MATH 308, Spring 2016
QUIZ # 1
SECTION #:
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INSTRUCTOR: Dr. Marco A. Roque Sol
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1. (10 pts.) Verify that the given function is a solution of the dierential equation.
0
y − 2ty = 1;
y=e
t2
Z
t
2
e−s ds + et
2
0
.
Solution
2 0 R
R
0
2
t −s2
t −s2
t2
y 0 = et
e
ds
+
e
e
ds
+ (et )0
0
0
2
y 0 = 2tet
2
y 0 = 2tet
Rt
0
Rt
0
2
2
2
2
e−s ds + et e−t + 2tet
2
2
e−s ds + 2tet + 1
2R
2
2
t
y 0 = 2t et 0 e−s ds + et + 1
y 0 = 2ty + 1 =⇒ y − 2ty = 1
2. (10 pts ) Determine the values of r for which the given dierential equation has a solution of the form y = ert .
y 00 + y 0 − 6y = 0
.
Solution
y = ert =⇒ y 0 = rert
and
y 00 = r2 ert
Substituing y, y 00 , and y 00 in the ODE
y 00 + y 0 − 6y = (r2 ert ) + (rert ) − 6(ert ) = 0
we obtain
y 00 + y 0 − 6y = (r2 + r − 6)ert = 0
and since ert 6= 0
r2 + r − 6 = (r + 3)(r − 2) = 0
Therefore r = −3, 2 .
3. (10 pts.) Determine the long term behavior of the solution of the given directional eld.
.
Solution
From the graph we can conclude that
1) If the initial y0 is in −∞ < y < 0 or 0 < y < 3 the solution, y → y = 0
1) If the initial y0 is in 3 < y < ∞ the solution, y → y = ∞