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22M:100 (MATH:3600:0001) Final Exam \hfil \break
May 15, 2013 \hfill SHOW ALL STEPS \hfill


[13]~ 1.)   Solve:  $ty' + 6t^2y - t^2 = 0$, $y(1) = 3$


$y' + 6ty  = t$, $\int 6tdt = 3t^2$


$e^{3t^2}y' + 6te^{3t^2} =  te^{3t^2}$
~~implies~~
$[e^{3t^2}y]' = te^{3t^2}$
~~implies~~
$\int [e^{3t^2}y]'dt = \int te^{3t^2}dt$

implies
$e^{3t^2}y = \frac{e^{3t^2}}{6} + C$ (via $u$-substitution).~~~
Thus $y = \frac{1}{6} +  Ce^{-3t^2}$

$y(1) = 3$:~~~~  $3 = \frac{1}{6} + Ce^{-3}$
implies $C = \frac{17}{6}e^3$
\vfill
\centerline{Answer: \underline{~~$y = \frac{1}{6} +  \frac{17}{6}e^{-3t^2+3}$~~}}

\vskip 10pt \hrule

[13]~ 2.)  Solve:  $(2x^3y^2 - 3x)dx + (x^4y - y^{-1}) dy = 0$


Note:  $\frac{\partial}{\partial y} (2x^3y^2 - 3x) = 4x^3y
= \frac{\partial}{\partial x}(x^4y - y^{-1})$.  Thus equation is exact.



Let $\psi_x = 2x^3y^2 - 3x$.  Then $\psi = \frac{x^4y^2 - 3x^2}{2} + h(y)$
~~and~~
$\psi_y = x^4y + h'(y) = x^4y - y^{-1}$

$\int h'(y)dy = \int -y^{-1}dy = -ln|y|$ 
~~and~~
$ \frac{x^4y^2 - 3x^2}{2} -ln|y| = C$


\vfill

\centerline{Answer: \underline{~~$ \frac{x^4y^2 - 3x^2}{2} -ln|y| = C$~~}}
\hr
[14]~ 3.)  Use the infinite series solution method (ch 5) about the point $x_0 = 0$ to solve the differential equation, $(x+1) y' - 2y = 0$.

%Note:  You must use the infinite series method.  However, this is still a relatively short problem as most terms will cancel out (so no need for induction proof/ratio test).


Suppose $y = \So a_nx^n$.  Then $y' = \S na_n x^{n-1}$


 $(x+1)  \S na_n x^{n-1} - 2\So a_nx^n = 0$.


 $  \S (x+1)na_n x^{n-1} - 2\So a_nx^n = 0$.

 $\S na_n x^{n} + \S na_n x^{n-1} - 2\So a_nx^n = 0$.

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

$\S na_n x^{n} + \S (n+1)a_{n+1} x^{n} - 2\S a_nx^n  + a_1 - 2a_0= 0$.

$\S [na_n + (n+1)a_{n+1}  - 2a_n]x^n  + a_1 - 2a_0= 0$.

hence $a_1 - 2a_0= 0$ and $na_n + (n+1)a_{n+1}  - 2a_n = 0$ for all $n \geq 1$

hence $a_1 = 2a_0$ and $a_{n+1} = \frac{(2-n)a_n}{n+1}$ for all $n \geq 1$
\vskip 10pt
\centerline{Thus $a_2 = \frac{(2-1)a_1}{2} =  \frac{a_1}{2} =  \frac{2a_0}{2} = a_0 $.  And $a_3 = \frac{(2-2)a_2}{2+1} = 0$.  Thus $a_4 = \frac{(2-3)a_3}{3+1} = 0$.  Thus $a_n = 0~ \forall n \geq 3$.}



Thus $y = a_0x^2 + 2a_0x + a_0$

Check:   $y' = 2a_0x + 2a_0$

 $(x+1) ( 2a_0x + 2a_0) - 2(a_0x^2 + 2a_0x + a_0)$.
 %$(x+1) ( a_0x + a_0) - (a_0x^2 + 2a_0x + a_0)$.
  $= 2(a_0x^2 + a_0x + a_0x + a_0  - a_0x^2 - 2a_0x - a_0) = 0$.
  

\vskip 10pt
\centerline{Answer: \underline{~~$y = a_0x^2 + 2a_0x + a_0$~~}}
\eject

[20]~ 4.)  Solve:  $ y'' - 3y' - 4y = 8t$, $y(0) = 5$, $y'(0) = 3$


$r^2 - 3r - 4 = 0$.  
~~~~~~~~~
$(r - 4)(r + 1) = 0$. ~~~~ Thus $r = 4, -1$.

general homogeneous solution:  $y = c_1e^{4t} + c_2e^{-t}$


Let $y = At + B$.  Then $y' = A$ and $y'' = 0$

$-3A - 4(At + B) = 8t$

$A = -2$, $6 - 4B = 0$.  Thus $B = \frac{3}{2}$


Thus $\psi = -2t + \frac{3}{2}$ is a non-homogeneous solution.

General solution to non-homogeneous DE:   $y = c_1e^{4t} + c_2e^{-t} - 2t + \frac{3}{2}$

$y' =  4c_1e^{4t} - c_2e^{-t} - 2$

$5 = c_1 + c_2 + \frac{3}{2}$

$3 = 4c_1 - c_2 - 2$


Thus $c_1 =  \frac{17}{10}$ and $c_2 =  \frac{9}{5}$


\centerline{Answer: \underline{~~ $y = \frac{17}{10}e^{4t} +   \frac{9}{5}e^{-t} - 2t + \frac{3}{2}$~~}}
\hr

[20]~ 5.)  Suppose $\frac{dx}{dt} = (x-3)(y+2)$ and $\frac{dy}{dt} = 2x - y $. 

(a.)  State all critical points of this system of differential equations  \underline{~~(3, 6), (-1, -2)~~} 

$ (x-3)(y+2) = 0$ implies $x = 3$ or $y = -2$.

$2x - y = 0$ implies if $x = 3$, then $y = 6$  and if $y = -2$, then $x = -1$.

(b.)  State two solutions to this system of differential equations.

$\begin{pmatrix}
x \\ y
\end{pmatrix} = $
$\begin{pmatrix}
3 \\ 6
\end{pmatrix}  $ and $\begin{pmatrix}
x \\ y
\end{pmatrix} = $
$\begin{pmatrix}
-1 \\ -2
\end{pmatrix}  $
  
(c.)  For the critical point with negative $x$ coordinate, find the corresponding linear system.\hb

$\begin{pmatrix}
y+ 2 & x - 3 \\
2 & -1
\end{pmatrix}  $.  For $(x, y) = (-1, -2)$, the Jacobian matrix is $\begin{pmatrix}
0 & -4 \\
2 & -1
\end{pmatrix}.$ 

Thus ${\bf x}' = \begin{pmatrix}
0 & -4 \\
2 & -1
\end{pmatrix} {\bf x}$ is the corresponding linear system where the critical point (-1, -2) has been translated to the origin.


(d.)  Based on the linear translated approximation found in part b, what conclusions regarding stability can you draw about the nonlinear system near the critical point with negative $x$ coordinate?  

$( - \lambda)(-1 - \lambda) + 8 = ( \lambda + \lambda^2) + 8 = 
\lambda^2 + \lambda +8 
%= (\lambda - 1)(\lambda - 4) 
= 0$.  Thus the Jacobian matrix at 
$(x, y) = (-1, -2)$ has two complex eigenvalues $\lambda = 
\frac{-1 \pm \sqrt{1 - 32}}{2}$ that are neither real nor imaginary where the real part is negative.

Hence the (-1, -2) is an asymtotically stable spiral source.

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\hr


[20]~  6.)  Choose 2 from the following three problems (6A, 6B, 6C).  Clearly indicate your TWO choices. You may do more than 2
problems in which case I may substitute your unchosen problem for one of your two choices
(with a penalty) if it improves your grade.

Your two choices:   \underline{\hskip 3in}



6A.)  Suppose $y = \phi_1(t)$, $y = \phi_2(t)$, and $y = \phi_3(t)$ are all solutions to the homogeneous \break
differential equation $y''' + p(t)y'' + q(t)y' + r(t)y = 0$ and suppose that $y = \psi(t)$ is a solution to  $y''' + p(t)y'' + q(t)y' + r(t)y = g(t)$.  Prove that $y = c_1\phi_1(t) + c_2\phi_2(t) +  c_3\phi_3(t) + \psi(t)$ is a solution to  $y''' + p(t)y'' + q(t)y' + r(t)y = g(t)$.


6B.)  Suppose $a_n = \frac{-2a_{n-1}}{n+1}$ for all $n \geq 1$.  Use induction to prove that $a_n = 
\frac{(-2)^n a_0}{(n+1)!}$ for all $n \geq 0$

6C.)  Transform the given initial value problem into an equivalent problem with initial point at the origin:
\vskip -5pt
\centerline{$\frac{dy}{dt} = 2t - 3(y+1)^2$, $y(8) = 10$}


[5 ~ Extra Credit]  Prove the following theorem for up to 5 points extra credit.


Suppose for each $i = 1, ..., n$, $y =  \phi_i(t) $ are solutions to the homogeneous linear differential equation $\Sno p_i(t) y^{(i)}(t) = 0$.  Suppose also that $\{\phi_1, ..., \phi_n\}$ are linearly independent.  Show there is a unique choice of constants $c_i$ such that $\Sn c_i \phi_i$ is a solution to the initial value  problem $\Sno p_i(t) y^{(i)} = 0$, $  y^{(i)}(t_0) = y_i$, $i = 0, ..., n-1$.  

Note you may earn partial credit by proving special cases (eg $n = 1$ or $n = 2$) or just for starting the proof, stating definitions, etc..

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Suppose $M(x, y) + N(x, y)y' = 0$ is an exact differential equation.  Prove that this equation has a solution if $M, N, M_y, N_x$ are continuous.
