Math 331 Solution of Midterm 1 Fall 2003

1. (25 points) a) (20 points) Solve the initial value problem $\begin{array}{rcl} y'' - y' -6y & = & 0, \\ y(0) & = & 1, \\ y'(0) & = & y'_0 \end{array}$

   Answer: The two roots of the characteristic equation $\lambda^2-\lambda-6=0$ are $\lambda_1=3$ and $\lambda_2=-2$. Hence, the general solution is

   $$\begin{eqnarray*} y(t) & = & c_1e^{3t} + c_2 e^{-2t} \ \ \mbox{and its derivative is} \\ y'(t) & = & 3c_1e^{3t} -2c_2e^{-2t} \end{eqnarray*}$$
   
   

   Plugging in $t=0$, the initial value problem translates to the system of two linear equations (where we regard $y'_0$ as a given constant and $c_1$ and $c_2$ as the unknowns):

   $$\begin{eqnarray*} 1 & = & c_1+c_2 \\ y'_0 & = & 3c_1 - 2c_2. \end{eqnarray*}$$
   
   

   Solving for the coefficients, we get ${\displaystyle c_1=\frac{2+y'_0}{5}}$ and ${\displaystyle c_2=\frac{3-y'_0}{5}}$. The solution is thus
   

   $$y(t) \ = \ \frac{2+y'_0}{5}\cdot e^{3t} \ + \ \frac{3-y'_0}{5}\cdot e^{-2t}.$$
   
   

   b) (5 points) Find a condition on the slope $y'_0$ which guarantees that ${\displaystyle \lim_{t\rightarrow \infty}y(t) = \infty. }$ Explain your answer!

   
   Answer: The limit ${\displaystyle \lim_{t\rightarrow \infty}y(t) }$ is $+\infty$ if and only if the coefficient $c_1=\frac{2+y'_0}{5}$ is positive, i.e., if and only if $y'_0 > -2$.

2. (25 points) a) (4 points) Given below is the direction field
   

   $$y' \ = \ y+e^{x/2}-3.$$ (1)

   
   
   Use the direction field in order to sketch the graphs of the following two solutions.

   The solution satisfying $y(0)=-1.$

   The solution satisfying $y(0)=4.$

   Answer: We sketch the graph $y=\phi(x)$, of the solution $\phi(x)$ satisfying $y(0)=\phi(0)=-1$, by starting at the initial point $(x_0,y_0)=(0,-1)$ and ``flowing'' along the direction field (in both the negative and positive direction of $x$). The resulting solution is decreasing. We do the same for the solution through $(x_0,y_0)=(0,4)$. The second solution is increasing.

   b) (4 points) Based on the direction field, describe how the behavior of the solution, as $x$ gets larger, depends on the value $y_0$ in the initial condition
   

   $$y(0)=y_0.$$ (2)

   
   

   Answer: The slope is zero along the curve $y=3-e^{x/2}$ and in particular at $(x_0,y_0)=(0,2)$. If $y_0>2$, then the solution $y(x)$ is increasing and positive and its slope is increasing with $x$. Hence, the solution would satisfy ${\displaystyle \lim_{x\rightarrow \infty}y(x)=\infty}$.

   There is a range $y_{critical}< y_0 < 2$, for which the solution is first decreasing, but eventually increasing. In part d we will see, that this happens for $1\leq y_0<2$. At this point, it is not obvious what $y_{critical}$ is (and even, that it is a finite number and not $-\infty$).

   It seems plausible, that $y_{critical}$ exists as a finite number $<2$, and solutions with $y_0<y_{critical}$ are decreasing and satisfy ${\displaystyle \lim_{x\rightarrow \infty}y(x)=-\infty}$.

   c) (13 points) Find the general solution of the differential equation (1).

   
   Answer: The equation is linear and its normal form is
   

   $$y'+p(x)y \ \ = \ \ q(x)$$
   
   
   with $p(x)=-1$ and $q(x)=e^{x/2} - 3$. We multiply both sides by the integrating factor $\mu(x)=e^{\int p(x)dx}=e^{-x+{\rm constant}}= {\rm (constant)}\cdot e^{-x}$ to get
   
   $$e^{-x}(y'-y) \ \ = \ \ e^{-x/2}-3e^{-x}.$$
   
   
   Integrating both sides, we get the equations
   $$\begin{eqnarray*} e^{-x}y & = & -2e^{-x/2}+3e^{-x}+C \ \ \ \mbox{or} \\ y & = & -2e^{x/2}+3+Ce^{x}. \end{eqnarray*}$$
   
   

   d) (2 points) Solve the initial value problem (1) and (2) (express your solution in terms of $y_0$).

   
   Answer: Plug $0$ for $x$ and $y_0$ for $y$ to get that $C=y_0-1$ and the solution is:
   

   $$y = -2e^{x/2}+3+(y_0-1)e^{x}.$$
   
   

   e) (2 points) Determine the range of values of $y_0$ for which the solution $y(x)$ in part (d) satisfies
   

   $$\lim_{x\rightarrow \infty}y(x) = +\infty.$$
   
   

   Factor out $e^{x/2}$ to write $y = e^{x/2}[-2+3e^{-x/2}+(y_0-1)e^{x/2}]$. We see, that $\lim_{x\rightarrow \infty}y(x) = +\infty$ if and only if $y_0>1$.

3. (25 points) (a) (20 points) Solve the initial value problem ${\displaystyle \begin{array}{rcl} y' & = & {\displaystyle\frac{y^2}{x}}, \\ y(1) & = & y_0, \end{array}}$ where $y_0$ is an arbitrary real number.

   
   Answer: If $y_0=0$, then the solution is the constant function $y\equiv0$. The equation is separable. If $y_0\neq 0$, then we separate variables by dividing by $y^2$ to get:
   

   $$y^{-2}\cdot y' \ \ = \ \ \frac{1}{x}.$$
   
   
   Integrating both sides with respect to $x$ (using the substitution $y'dx=dy$ on the left hand side), we get
   $$\begin{eqnarray*} -y^{-1} & = & \ln\mid\!x\!\mid +C \ \ \ \mbox{or} \\ y & = & \frac{-1}{\ln\mid\!x\!\mid +C}. \end{eqnarray*}$$
   
   
   The initial condition determines $C=-1/y_0$ and $x$ is positive in the interval of definition containing $x=1$ (since $\ln\mid\!x\!\mid$ is not defined for $x=0$). We get the solution
   
   $$y \ \ = \ \ \left\{ \begin{array}{lcl} \frac{1}{(1/y_0)-\ln... ...if} & y_0\neq 0, \\ 0 & \mbox{if} & y_0=0. \end{array}\right.$$
   
   

   (b) (5 points) Determine the largest interval, on which the solution of the initial value problem is defined, in terms of the constant $y_0$. For which value of $y_0$, will the solution be defined for all $x\geq 1$ (as well as some other values of $x$)?

   
   Answer: If $y_0\neq 0$, then the solution is not defined for $x=0$ and $x=e^{1/y_0}$. Hence, the largest interval of definition, containing $x=1$, is:
   

   $$\left\{ \begin{array}{lcl} 0<x<e^{1/y_0} & \mbox{if} & y_0>0... ..., \\ -\infty<x<\infty & \mbox{if} & y_0=0. \end{array}\right.$$
   
   
   We see, that the solution is defined, for all $x\geq 1$, if and only if $y_0\leq 0$.

4. (25 points) (a) (20 points) Solve the initial value problem
   
   $$y' \ = \ \frac{6x+\frac{y}{x}}{2y-\ln(x)}, \ \ \ y(1)=1$$
   
   

   Answer: The equation becomes exact, when written in the form
   

   $$-\left(6x+\frac{y}{x}\right) + \left(2y-\ln(x)\right)y' \ \ \ = \ \ \ 0.$$
   
   
   Set $M(x,y):=-\left(6x+\frac{y}{x}\right)$ and $N(x,y):=2y-\ln(x)$. Exactness holds, since $M_y=-1/x=N_x$. We look for a function $\Psi(x,y)$ with partials
   

   $$\Psi_x = M(x,y) \ \ \ \mbox{and}$$ (3)

   $$\Psi_y = N(x,y).$$ (4)

   
   
   Integrating both sides of equation (3), we get
   
   $$\Psi(x,y) \ \ = \ \ \int -\left(6x+\frac{y}{x}\right)dx \ = \ -3x^2 -y\ln\mid\!x\!\mid +h(y)$$
   
   
   for some function $h(y)$ of $y$. We determine $h(y)$ using equation (4), which yields
   
   $$-\ln\mid\!x\!\mid + h'(y) \ \ \ = \ \ \ 2y -\ln\mid\!x\!\mid .$$
   
   
   Hence, $h'(y)=2y$ and we can choose $h(y)=y^2$. The general solution is given implicitly by
   
   $$-3x^2 -y\ln\mid\!x\!\mid +y^2 \ \ \ = \ \ \ C.$$
   
   
   The initial condition determines, that $C=-2$ and the above equation is equivalent to
   

   $$y^2-\ln\mid\!x\!\mid y+(2-3x^2) \ \ \ = \ \ \ 0.$$ (5)

   
   

   (b) (5 points) Graph your solution in the window $0<x<2$ and $0\leq y \leq 5$.

   
   Answer: The solution of part a) was given by the quadratic equation (5). The solutions for a general quadratic equation $ay^2+by+c=0$ are given by the formula $y_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Use the formula with $a=1$, $b=-\ln\mid\!x\!\mid$ and $c=2-3x^2$, to obtain
   

   $$y(x) \ \ = \ \ \frac{\ln\mid\!x\!\mid +\sqrt{\ln^2\mid\!x\!\mid +12x^2-8}}{2}.$$
   
   
   The sign in the numerator is determined by the initial condition $y(1)=1$. Now, use your graphing calculator, to obtain the graph below. Students who obtained the graph using the software DEgph on the TI-85/86 or the built-in software on their calculator, got full credit.