FINAL EXAM REVIEW SOLUTIONS
1. Equations of first order
(1) y = t−2 [(π 2 /4) − 1 − t cos(t) + sin(t)]
(2) y = (t − 1 + 2e−t )/t,
t 6= 0
√
(3) y = −(3/4) + (1/4) 65 − 8ex − 8e−x
(4) x2 y 2 + 2xy = c
(5) y = −(1 + t)e−t /t4 ,
t 6= 0
(6) x3 − x2 y + 2x + 2y 3 + 3y = c
(7) x = (c/a)y +
(ad−bc)
a2
ln |ay + b| + k,
a 6= 0, ay + b 6= 0
(8) y = (et − e + a sin(1))/ sin(t),
(9) your answer here
p
(10) y = − (x2 + 1)/2
(11) y = ex sin(y) + 2y cos(x) = c,
2
ory = 0.
2
(12) y = t2 e−t + ce−t
(13) y = −(2 ln(1 + x2 ) + 4)1/2
(14) (3x2 y + y 3 )e3x = c
(15) y = cex + 1 + e2x
2. Equations of second order
(1) y = c1 et sin(t) + c2 et cos(t)
(2) y2 (t) = tet
1
2
FINAL EXAM REVIEW SOLUTIONS
(3) t = −e−t/3 cos(3t) + (5/9)e−t/3 sin(3t)
(4) y = c1 et/3 + c2 e−4t/3
(5) y2 (x) = cos(x2 )
(6) y = e−t cos(2t) + (1/2)e−t sin(2t) + te−t sin(2t)
(7) y = e−2t cos(t) + 2e−t sin(t)
(8) y = c1 et + c2 tet − (1/2)et ln(1 + t2 ) + tet tan−1 (t)
√
√
√
√
(9) t = (2/ 33) exp[(−1 + 33)t/4] − (2/ 33) exp[(−1 − 33)t/4]
(10) y = c1 cos(t) + c2 sin(t) − cos(t) ln(tan(t) + sec(t))
(11) y = c1 cos(t) + c2 sin(t) − (1/3)t cos(2t) − (5/9) sin(2t)
(12) y = e3t + (2/3)e−t − (2/3)e2t − te2t
(13) particular solution Y (t) = (1/2)(t − 1)e2t
3. Series solutions
(1) an+2 =
−(n2 −2n+4)
2(n+1)(n+2) an ,
n ≥ 2,
a2 = −a0 ,
a3 = −(1/4)a1
(2) y = c1 x3 + c2 x3 ln(|x|)
(3) (n + 2)(n + 1)an+2 + (n − 2)(n − 3)an = 0,
n≥0
(4) indicial equation: 2r2 + r − 1, recursion (2n + 2r − 1)(n + r + 1)an + 2an−2 = 0.
(5) (n + 2)(n + 1)an+2 − (n + 1)nan+1 + (n − 1)an = 0,
n ≥ 0.
(6) indicial equation r2 − 4r + 3, recursion (n + r − 3)(n + r − 1)an − (n + r − 2)an−1 = 0
(7) y1 = x + 23 x2 +
9/4 3
x
+
51 4
16 x
y2 = 3y1 (x) ln(x) + 1 −
+ ...
21 2
4 x
−
19 3
4 x
+ ...
FINAL EXAM REVIEW SOLUTIONS
3
4. The Laplace transform
(1) y = e−t sin(t)+ 12 uπ (t)[1+e−(t−π) cos(t)+e−(t−π) sin(t)]− 21 u2π (t)[1+e−(t−2π) cos(t)+
e−(t−2π) sin(t)]
(2) y = cos(t) + u3π (t)[1 − cos(t − 3π)]
(3) y = h(t) − uπ/2 (t)h(t − π/2),
(4) y =
1
4
h(t) =
4
25 [−4
+ 5t + 4e−t/2 cos(t) − 3e−t/2 sin(t)]
√
√
√
sin(t) − 14 cos(t) + 14 e−t cos( 2t) + (1/ 2)u3π (t)e−(t−3π) sin( 2(t − 3π))
(5) y = uπ/2 (t)[1 − cos(t − π/2)] + 3u3π/2 (t) sin(t − 3π/2) − u2π (t)[1 − cos(t − 2π)]