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22M:100 (MATH:3600:0001) Exam 2 [April 27, 2013] Answers


1.)  Circle T for True and F for false.


[4]~ 1a.)  If $y = f(t)$ and $y = g(t)$ are solutions to $y'' + p(t)y' + q(t)y = 0$, then $y = c_1f(t) + c_2g(t)$ is also a solution to this differential equation for any constants $c_1$ and $c_2$.

\rightline{T \hskip 0.8in }

\v

[4]~ 1b.)  If ${\bf x} = {\bf f}(t)$ and ${\bf x} = {\bf g}(t)$ are solutions to ${\bf x}' = A{\bf x}$, then ${\bf x} = c_1{\bf f}(t) + c_2{\bf g}(t)$ is also a solution to this system of differential equation for any constants $c_1$ and $c_2$.

\rightline{T \hskip 0.8in }

\v\v

[12]~ 2.)  The phase portrait for  $\frac{dx}{dt} =x(y-2) $ and $\frac{dy}{dt} = x + 3$
is drawn below. Find all equilibrium solutions and determine whether the critical point is asymptotically stable, stable, or unstable.  Also classify it as to type (nodal source, nodal sink, saddle point, spiral source, spiral sink, center).

 equilibrium solutions: $\frac{dx}{dt} =x(y-2)= 0 $ and $\frac{dy}{dt} = x + 3 = 0$

Hence [$x = 0$ or $y = 2$] and [$x = -3$].  Thus $(x, y) = (-3, 2)$ is the only solution 
satisfying both 
equations,



\centerline{Equilibrium solution: $\underline{~~ (x, y) = (-3, 2) ~~}$}
\v

\centerline{~~~Stability : $\underline{~~stable~~}$ \hfill Type: $\underline{~~center~~}$~~~}

\hr

[20]~ 3.) Find a recursive formula for the constants of the series solution to
$y'' + 2y = 0$ near $x_0 = 0$.


Let $y = \So a_n x^n$, then $y' = \S n a_n x^{n-1}$, and $y'' = \St n(n-1) a_n x^{n-2}$

$ \St n(n-1) a_n x^{n-2} + 2\So a_nx^n = 0$


$ \So (n+2)(n+1) a_{n+2} x^{n} + 2\So a_nx^n = 0$

$ \So [(n+2)(n+1) a_{n+2}  + 2 a_n]x^n = 0$

Thus $(n+2)(n+1) a_{n+2}  + 2 a_n = 0$


Hence $a_{n+2} = \frac{-2a_n}{(n+2)(n+1)}$


\centerline{Answer: $\underline{~~a_{n+2} = \frac{-2a_n}{(n+2)(n+1)}~~}$}

\hr

[20]~ 4.) Suppose that $\mathlarger{a_{2k} = \frac{-a_{2k-2}}{4k^2}}$.  Use induction to prove that
$\mathlarger{a_{2k} = \frac{(-1)^ka_0}{4^k(k!)^2}}$ for all $k \geq 0$.



For $k = 0$, $\frac{(-1)^0a_0}{4^0(0!)^2}  = a_0 = a_{2(0)}$


Induction hypothesis:  Suppose that for $k = n$, $a_{2n} = \frac{(-1)^n a_0}{4^n(n!)^2}$


Claim:  for $k = n+1$:  $a_{2(n+1)} = \frac{(-1)^{n+1} a_0}{4^{n+1}((n+1)!)^2}$

$a_{2(n+1)} = a_{2n+ 2} = \frac{-a_{2n}}{4(n+1)^2}$ by hypothesis.
 
 
\hskip 1in $ =  \frac{-(-1)^n a_0}{[4^n(n!)^2][4(n+1)^2]}$ by the induction hypothesis.

\hskip 1in $ =  \frac{(-1)^{n+1} a_0}{4^{n+1}((n+1)!)^2}$

\eject

[20]~ 5.) Solve ${\bf x}' = \left(\matrix{0 & 3 \cr -3 & 0}\right){\bf x}$


Eigenvalues:  $det(A - rI) = \left|\matrix{-r & 3 \cr -3 & -r}\right| = r^2 + 9 = 0$.  Thus $r^2 = -9$ and $r = \pm 3i$

Eigenvectors:  $A - rI =  \left(\matrix{\mp 3i & 3 \cr -3 & \mp 3i}\right)  
\left(\matrix{x_1 \cr x_2}\right) = \left(\matrix{0 \cr 0}\right)$

Thus $\mp 3i x_1 + 3x_2 = 0$.  Let $x_1 = \pm 1, x_2 = i$.  Note these values also satisfy the 
second 
equation, $-3x_1 \mp 3i x_2$.

Thus an e. vector corresponding to e. value $\pm 3i$ is any nonzero multiple of 
$\left(\matrix{\pm 1 \cr i}\right) = \left(\matrix{\pm 1\cr 0}\right) + i\left(\matrix{0 \cr 
1}\right)$  


If you forgot the formula, you can  use Euler's formula to derive it:

General solution:  $\left(\matrix{x_1 \cr x_2}\right) $
$= c_1 e^{3it} \left(\matrix{1 \cr i}\right) +  c_2 e^{-3it} \left(\matrix{-1 \cr i}\right) $

$= c_1 (cos (3t) + isin(3t) \left(\matrix{1 \cr i}\right) +  c_2 (cos(-3t) + isin(-3t)) 
\left(\matrix{-1 \cr 
i}\right) $

$= c_1 (cos (3t) + isin(3t)) \left(\matrix{1 \cr i}\right) +  c_2 (cos(3t) - isin(3t)) 
\left(\matrix{-1 \cr 
i}\right) $

$= c_1  \left(\matrix{cos (3t) + isin(3t) \cr icos (3t) - sin(3t)}\right) +  
c_2 \left(\matrix{-cos (3t) + isin(3t) \cr icos (3t) + sin(3t)}\right) $
$=   \left(\matrix{(c_1 - c_2)cos (3t) + i(c_1 + c_2)sin(3t) 
\cr i(c_1+ c_2)cos (3t) - (c_1 - c_2) sin(3t)}\right) $

$=   \left(\matrix{k_1cos (3t) + k_2sin(3t) 
\cr k_2cos (3t) - k_1 sin(3t)}\right) $
$=   k_1 \left(\matrix{cos (3t)  
\cr -sin(3t)}\right) 
+   k_2 \left(\matrix{sin(3t) 
\cr cos (3t) }\right) $
\vskip 10pt

\centerline{Answer: $\underline{~~\left(\matrix{x_1 \cr x_2}\right) =
 c_1 \left(\matrix{cos (3t)                
\cr -sin(3t)}\right)                   
+   c_2 \left(\matrix{sin(3t)                 
\cr cos (3t) }\right) ~~}$}

[20]~ 6.)  Suppose the matrix $A$ has eigenvalue -1 with eigenvector $\left(\matrix{0  \cr 3}\right)$ and \hfil \break
eigenvalue 4 with eigenvector $\left(\matrix{2  \cr 1}\right)$.  Draw the phase portrait for this system of differential equations.  
Identify the equilibrium solution.  Also state the general solution.
\v\v

Equilibrium solution is $\underline{ \left(\matrix{x
\cr y }\right) =  \left(\matrix{0
\cr 0 }\right)}$
\v\v
General solution is $\underline{ \left(\matrix{x        
\cr y }\right) = c_1e^{-t}\left(\matrix{0  \cr 3}\right) + c_2e^{4t}\left(\matrix{2  \cr 
1}\right)}$
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