18.034 Solutions to Problemset 1
Spring 2009
1. (b) a = 1 or a = 4
2. (a) y(π/6) = e2
(b) Same as part (a)
(c) Because dx is not integrable on any interval containing the point
x
x = 0.
3. (a) y is increasing because y� = y2 + 1 > 0. The formula is obtained
by the separation of variables.
(b) y(x) = tan(x − c) is defined on the interval (c − π/2, c + π/2).
4. (b) Let c = sup{x : y(x) = 0}.
If c = +∞, then y = y1. If c < +∞, then y(x) = 0 for x ≤ c
and y(x) = (x − c)3/2 for x > c by the separation of variables and
uniqueness.
5. Let q(x) = anxn + an−1xn−1 + . . . + a0, an = 0 �
an n + an−1
The general solution y(x) = c/x+ n+1 xn xn−1 +. . .+a0, (x = 0) �
where c is constant admits the only polynomial solution when c = 0.
6. (a) 1−
1 n u� + p(x)u = q(x).
(b) y = u−1/2 where u� − 2u = −2x for y = 0. �
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