MATH 308, Spring 2016 Take Home QUIZ # 6

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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
"An Aggie does not lie, cheat, or steal, or tolerate those who do." 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 > 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 > 0, y1 (t) = t2
.
t2 y 00 + 3ty 0 + y = 0;
.
t > 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 > 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');
% Add an x label
ylabel('y');
% Add an y label
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
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