@

title

.75em
author

date

Show all of your work. Full credit may not be given for an answer alone. This is a non-calculator exam. You may use the backs of the pages for scratch work. Do not unstaple or remove pages.

By taking this exam you are agreeing to abide by Rutgers University's Academic Integrity Policy.

Problem Possible Points
Number Points Earned
$ \displaystyle\vphantom{\int}$ 1 15  
$ \displaystyle\vphantom{\int}$ 2 15  
$ \displaystyle\vphantom{\int}$ 3 15  
$ \displaystyle\vphantom{\int}$ 4 20  
$ \displaystyle\vphantom{\int}$ 5 15  
$ \displaystyle\vphantom{\int}$ 6 20  
$ \displaystyle\vphantom{\int}$ Total 100  

  1. (15 points)
    1. (10 points) Compute $ {\mathfrak{L}\left[1\right]}(s)$, using only the definition of Laplace transform.

      Proof. [Solution]

      $\displaystyle {\mathfrak{L}\left[1\right]}(s)$ $\displaystyle = \int_0^\infty e^{-st} \cdot 1 dt$    
        $\displaystyle = \lim_{k\to \infty} \left.\frac{e^{-st}}{-s}\right\vert^k_0$    
        $\displaystyle = \lim_{k \to\infty} \frac{1}{s} - \frac{e^{-sk}}{s}$    
        $\displaystyle = \frac{1}{s}.$    

      $ \qedsymbol$

    2. (5 points) Compute $ {\mathfrak{L}\left[t\right]}(s)$, using either the definition or by differentiating part (a).

      Proof. [Solution]One can do integration by parts to find $ \int_0^\infty t e^{-st}  dt$, but the easiest way is with differentiation.

      $\displaystyle {\mathfrak{L}\left[t\cdot1\right]}(s)$ $\displaystyle = -\frac{d}{ds} {\mathfrak{L}\left[1\right]}(s)$    
        $\displaystyle = - \frac{d}{ds}\left(\frac{1}{s}\right)$    
        $\displaystyle = \frac{1}{s^2}.$    

      $ \qedsymbol$

  2. (15 points) Compute $ {\mathfrak{L}^{-1}\left[\frac{s}{(s-1)^{3/2}}\right]}$.

    Hint 1   $ s=s-1 + 1$.

    Proof. [Solution]Using the hint, we have

    $\displaystyle \frac{s}{(s-1)^{3/2}}$ $\displaystyle = \frac{s-1 + 1}{(s-1)^{3/2}} = \frac{s-1}{(s-1)^{3/2}} + \frac{1}{(s-1)^{3/2}}$    
      $\displaystyle = \frac{1}{(s-1)^{1/2}} + \frac{1}{(s-1)^{3/2}}$    

    So

    $\displaystyle {\mathfrak{L}^{-1}\left[\frac{s}{(s-1)^{3/2}}\right]} = e^t{\mathfrak{L}^{-1}\left[\frac{1}{s^{3/2}} +
\frac{1}{s\sqrt s}\right]}.
$

    According to the table, $ {\mathfrak{L}^{-1}\left[\frac{1}{\sqrt s}\right]} = \frac{1}{\sqrt{\pi
t}}$. As for $ {\mathfrak{L}^{-1}\left[\frac{1}{s^{3/2}}\right]}$, write it as $ \frac{1}{s \sqrt
s}$, so we can just integrate $ {\mathfrak{L}^{-1}\left[\frac{1}{\sqrt s}\right]}$. Thus

    $\displaystyle {\mathfrak{L}^{-1}\left[\frac{s}{(s-1)^{3/2}}\right]} = \frac{e^{t}}{\sqrt \pi}\left(\frac{1}{\sqrt{t}} +
2\sqrt{t}\right).
$

    $ \qedsymbol$

  3. (15 points) Use the Laplace transform to solve the ordinary differential equation

    $\displaystyle y'' + 5y' + 6y = \delta(t),
\qquad y(0) = 3, \qquad y'(0) = 0.
$

    Proof. [Solution]The Laplace transform of the equation is

    $\displaystyle s^2Y - 3s + 5(Y-3) + 6Y$ $\displaystyle = 1$    
    $\displaystyle Y = \frac{3s + 16}{s^2 + 5s + 6}.$    

    We factor $ s^2 + 5s + 6 = (s+3)(s+2)$ and thus we have a partial fraction decomposition

    $\displaystyle \frac{3s + 16}{s^2 + 5s + 6}$ $\displaystyle = \frac{A}{s+3} + \frac{B}{s+2}$    
    $\displaystyle 3s + 16$ $\displaystyle = A(s+2) + B(s+3)$    

    This solves to give $ A = -7$, $ B = 10$. Thus

    $\displaystyle Y(s)$ $\displaystyle = \frac{-7}{s+3} + \frac{10}{s+2}$    
    $\displaystyle y(s)$ $\displaystyle = -7 e^{-3t} + 10 e^{-2t}.$    

    $ \qedsymbol$

  4. (20 points) Let $ a \in {\mathbb{R}}$ and suppose $ y$ is a solution to the ordinary differential equation

    $\displaystyle ty'' + y' + ta^2y = 0,
\qquad y(0) = 1, \qquad y'(0) = 0.
$

    1. (15 points) Show $ Y(s) = \frac{k}{\sqrt{s^2 + a^2}}$ for some constant $ k$.

      Proof. [Solution]Remember that $ t$ transforms to $ -\frac{d}{ds}$, so the transformed equation is

      $\displaystyle -\frac{d}{ds}(s^2Y - s) + (sY - 1) - \frac{d}{ds}{a^2Y}$ $\displaystyle =0$    
      $\displaystyle -(2s Y + s^2 Y' - 1) + (sY - 1) - a^2Y'$ $\displaystyle =0$    
      $\displaystyle Y' + \frac{s}{s^2 + a^2}Y$ $\displaystyle =0$    

      The integrating factor for this ODE in $ Y$ is

      $\displaystyle e^{\int \frac{s ds}{s^2 + a^2}} = e^{\frac12\ln(s^2 + a^2)} = (s^2 + a^2)^{1/2}.$    

      Thus

      $\displaystyle \left(\sqrt{s^2 + a^2} Y\right)'$ $\displaystyle =0$    
      $\displaystyle \sqrt{s^2 + a^2} Y$ $\displaystyle = k$    
      $\displaystyle Y$ $\displaystyle = \frac{k}{s^2 + a^2}.$    

      $ \qedsymbol$

    2. (5 points) Suppose $ \int_0^\infty y(t) dt = \alpha$. What is $ k$ in terms of $ \alpha$?

    Proof. [Solution]Notice that

    $\displaystyle \alpha = \int_0^\infty y(t) dt = Y(0) = \frac{k}{\sqrt{0^2 +
a^2}}
= \frac{k}{\alpha}.
$

    Thus $ k = a \alpha$. $ \qedsymbol$

  5. (15 points) Let $ f(x) = \vert x\vert$ on $ [-\pi,\pi]$. Calculate the Fourier coefficients of $ f$ and write down the Fourier series.

    Proof. [Solution]This function is even, so all of its sine Fourier coefficients $ b_n$ are zero. So we need only integrate between 0 and $ \pi$, and count it twice as much. Moreover, on $ [0,\pi]$, $ \vert x\vert = x$. So

    $\displaystyle a_0 = \frac{2}{\pi}\int_0^\pi x dx
= \left.\frac{2}{\pi}{x^2}{2}\right\vert^{\pi}_0 = \pi.
$

    Furthermore, if $ n > 0$,

    $\displaystyle a_n$ $\displaystyle = \frac{2}{\pi}\int_0^\pi x \cos(nx) dx$    
      $\displaystyle = \frac{2}{n\pi}\left(\left.x\sin(nx)\right\vert^\pi_0 - \int_0^\pi \sin(nx) dx\right)$    
      $\displaystyle = \frac{-2}{n\pi}\int_0^\pi \sin(nx) dx$    
      $\displaystyle = \left.\frac{2}{\pi n^2}\cos(nx)\right\vert^\pi_0$    
      $\displaystyle = \frac{2}{(-1)^n - 1}{\pi n^2}.$    

    Therefore the Fourier series is

    $\displaystyle Sf = \frac{\pi}{2} + \frac{2}{\pi}\sum_{n=1}^\infty \frac{(-1)^n
-1}{n^2}\cos(nx).
$

    $ \qedsymbol$

  6. (20 points) Let $ 0 < L \leq \pi$ and define $ f(x) = \cos(x)$ on $ [-L,L]$. Calculate the Fourier coefficients of $ f$ and write down the Fourier series.

Proof. [Solution]$ L$ is intended to be an arbitrary number between 0 and $ \pi$. I should have excluded the case $ L = \pi$, but many people ignored $ L$ altogether and integrated between $ -\pi$ and $ \pi$. As I said in class, in fact during the review, the Fourier series for $ \cos(x)$ on $ [-\pi,\pi]$ is $ \cos(x)$. You will get $ a_0 = 1$ and all other $ a_n = 0$.

Assume $ 0 < L < \pi$ then. Notice $ \cos(x)$ is an even function so all $ b_n$ are zero, and we can integrate over half the interval.

$\displaystyle a_0 = \frac{2}{L}\int_0^L \cos(x) dx =
\left.\frac{2}{L}\sin(x)\right\vert^L_0 = \frac{2\sin(L)}{L}.
$

Moreover, if $ n > 0$,

$\displaystyle a_n$ $\displaystyle = \frac{2}{L}\int_{0}^L \cos(x) \cos\left(\frac{\pi n x}{L}\right) dx.$    

Let $ u = \cos(x)$, $ dv = \cos\left(\frac{\pi n
x}{L}\right) dx$. Then

$\displaystyle a_n$ $\displaystyle \stackrel{\text{IBP}}{=} \frac{2}{L}\left(\frac{L}{n\pi}\right)\l...
...\right\vert^L_0 + \int_0^L\sin(x)\sin\left(\frac{\pi n x}{L}\right) dx\right\}$    
  $\displaystyle = \frac{2}{L}\left(\frac{L}{n\pi}\right) \int_0^L\sin(x)\sin\left(\frac{\pi n x}{L}\right) dx.$    

Now let $ u = \sin(x)$, $ dv = \sin\left(\frac{\pi n
x}{L}\right) dx$. We have

$\displaystyle a_n$ $\displaystyle \stackrel{\text{IBP}}{=} \frac{2}{L}\left(\frac{L}{n\pi}\right)^2...
...right\vert^L_0 + \int_0^L \cos(x)\cos\left(\frac{\pi n x}{L}\right) dx\right\}$    
  $\displaystyle = \frac{2}{L}\left(\frac{L}{n\pi}\right)^2 (-1)^{n+1}\sin(L) + \left(\frac{L}{n\pi}\right)^2 a_n.$    

Thus

$\displaystyle \frac{\pi^2 n^2 - L^2}{\pi^2 n^2}a_n$ $\displaystyle = \frac{2L(-1)^{n+1} \sin(L)}{\pi^2 n^2}$    
$\displaystyle a_n$ $\displaystyle = \frac{2L(-1)^{n+1}\sin(L)}{\pi^2 n^2 - L^2}.$    

Therefore

$\displaystyle Sf = \frac{\sin L}{L} + 2L\sin(L) \sum_{n=1}^\infty
\frac{(-1)^{n+1}}{\pi^2 n^2 - L^2}\cos\left(\frac{\pi n x}{L}\right).
$

$ \qedsymbol$



Matthew Leingang 2002-10-10