ln(3/2)}}={\displaystyle \frac{3}{2}} \).
Problem3. % latex2html id marker 415 \( (\alph{enumi}) \) Determine whether the given sequence converges or diverges. If it converges, find the limit. Show all steps leading to your solution.
1.
\( {\displaystyle a_{n}= \frac{2n^{2}-3n+1}{3n^{2}+2n-1}} \)Solution The simplest solution to this sort of limit is to remove the highest power of the variable \( n \) from numerator and denominator by factoring and then canceling factors of \( n \):

2.
\( {\displaystyle b_{n}= \frac{1+n^{2}}{1+n}} \)
Problem4. % latex2html id marker 420 \( (\alph{enumi}) \) Determine whether the given series is convergent or divergent. If it is convergent, find the exact value of its sum. Show all work.
1.
\( {\displaystyle \sum_{n=1}^{\infty} \frac{8}{9^{n-1}}=8+\frac{8}{9}+ \frac{8}{9^{2}}+\cdots} \)
2.
\( {\displaystyle \sum_{n=1}^{\infty} \frac{(-5)^{n-1}}{4^{n}}=\frac{1}{4}-\frac{5}{4^{2}}+ \frac{5^{2}}{4^{3}}-\cdots} \)
Problem5. [Do only one]% latex2html id marker 423 \( (\alph{enumi}) \) Use the integral test to determine which of the following series is convergent and which is divergent. Show all work.
1.
\( {\displaystyle \sum_{n=1}^{\infty} \frac{n}{n^{2}+1}} \)
2.
\( {\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\left(\ln(n)\right)^{2}}} \)
Problem6. [Do only one]% latex2html id marker 426 \( (\alph{enumi}) \) Test the series for convergence or divergence. In each case, name the test you used: divergence test; alternating series test; integral test; comparison test; limit comparison test; ratio test.
1.
\( {\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{1+\sqrt{n}}= \frac{1}{2}-\frac{1}{1+\sqrt{2}}+\frac{1}{1+\sqrt{3}}-\cdots} \)
2.
\( {\displaystyle \sum_{n=1}^{\infty} \frac{(-n)^{n+1}}{1+3n}= \frac{1}{4}-\frac{8}{7}+\frac{81}{10}-\cdots} \)
Problem7. % latex2html id marker 429 \( (\alph{enumi}) \) Determine whether the given series is absolutely convergent. In each case, name the test you used: divergence test; alternating series test; integral test; comparison test; limit comparison test; ratio test.
1.
\( {\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n+1}} \)
2.
\( {\displaystyle \sum_{n=1}^{\infty} \frac{n}{2^{n}}} \)