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Unformatted text preview: MATH 251
Fall 2001
Exam I
September 25, 2001
ANSWERS:
1. There are inﬁnitely many correct answers for each part. A few examples are given.
(a) e.g., y = y 2 or y = ey
(b) e.g., y = 0 or y + y + y = 0
2. y (t) = √ 2et − 1 3. (First move everything to the lefthand side of the equation and check to see that this is an exact
equation.)
Solution: −2x − exy + y 2 = C
4.
The initial value problem is
Q =8−
−t
The solution is Q(t) = 400 − 300e 50 .
Finally, lim Q(t) = 400 1
50 Q, Q(0) = 100. t→∞ 3
5. First rewrite the equation as y − y = 1, y (4) = −1.
t
(a) The guaranteed solution interval is (0, ∞).
t3
−t
+
(b) y (t) =
3
64
6. (a) The equilibrium solutions are: y = −1 (stable), y = 1 (unstable), and y = 2 (stable).
7. (a) y (t) = C1 e2t cos t + C2 e2t sin t
(b) y (t) = C1 e−3t + C2 te−3t
8. (a) y (t) = 3et − e4t
(b) lim y (t) = lim et (3 − e3t ) = −∞
t→∞ t→∞ 9. First substitute y1 (t) and y2 (t) into the equation to verify that they both satisfy it. This shows that both
functions are indeed solutions of the given equation. Then calculate their Wronskian, W (y1 , y2 ) = t4 = 0
when t > 0. This shows that they are linearly independent. Therefore, the two functions do, in fact, form a
fundamental set of solutions for the given equation. ...
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 Spring '08
 CHEZHONGYUAN
 Math, Differential Equations, Equations, Partial Differential Equations

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