Test 2(Mock Exam)
( \newcommand{\kernel}{\mathrm{null}\,}\)
These mock exams are provided to help you prepare for Term/Final tests. The best way to use these practice tests is to try the problems as if you are taking the test. Please don't look at the solution until you have attempted the question(s). Only reading through the answers or studying them, will typically not be helpful in preparing since it is too easy to convince yourself that you understand it.
Exercise 1
Solve the following Initial Value Problems:
- 3y2y′−xy3=x,y(0)=0.
- xy′−y=x,y(1)=−1
- Answer
-
- y=(ex22−1)1/3
- y=x2−x.
- Solution
-
1. 3y2y′−xy3=x⟹3y2y′=xy3+x=x(y3+1).
∫3y2dyy3+1=∫xdxln|y3+1|=x22+Cy3+1=Aex22y=(Aex22−1)1/3
Now, x=0,y=0⟹A=1.
Hence y=(ex22−1)1/3.
2.
xy′−y=x,y(1)=−1⟹y′−1xy=1,y(1)=−1.
Hence the integrating factor =e∫−1xdx=e−ln(x)=eln(x−1)=x−1.
x−1y′−x−11xy=x−1ddx(x−1y)=x−1x−1y=∫x−1dxx−1y=ln|x|+Cy=xln|x|+Cx
Since y(1)=−1,−1=(1)(ln(1))+C. Hence C=−1.Thus the solution is y=x2−x.
Exercise 2
Use an appropriate test to determine if the following series converges or diverges. Justify your answer.
- ∑∞k=2k7k
- ∑∞k=1k+2(2k+1)
- ∑∞k=2kk(ln(k))k
- Answer
-
Converges, diverges, diverges.
- Solution
-
1. By Ratio Test,
ρ=limk→∞(k+17k+1)(k7k)=limk→∞k+1k17=17<1. Hence the series converges.
2. By divergence test limk→∞k+2(2k+1)=12≠0, the series diverges.
3. ρ=limk→∞k√kk(ln(k))k=limk→∞kln(k)=limk→∞11k( by L'H\^{o}pital's rule)=∞.
By the Root test, the series diverges.
Exercise 3
Determine whether each of the following series converges or diverges, and if it converges, find its sum. Justify your answer.
- ∑∞k=11(k+1)(k+2)
- 1+πe+π2e2+π3e3+⋯
- Answer
-
- Convergent telescoping series with sum is 12.
- Divergent geometric series.
- Solution
-
1. By using partial fraction decomposition, we get 1(k+1)(k+2)=1k+1−1k+2.
Thus ∑∞k=11(k+1)(k+2)=∞∑k=1(1k+1−1k+2). The second part of each term cancels with the first part of the succeeding term, hence Sn=12−1n+2. Therefore, limn→∞Sn=12.
2. This is a geometric series with first term a=1 and the ratio r=πe>1. Thus this series diverges.
Exercise 4
A glass of juice with a temperature of 40∘F is placed in a room with a constant temperature of 75∘F, and 1 hour later its temperature is 55∘F. Using Newton’s Law of Cooling, show that t hours after the juice is placed in the room its temperature is approximated by T=75−35e−.56t.
- Solution
-
The ambient temperature (surrounding temperature) is 75°F, so T_s=75. The temperature of the glass of juice when it placed in a room is 40°F, which is the initial temperature (i.e., initial value), so T_0=40. Therefore Equation becomes
\dfrac{dT}{dt}=k(T−75) \nonumber
with T(0)=40.
Rewrite the differential equation by multiplying both sides by dt and dividing both sides by T−75:
\dfrac{dT}{T−75}=k\,dt. \nonumber
Integrate both sides:
\begin{align*} ∫\dfrac{dT}{T−75} &=∫k\,dt \\ \ln|T−75| &=kt+C.\end{align*}
Solve for T by first exponentiating both sides:
\begin{align*}e^{\ln|T−75|} &=e^{kt+C} \\ |T−75| &=C_1e^{kt}, & & \text{where } C_1 = e^C. \\ T−75 &=\pm C_1e^{kt} \\ T−75 &=Ce^{kt}, & & \text{where } C = \pm C_1\text{ or } C = 0.\\ T(t) &=75+Ce^{kt}. \end{align*}
Solve for C by using the initial condition T(0)=40:
\begin{align*}T(t) &=75+Ce^{kt}\\ T(0) &=75+Ce^{k(0)} \\ 40 &=75+C \\ C &=-35.\end{align*}
Therefore the solution to the initial-value problem is
T(t)=75-35e^{kt}.\nonumber
To determine the value of k, we need to use the fact that after 1 hour the temperature of the juice is 52°F. Therefore T(1)=55. Substituting this information into the solution to the initial-value problem, we have
T(t)=75-35e^{kt}\nonumber
T(1)=55=75-35e^{k}\nonumber
-20=-35e^{k}\nonumber
e^{k}=\dfrac{4}{7}\nonumber
\ln e^{k}=\ln(\dfrac{4}{7})\nonumber
k=\ln(\dfrac{4}{7})\nonumber
k≈−0.56.\nonumber
So now we have T(t)=75-35e^{−0.56t}.
Exercise \PageIndex{5}
For each of the following series determine whether the series converges absolutely, converges conditionally, or diverges. Justify your answer.
- \sum_{k=1}^\infty \displaystyle (-1)^{k+1}{3^k \over k!}
- \sum_{k=1}^\infty \displaystyle (-1)^{k+1}{\sqrt{k+3} \over k}
- \sum_{k=1}^\infty \displaystyle (-1)^{k+1} \left( \frac{k+1}{k} \right)^k
- Answer
-
Converges absolutely, Converges conditionally, diverges
- Solution
-
1. \begin{align*}\rho & = \lim_{k\to\infty}\left|a_{k+1}\over a_k\right|\\&=\lim_{k\to\infty} \dfrac{\left|(-1)^{k+2} {3^{k+1}\over (k+1)!} \right|}{\left|(-1)^{k+1} {3^k \over k!} \right|}\\&= 3 \lim_{k\to\infty} \dfrac{1}{(k+1)}\\&= 0<1. \end{align*}
Hence by the Ratio test, the series converges absolutely.
2. Converges conditionally by alternating series test, since \displaystyle \sqrt{k+3}/k is decreasing. Does not converge absolutely by comparison with p-series, \displaystyle p=1/2
3. Diverges by divergence test since \displaystyle \lim_{k→∞}|a_k|=e.