Practice Problem Answers
Exam 1

1.

$${y + \frac{y^2}{2} = e^t + 3}$$

2.

(a)

y(t) = C e^-rt

(b)

As $t \to \infty$, $y \to 0.$

3.

(a)

$${y(x) = -\frac{1}{2}x + \frac{1}{4}}$$ [Note: no arbitrary constant!]

(b)

Solving y'' - y' - 2y = 0 yields y(x) = c₁e^2x + c₂e^-x.

4.

$y_1(t) = e^{5t};\quad y_2(t) = e^{-5t}.$ (There are other possible answers)

5.

y(t) = 4e^2t - 3e^3t

6.

$\lim_{n \to \infty} a_n = 0$
To prove this, when given an $\epsilon$, we must find an M such that when n > M, $\vert a_n - 0\vert = \vert a_n\vert < \epsilon$. If we assume that $\vert a_n\vert < \epsilon$, we can compute M:

$$\begin{eqnarray*} & &\vert a_n\vert < \epsilon\\ [.1in] \Longrightarrow & \quad ... ...in] \Longrightarrow & \quad & n > \sqrt{\frac{1}{\epsilon} - 1}. \end{eqnarray*}$$

So, if we define M as

$$M = \sqrt{\frac{1}{\epsilon} - 1}$$

then the preceding steps, when followed backwards, will show that if n > M, we are guaranteed that $\vert a_n\vert < \epsilon$.

7.

(a)

$${e^{x^2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}}$$

(b)

$${\ln{(5x - 4)} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 5^n}{n} (x-1)^n; \quad R = \frac{1}{5}}$$

[Note: it's a good idea to double check that the summation you write down generates the same terms that you computed for the Taylor series. Sometimes it takes a little fiddling to get the factorials, exponents and negative signs to line up. Don't forget that we define 0!=1.]

8.

(a)

The Taylor Series is, of course, f(t) = t² + 3t.

(b)

Let's first compute the various derivatives of f(t) evaluate them at 0.

$$\begin{eqnarray*} f(0) & = & a_0\\ f'(0) & = & 1\cdot a_1\\ f''(0) & = & 2\cdo... ...\cdot a_3\\ \vdots\\ f^k(0) & = & k! a_k \quad 0 \leq k \leq n \end{eqnarray*}$$

Now, when we form the Taylor polynomial coefficients we get

$$\frac{f^k(0)}{k!} = a_k, \quad 0\leq k\leq n, \quad \frac{f^k(0)}{k!} = 0, \quad k > n.$$

Since a_k = 0 for k > n (because the (n+1)st and higher derivatives of an nth degree polynomial are zero), we get a finite sum for our Taylor series, and as we just computed, the non-zero coefficients are equal to the coefficients of f(t), hence $${ f(t) = \sum_{k=0}^{n} a_k t^k}.$$

9.

(a)

$${\sum_{n=0}^{\infty} \frac{1}{2^n}} = 2$$

(b)

$${\sum_{n=0}^{\infty} \frac{2^{n+3}}{3^{n+2}}} = \frac{8}{3}$$

(c)

$${\sum_{n=1}^{\infty} \frac{1}{10\sqrt{n}}}$$ diverges

(d)

$${\sum_{n=1}^{\infty} \frac{1}{(1.1)^n \sqrt{n}}}$$ converges

(e)

$${\sum_{n=1}^{\infty} \frac{n^5 + 4n^4 - 1}{n^{10} - 17n^2}}$$ converges

(f)

$${\sum_{n=1}^{\infty} \frac{1+\frac{1}{n}}{1+\frac{1}{n^2}}}$$ diverges

(g)

$${\sum_{n=1}^{\infty} \frac{(-1)^n}{n}}$$ converges

(h)

$${\sum_{n=3}^{\infty} \frac{1}{(n-1)(n-2)}} = 1$$

(i)

$${\sum_{n=1}^{\infty} \frac{1}{(2n+2)(n+2)}} = \frac{1}{4}$$

10.

(a)

$${\frac{dQ}{dt} = -rQ }$$

(b)

You should obtain $${Q(t) = .05e^{-.015t}}.$$ The time for the tank to reach a concentration of .1% is 261 minutes.

(c)

r = 65.2 gal/min.

11.

(a)

Let $\alpha$ denote the coefficient of air resistance. Then the D.E. is $${m \frac{dv}{dt} = mg - \alpha v}.$$ Setting $${\frac{dv}{dt}=0}$$ gives the single critical point $${v_0 = \frac{mg}{\alpha}}$$ (the terminal velocity). By plotting $g - \frac{\alpha}{m} v$ and drawing the left and right pointing arrows, you can determine that the critical point is asymptotically stable (which should check with our intuition).

(b)

An object will accelerate until it reaches some terminal velocity, after which it will fall at a constant speed.

12.

$${\Theta(t) = 70 + .192 \cos{t} + .96 \sin{t} + 79.808 e^{-.2t}}$$
This was a fairly challenging problem, so give yourself a pat on the back if you were able to work it out...