MATH 2110/3405 Assignment
Chapter 4:
1. Solution. Rewrite the equation as:
y (3) +
et
4t2
ln t
y ′′ +
y=
, y(t0 ) = 1, y ′ (t0 ) = 0, y ′′ (t0 ) = 0.
(t − 1)(t − 2)
(t − 1)(t − 2)
(t − 1)(t − 2)
t
2
e
4t
ln t
The functions (t−1)(t−2)
, (t−1)(t−2)
, and (t−1)(t−2)
are continuous on (0, 1), (1, 2), and (2, ∞).
Then from Theorem 4.1.1, a solution exists if t0 ∈ (0, 1), or t0 ∈ (1, 2), or t0 ∈ (2, ∞).
2. Solution. If there are constants k1 , k2 , k3 , and k4 , such that
k1 f1 (t) + k2 f2 (t) + k3 f3 (t) + k4 f4 (t) = 0, ∀t,
or
k1 (2t − 3) + k2 (t3 + 1) + k3 (2t2 − t) + k4 (t2 + t + 1) = 0, ∀t.
Then we have (coefficients of t3 , t2 , t1 , t0 )
k2 = 0, 2k3 + k4 = 0, 2k1 − k3 + k4 = 0, −3k1 + k2 + k4 = 0.
Solve this we get k1 = k2 = k3 = k4 = 0. Thus, f1 (t), f2 (t), f3 (t), and f4 (t) can not be linearly
dependent, they are linearly independent.
3. Solution. y = c1 et + c2 t + c3 tet .
4. Solution. (a). y = c1 + c2 t + c3 e2t +√c4 te2t .
√
(b). y = c0 + c1 t + c2 e2t + e−t (c3 cos 3t + c4 sin 3t)
5. Solution. y = c1 et + c2 e−t + c3 cos t + c4 sin t − 3t − (1/4) sin t + (1/4)t cos t.
√
√
6. Solution. y = c1 + c2 t + c3 t2 + c4 e−t + et/2 [c5 cos( 3t/2) + c6 sin( 3t/2)] + (1/24)t4 .
7. Solution. (a). y = t(A0 t3 + A1 t2 + A2 t + A3 ) + Bt2 et .
(b). y = t(A0 t2 + A1 t + A2 ) + (B0 t + B1 ) cos t + (C0 t + C1 ) sin t.
8. Solution. y = c1 et + c2 cos t + c3 sin t − (1/5)e−t cos t.

HW 4, Due on Oct. 20

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