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\centerline{\bf Solutions to Math. 226, Exam 2}
\bigskip
\noindent
{\bf 1a.} 
The characteristic equation is $4r^2 - 8r + 3 = 0$ and has roots
$r= \frac{1}{2} or \frac {3}{2}.$  Therefore the general solution to the DE is
given by $\boxed{y = c_1e^{\frac{1}{2}x} + c_2e^{\frac{3}{2}x}}$. 
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\noindent
{\bf 1b.} 
The characteristic equation is $r^2 + 6r + 25 =0$ and has roots $r=-3 \pm 4i.$
Therefore, the general solution of the D.E. is given by 
$\boxed {y= e^{-3x}[c_1 \cos 4x + c_2 \sin 4x]}$.
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\noindent
{\bf 2.} 
Another solution to the DE is given by $y=e^xv$, $v$ to be determined. 
We must have
 $$x(e^xv)'' - (2x +1)(e^xv)' + (x +1)e^xv =0$$ or 
$$x(e^xv'' + 2e^xv' + e^xv) - (2x + 1)(e^xv' +
e^xv) + (x +1)e^xv=0$$ or $$xe^xv'' - e^xv' = 0 \text{ or }v'' - \frac{1}{x}v' = 0.$$ 
If we let $u=v'$ then we obtain the equation $u' - \frac{1}{x}u = 0$ or $\frac{du}{u} =
\frac{dx}{x}$ or $\ln u = \ln x$
 or $u=x$ or $v' = x$ or $v= \frac{1}{2}x^2$.
Therefore $y_2 = \frac{1}{2}x^2e^x$ is a second solution which is independent from
$y_1 = e^x.$  Thus the general solution to the above D.E. is given by
$\boxed{y=e^x[c_1 +c_2x^2]}$.
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\noindent
{\bf 3.} 
The general solution to the DE is given by $$y=c_1x + c_2 \frac{1}{x} + y_p,$$
where $y_p = y_1u_1 + y_2u_2$ and $u_1$ and $u_2$ satisfy the system
$$\cases y_1u_1' + y_2u_2' = 0\\
y_1'u_1' + y_2u_2' = 2e^x \endcases \, or \, \, \cases xu_1' + \frac{1}{x}u_2' = 0\\
u_1' - \frac {1}{x^2}u_2' = 2e^x. \endcases$$
We have:
$$u_1' = \frac{\left| \matrix 0 & \frac{1}{x} \\2e^x & \frac{-1}{x^2} \endmatrix
\right|}{\left| \matrix x & \frac{1}{x} \\ 1 & \frac{-1}{x^2} \endmatrix \right|} =
\frac{\frac{-2}{x}e^x}{\frac{-2}{x}} = e^x,\text{ and }
u_2' = \frac{\left| \matrix x & 0 \\1 & 2e^x \endmatrix
\right|}{\frac{-2}{x}} = -x^2e^x.$$
Thus  $u_1 = e^x$ and $u_2 = - \int x^2e^x dx$
  or $u_2 = - \int x^2d(e^x) = -x^2e^x + \int 2xe^x = -x^2e^x + 2xe^x - 2e^x.$
Therefore, $y_p = xe^x + \frac{1}{x}e^x(-x^2 + 2x -2)$ and the general solution is
given by
$$\boxed{y=c_1x + c_2 \frac{1}{x} + xe^x + \frac{1}{x}e^x(-x^2 + 2x -2)}$$
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\noindent
{\bf 4.} 
The motion of the center of mass of the bridge
is described by the solution of the following initial value problem:
$$\aligned\ddot u + 16u&= \frac{1}{2} \cos 4t \\
u(0) &=0 \\
\dot u(0) &=0 \endaligned$$
The general solution of $\ddot u + 16u = 0$ is $u_c = c_1 \cos 4t + c_2 \sin 4t.$  A
particular solution of $\ddot u + 16u = \frac{1}{2} \cos 4t$ is of the form
 $u_p = t (A \cos 4t + B \sin 4t).$  We must have
 $$A[-8 \sin 4t - 16t \cos 4t + 16t
\cos 4t] + B[8 \cos 4t - 16t \sin 4t + 16t \sin 4t] = \frac{1}{2} \cos 4t.$$
Therefore $A = 0$ and $B = \frac{1}{16}.$  Thus, the general solution is given by
$$u(t) = c_1 \cos 4t + c_2 \sin 4t + \frac{1}{16}t \sin 4t.$$
Since $0=u(0)=c_1$ we have $c_1 =0.$  Since $\dot u(t) = 4c_2 \cos 4t + \frac{1}{16}
\sin 4t + \frac{1}{4}t \cos 4t$ and $0 = u(0) = 4c_2$ we have $c_2 = 0$ and
therefore $\boxed{u(t)= \frac{1}{16}t \sin 4t}.$  Since $|u(t)| \rightarrow \infty$ as
$t \rightarrow \infty$ the bridge will collapse when $t$ gets large.
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\noindent
{\bf 5a.} 
We must have $$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}-2x
\sum_{n=1}^{\infty}na_nx^{n-1}-4 \sum_{n=0}^{\infty}a_nx^n=0$$
or$$\sum_{n=0}^{\infty}(n+2)(n+1)a_n +2x^n - 2 \sum_{n=0}^{\infty} na_nx^n-4
\sum_{n=0}^{\infty}a_nx^n=0$$
or$$ \sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} - 2na_n-4a_n]x^n=0
\text{ or }
(n+2)((n+1)a_{n+2}-2na_n-4a_n=0$$

or $$a_{n+2} = \frac{(n+2)a_n}{(n+2)(n+1)} \, \, or\,\, \boxed{a_{n+2}= \frac{2}{n+1}a_n,
n=0,1,2...}$$
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\noindent
{\bf 5b.} 
Since $y(0)=0$ we have $a_0 =0$ and since $y'(0) =1$ we have $a_1 = 1$.  By the
recurrence relation we see that 
$\boxed{a_{2n} =0 \, \, \text{and} \, \, a_{2n+1} = \frac{1}{n!}}$.
\bigskip
\noindent
{\bf 5c.} 
The solution is  $\boxed {y= \sum_{n=0}^{\infty} \frac 1{n!} x ^{2n+1}}$.  Since 
$\frac{\frac{1}{(n +1)!}|x|^{2(n+1)+1}}{\frac{1}{n!}|x|^{2n+1}} =
\frac{x^2}{n+1}\longrightarrow 0 \text{ as  } n \rightarrow \infty ,$ for all $x \in
\Bbb R$ we have that the power series converges for all $x \in \Bbb R$, i.e. its
radius of convergence is $\boxed{R= \infty}$.          



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