TMA4110 Calculus 3
Fall 2012
Solutions to exercise set 2
Norwegian University of Science
and Technology
3 Nonlinear.
7 Nonlinear.
8 Linear and homogeneous
16 y100 (t) + 4y10 (t) + 4y1 (t) = 4e−2t − 8e−2t + 4e−2t = 0,
y200 (t) + 4y20 (t) + 4y2 (t) = −2e−2t − 2e−2t + 4te−2t + 4e−2t − 8te−2t + 4te−2t = 0, and
d2
d2
d
−2t +
(C1 y1 (t) +C2 y2 (t)) +4(C1 y1 (t) +C2 y2 (t)) = dt
(C1 y1 (t) +C2 y2 (t)) +4 dt
2 (C1 e
dt2
2
d
d
−2t ) +
(C1 e−2t + C2 te−2t ) + 4(C1 e−2t + C2 te−2t ) = dt
C2 te−2t ) + 4 dt
2 ((C1 + C2 t)e
d
−2t
−2t
−2t
−2t
− 2C2 e
+ 4(C1 + C2 t)e−2t +
4 dt ((C1 + C2 t)e ) + 4((C1 + C2 t)e ) = −2C2 e
4C2 e−2t − 8(C1 + C2 t)e−2t + 4(C1 + C2 t)e−2t = 0.
18 Assume that C1 y1 (t) + C2 y2 (t) = 0 for all t. Then C1 = C1 y1 (0) + C2 y2 (0) = 0 and
C2 = C1 y1 (π/6)+C2 y2 (π/6) = 0. This shows that y1 and y2 are linearly independent.
y100 (t)+9y1 (t) = −9 cos(3t)+9 cos(3t) = 0 and y200 (t)+9y2 (t) = −9 sin(3t)+9 sin(3t) =
0, so y1 and y2 are solutions to the equation y 00 (t) + 9y(t) = 0.
W (t) = y1 (t)y20 (t) − y10 (t)y2 (t) = 3 cos2 (t) + 3 sin2 (t) = 3
for all t, so it follows that y1 and y2 are linearly independent.
21 Assume that C1 y1 (t)+C2 y2 (t) = 0 for all t in the interval (−∞, +∞). Then C1 +C2 =
C1 y1 (1) + C2 y2 (1) = 0 and C1 − C2 = C1 y1 (−1) + C2 y2 (−1) = 0. It follows that
C1 = C2 = 0. So y1 and y2 are linearly independent on (−∞, +∞).
W (t) = y1 (t)y20 (t) − y10 (t)y2 (t) = −2t2 |t| − 2t2 |t| = 0
for all t. This does not contradict Proposition 1.27 because y2 is not two times
differentiable and thus not a solution to a second order differential equation.
22 y100 (t)+2y10 (t)−3y1 (t) = et +2et −3et = 0 and y200 (t)+2y20 (t)−3y2 (t) = 9e−3t −6e−3t −
3e−3t = 0, so y1 and y2 are solutions to the differential equation y 00 (t)+2y 0 (t)−3y(t) =
0.
W (t) = y1 (t)y20 (t) − y10 (t)y2 (t) = −3et e−3t − e−3t et = −4e−2t 6= 0
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Solutions to exercise set 2
for all t, so y1 and y2 are linearly independent and therefore form a fundamental set
of solutions for the differential equation y 00 (t) + 2y 0 (t) − 3y(t) = 0.
y(t) = 1/4et +3/4e−3t is a solution to the differential equation y 00 (t)+2y 0 (t)−3y(t) = 0
and satisfies that y(0) = 1 and y 0 (0) = −2.
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