# MATH 308, Spring 2016 Take Home QUIZ # 6 ```MATH 308, Spring 2016
Take Home QUIZ # 6
Due: Thursday 03 / 07 / 2016. In my oce 10:00 a.m.
SECTION #:
Print name (LAST, First):
INSTRUCTOR: Dr. Marco A. Roque Sol
THE AGGIE CODE OF HONOR
&quot;An Aggie does not lie, cheat, or steal, or tolerate those who do.&quot; By signing below, you indicate that all
work is your own and that you have neither given nor received help from any external sources.
SIGNATURE:
1. (10 pts.) Find the general solution to
y 00 + 3y 0 + 2y = et
.
Using MATLAB* plot the graph of y(t) for dierent values of the constants c1 and c2 .
2. (10 pts.) Find the general solution to
y 00 + 4y 0 + 5y = sin(2t)
.
Using MATLAB* plot the graph of y(t) for dierent values of the constants c1 and c2 .
3. (10 pts.) Solve the initial value problem
y 00 + 5y 0 + 6y = e−3t ;
y(0) = 1, y 0 (0) = 0
.
Using MATLAB* plot the graph of y(t) .
4. (10 pts.) Consider the initial value problem
9y 00 + 12y 0 + 4y = 0;
y(0) = a &gt; 0, y 0 (0) = −1
.
(a) Solve the initial value problem. (b) Find the critical value of a that separates solutions that become negative
from those that are always positive. Using MATLAB* plot the graph of y(t) for dierent values of a, showing
this transition.
5. ( 10 pts. ) Use the method of reduction of order to nd a second solution of the given dierential equation.
t2 y 00 − 4ty 0 + 6y = 0;
t &gt; 0, y1 (t) = t2
.
t2 y 00 + 3ty 0 + y = 0;
.
t &gt; 0, y1 (t) = t−1
6. ( 20 pts. )
(a) Consider the equation y 00 +2ay 0 +a2 y = 0. Show that the roots of the characteristic equation are r1 = r2 = −a,
so that one solution of the equation is e−at .
(b) Use Abel's formula, to show that the Wronskian of any two solutions of the given equation is
W (t) = y1 (t)y20 (t) − y10 (t)y2 (t) = c1 e−2at ,
where c1 is a constant.
(c) Let y1 (t) = e−at and use the result of part (b) to obtain a dierential equation satised by a second solution
y2 (t). By solving this equation, show that y2 (t) = te−at .
(d) If y1 is any known nonvanishing solution of y 00 + p(t)y + q(t)y = 0, show that a second solution y2 satises
(y2/y1)0 = W (y1 , y2 )/y12 (Hint: Use the quotient rule), where W (y1 , y2 ) is the Wronskian of y1 and y2 . Then
use Abel's formula to determine y2 .
7. ( 10 pts. ) Using the results from the previous problem to nd a second independent solution
ty 00 − y 0 + 4t3 y = 0;
t &gt; 0, y1 (t) = sin(t2 )
.
Using MATLAB* plot the graph of the general solution y(t) = c1 y1 (t) + c2 y2 (t) for dierent values of the
constants c1 and c2 .
*MATLAB
Use the following code in MATLAB to plott a function
[email protected](t) (-24/37)*cos(3*t) - (4/37)*sin(3*t) + (0.2)*exp(t/2); % Dene the function.
axis([0 8 -8 8]);
% Dene the axis.
hold on
%Hold on the gure.
grid on
% Show the grid.
fplot(r1,[0,8]);
% Plot the function on [ 0, 8 ]
xlabel('x');
ylabel('y');
title([ 'Function r1 (t)0 ]); %Addatitle
text(2,(-24/37)*cos(3*2) - (4/37)*sin(3*2) + (0.2)*exp(2/2), ...
'r1 = (−24/37)cos(3t) − (4/37)sin(3t) + (0.2)et/2 0 ); %Addatext
If you repeat the same program with dierent functions, without erasing the previous graphs, at the end of the
day you will have a plot with all of them.
% First function
..
.
..
.
clear t [email protected](t) (-24/37)*cos(3*t) - (4/37)*sin(3*t) + (0.2)*exp(t/2);;
axis([0 8 -8 8]);
hold on
grid on
fplot(r1,[0,8]);
xlabel('x');
ylabel('y');
title(['Functions r(t)']);
t=0.5;
text(t,(-24/37)*cos(3*t) - (4/37)*sin(3*t) + (0.2)*exp(t/2), ...
'r1 = (−24/37)cos(3t) − (4/37)sin(3t) + (0.2)et/2 0 );
% Second function
% Third function
```