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{\bf Name:}  \vrule height .4pt depth 0 pt width 1.75
in  \hskip .25 in {\bf Professor:} A.Alexandrou Himonas
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 \centerline{\bf {Mathematics 226, Solutions to Final Exam}}
\centerline{\bf {Differential Equations and Linear Algebra}}
\centerline{\bf Fall Semester }
\centerline{\bf {December 20, 1991, 8:00-10:00 AM}} 

{\bf 1.}  a)  Solve the initial value problem $$y' - xy + 2x = 0, y(0) = 0.$$

{\bf Solution:}  We have $y' - xy = -2x$ or $(e ^ \frac{x^2}{2})' = -2xe^\frac{-x^2}{2}$ or 
\newline
$e^{-x^2}{2}y = \int -2xe^ {\frac{-x^2}{2}} dx + c = 2e^ \frac{-x^2}{2} + c$ or
$y = 2 + ce^ \frac{-x^2}{2}.$ \newline
Since $0 = y(0) = 2 + c,$  we have $c = -2$ and $y=2-2e^{\frac{-x^2}{2}}$   

\bigskip
                                                                                                                                                                               
b)  Solve the DE $$(xe^{xy} + 4y^3) \frac{dy}{dx} = 2x - ye^{xy}$$

 {\bf Solution.}  We have $$\undersetbrace M \to{ye^{xy} - 2x} + \undersetbrace N
\to{(xe^{xy} + 4y^3)} = 0$$
Since $\frac{\delta M}{\delta y} = e^{xy} + xye^{xy} = \frac{\delta N}{\delta x}$ the DE is
exact.  Its solution is of the form $4(x,y) = C$
 where $\psi (x,y) = \int Mdx = \int(ye^{xy} - 2x) dx = e^{xy} - x^2 + h(y).$
We must have $\frac{\gamma \psi}{\delta y} = N$ or $xe^{xy} + \frac{dh}{dy} = xe^{xy} +
4y^3.$ \newline
Thus, $\frac {dh}{dy} = 4y^3$ or $h(y) = y^4.$  Therefore the solution of the DE is given by
$e^{xy} - x^2 + y^4 = C.$
  
 {\bf 2.}  Given that $y_1 = e^x$ and $y_2 =
x^2e^x$ are two solutions to the DE $$xy'' - (2x + 1)y' + (x + 1)y = 0,$$
find the general solution to the DE $$xy'' - (2x + 1)y' + (x + 1)y = x(x^2 - 2x -1)e^x.$$
{\bf Solution.}  A particular solution of the non-homogeneous equation is of the form
$y_p = y_1u_1 + y_2u_2$ where $u_1$ and $u_2$ satisfy the system $$y_1u_1' +
y_2u_2' = 0 \;\;\; \;\; e^xu_1' + x^2e^xu_2' = 0$$
or $$y_1'u_1' + y_2'u_2' = (x^2 - 2x - 1)e^x, \;\;\;\; e^xu_1' + (2x+1)e^xu_2' = (x^2 - 2x -
1)$$

or $\aligned u_1' + x^2u_2'  &= 0\\
u_1' + (2x + 1)u_2' &= x(x-2x-1) \endaligned$
We have $$u_1' = \frac{\vmatrix 0 & x^2\\(x^2-2x-1) & (2x+1) \endvmatrix}{\vmatrix 1
& x^2\\ 1 & 2x+1 \endvmatrix} = \frac{-(x^2 - 2x -1)x^2}{-x^2 + 2x +1} = x^2$$ and 
$$u_2' = \frac{\vmatrix 1 & 0\\ 1 & (x^2 - 2x - 1) \endvmatrix}{-x^2 + 2x + 1} = -1$$
Therefore we can take $u_1 = \frac{1}{3}x^3$ and $u_2 = -x$ and therefore $y_p =
- \frac{1}{3}x^3e^x - x^3e^x = - \frac{4}{3}x^3e^x.$  The general solution of the DE is
$y=(c_1 + c_2x^2 - \frac{4}{3}x^3)e^x.$

 {\bf 3.}  Let $y(x) = \sum_{n=0}^{\infty} a_n (x-3)^n$ be the power series expansion
of the solution to the initial value problem $$y'' - 2(x-3)y' - 2y =0, y(3) =0, y'(3) = 1.$$  

a)  Find the recurrence relation for the coefficients $a_n.$
 \newline
{\bf Solution.}  We must have 
$\displaystyle \sum_{n=2}^{\infty} n(n-1) a_n (x-3)^{n-2} -2(x-3)$
$\displaystyle \sum_{n=1}^{\infty} na_n (x-3)^{n-1} -2$ 
$\displaystyle \sum_{n=0}^{\infty} a_n (x-3)^n = 0$ \newline
 {\bf or} $\displaystyle \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2} (x-3)^n -2$
$\displaystyle \sum_{n=0}^{\infty} a_n(x-3)^n=0$ \newline
{\bf or} $\displaystyle \sum_{n=0}^{\infty} \left[(n+2)(n+1)a_{n + 2} - 2na_n - 2a_n
\right] (x-3)^n = 0$ \newline
{\bf or}  $(n+2)(n+1)a_{n+2} - 2na_n - 2a_n = 0, \;\;\; n = 0, 1, .....$ \newline
{\bf or} $(n+2)(n+1)a_{n+2} - 2(n+1)a_n = 0, \;\;\; n=0,1,2....$ \newline
{\bf or} $\boxed{a_{n+2} = \frac{2}{n + 2} a_n, \;\; \; n= 0,1,2,..}$
 \bigskip
b)  Then compute all $a_n$ and write the solution $y(x).$  What is its radius of
convergence? \newline
{\bf Solution.}  Since $a_0 = y(2) = 0$ and $a_1 = y'(2) = 1$, we have $a_2 = a_0 = 0, a_4 =
\frac{2}{4}a_2 = 0, ..., a_{2n} = 0, ...a_3 = \frac{2}{3}a, = \frac{2}{3}, a_5 = \frac {2}{5} a_3 =
\frac{2}{3} \frac{2}{5} = \frac{2^2}{3.5}, a_7 = \frac{2}{7}a_5 = \frac{2}{7} \frac{2^2}{3 \cdot 5}
= \frac{2^3}{3 \cdot 5 \cdot 7}, . . ., a_{2n + 1} = \frac{2^n}{3 \cdot 5 \cdots(2n+1)}$
 \newline
Thus $y(x) = \sum_{n = 0}{\infty} \frac {2^n}{3 \cdot 5 \cdots (2n + 1)}(x-3)^{2n+1}$  The radius
of convergence of $y$ is $R = \infty.$
 
{\bf 4.}  Let $$A=\pmatrix  1&3&-5&2&-7\\
-4&-9&17&-8&34\\
-3&6&12&-5&24\\2&3&-7&4&-8
\endpmatrix , x = \pmatrix x_1\\x_2\\x_3\\x_4\\x_5\endpmatrix$$

a)  Find the reduced echelon form of $A.$ \newline
{\bf Solution.}  
a)  By row elementary operations $A$ are reduced to
$$\pmatrix 1&3&-5&2&-7\\
0&3&-3&0&6\\
0&3&-3&1&3\\
0&-3&3&0&6
\endpmatrix     \to  \pmatrix 1&3&-5&2&-7\\
0&1&-1&0&2\\0&0&0&1&-3\\0&0&0&0&0 \endpmatrix$$

$$\pmatrix 1&3&-5&0&1\\
0&1&-1&0&2\\
0&0&0&1&-3\\
0&0&0&0&0 \endpmatrix     \to  \pmatrix 1&0&-2&0&-5\\
0&1&-1&0&2\\
0&0&0&1&0-3\\
0&0&0&0&0 \endpmatrix$$

b)  Find the solution space $W$ of the system 
$Ax = 0$ and a basis for $W.$ \newline
{\bf Solution.} The system $Ax = 0$ is reduced to
$$\aligned
x_1 \hphantom{x_2} -2x_3 \hphantom{x_4} -5x_5 &=0\\
\hphantom{x_1}x_2 -x_3 \hphantom{x_4}+2x_5&=0\\ 
\hphantom{x_1} \hphantom{x_2} \hphantom{x_3}x_4 -3x_5&=0 \endaligned$$

Let $x_5 = \alpha,$   $x_3 =\beta$  then $x_4 = 3\alpha,$ $x_2 =\beta - 2 \alpha$
and $x_1 = 5 \alpha + 2 \beta$ and

$$W =  \left\{ 
\pmatrix 
x_1\\
x_2\\
x_3\\
x_4\\
x_5 \endpmatrix = \pmatrix 5 \alpha + 2 \beta\\
\beta - 2 \alpha\\
3 \alpha\\
\beta\\
\alpha \endpmatrix = \alpha \pmatrix 5\\-2\\3\\0\\1 \endpmatrix + \beta \pmatrix
2\\1\\0\\1\\0 \endpmatrix : \alpha, \beta, \Bbb R \right\}$$

A basis for $W$ is $\left\{ \pmatrix 5\\ -2\\ 3\\ 0\\ 1 \endpmatrix , \pmatrix
2\\ 1\\ 0\\ 1\\ 0 \endpmatrix \right \}$ 

c)  Rank of $A = 3$  dim $(W) = 2$

{\bf 5.}  Use Cramer's Rule to solve the system: \newline

$$\aligned
3x_1 + x_2 \hphantom{ + x + 3} &= 1\\
-2x_1 -4x_2+3x_3 &= -7\\
5x_1 + 4x_2 -2x_3 &=6 \endaligned$$

{\bf Solution.}  The determinant $\Delta$ of the coefficient is $\Delta = \vmatrix 3 & 1 & 0 \\-2
& -4 & 3 \\ 5 & 4 & -2 \endvmatrix = 3 \vmatrix-4 & 3 \\ 4 & -2 \endvmatrix - 
\vmatrix -2 &3\\
5 & -2 \endvmatrix  
  = 3(8-12)-(4-15)
  = -12 + 11 = -1$

$$ \Delta_1 = \vmatrix 1 & 1 & 0\\-7 & -4 & 3\\6 & 4 & -2 \endvmatrix = \vmatrix -4 & 3\\4 &
-2 \endvmatrix - \vmatrix -7 & 3\\ 6 & -2 \endvmatrix = (8 - 12) -(14-18) = 0$$


$$ \Delta_2 = \vmatrix 3 & 1 & 0\\-2 & -7 & 3\\\\5 & 6 -2 \endvmatrix = 3 \vmatrix -7 & 3\\6
& -2 \endvmatrix - \vmatrix -2 & 3\\5 & -2 \endvmatrix = 3(14-18)-(4-15) = -1$$
$$ \Delta_3 = \vmatrix 3 & 1 & 1\\-2 & -4 & -7\\5 & 4 & 6 \endvmatrix = 3 \vmatrix -4 &
-7\\4 & 6 \endvmatrix - \vmatrix -2 & -7\\5 & 6  \endvmatrix + \vmatrix -2 & -4\\5 & 4
\endvmatrix$$
$=3(-24 + 28) - (-12 + 35) + (-8 +20) = 12 - 23 + 12 = 1.$
 \bigskip
Therefore $x_1 = \frac{\Delta_1}{\Delta} = \frac{-1}{-1} = 1, x_2 = \frac{\Delta_2}{\Delta} =
\frac{0}{-1} = 0, x_3 = \frac{\Delta_3}{\Delta} = \frac{1}{-1} = -1.$ 

{\bf 6.}  Find the general solution of the system of the DE
$$\frac{dx}{dt} = \pmatrix 1 & 2\\ 4 & 3 \endpmatrix x,  \;  x = \pmatrix x_1(t)\\ x_2(t)
\endpmatrix.$$

{\bf Solution.}  Let $x = \xi e^{\lambda t}$ where $\xi = (\frac{\xi_1}{\xi _2})$ is a constant
vector and $\lambda$ is a constant.  For $x$ to be a solution we must have
$$\lambda e^{\lambda t} \xi = e^{\lambda t} A \xi$$

The last system has non-zero solutions if $|A - \lambda I | = 0$ or $\vmatrix 1 - \lambda & 2\\
4 & 3 - \lambda \endvmatrix = 0$  {\bf or} $(1 - \lambda)(3 - \lambda) - 8 = 0$ {\bf or}
$\lambda = -1$ and $\lambda_2 = 5.$  An eigenvector $\xi$ with eigenvalue $\lambda = -1$
satisfies the system $\pmatrix 2 & 2\\4 & 4 \endpmatrix \pmatrix \xi_1\\\xi_2 \endpmatrix =
\pmatrix 0\\0 \endpmatrix$  {\bf or} $\xi_1 + \xi_2 = 0$ {\bf or} $\xi = \alpha \pmatrix 1\\-1
\endpmatrix ,  \alpha \in \Bbb R.$

If $\lambda = 5$ then $\xi$ satisfies the system $\pmatrix -4 & 2\\4 & -2 \endpmatrix
\pmatrix \xi_1\\ \xi_2 \endpmatrix = \pmatrix 0\\ 0 \endpmatrix$ 
\newline
{\bf or}  $$2 \xi_1 - \xi_2 = 0 \text{or} \pmatrix \xi_1\\  \xi_2 \endpmatrix = \alpha \pmatrix
1\\ 2 \endpmatrix.$$

 
Therefore, the general solution for our system of DE is
given by  $$\pmatrix x_1\\ x_2 \endpmatrix = c_1 \pmatrix 1\\ -1 \endpmatrix e^{-t} + c_2
\pmatrix 1\\ 2 \endpmatrix e^{5t}.$$



 {\bf 7.}  Assume that
the human population $p = p(t)$ of the earth obeys the logistic equation $$\frac{dp}{dt} = r(1 -
\frac{p}{k})p, \;  r = 0.029,  \; k = 10^{10}.$$ If the population now is $\frac{1}{2}10^{10}$, find
the population $p(t)$ at any time $t.$  \newline
Also find $\displaystyle \lim_{t \to \infty} p(t).$

{\bf 8.}  Find all terms up to degree four of the Taylor series expansion at $x_0 = 0$ of
the function $y = y(x)$, where $y$ is the solution to the following initial value problem

$$y'' - xy' - e^{2y} = 0, \; \; y(0) = 0, \;\; y'(0) = 1.$$
  

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