Indian Institute of Technology Roorkee
Mathematics-II (MAN-002)
Assignment-5
Laplace Transforms (II) Spring Semester-2015-16
Solve the following differential equations using Laplace transforms.
1. y 00 − 3y 0 + 2y=4e2t ; y(0) = −3, y 0 (0) = 5
2. y 00 + 2y 0 + 5y=e−t sin t; y(0) = 0, y 0 (0) = 1
3. y 000 − 3y 00 + 3y 0 − y=t2 et ; y(0) = 1, y 0 (0) = 0, y 00 (0) = −2
4. y 00 + 9y= cos 2t; y(0) = 1, y(π/2) = −1
5. y 00 − ty 0 + y=1; y(0) = 1, y 0 (0) = 2
6. x00 + y 0 + 3x = 15e−t ; x(0) = 35, x0 (0) = −48
y 00 − 4x0 + 3y = 15 sin 2t; y(0) = 27, y 0 (0) = −55
7. y iv − y = 1; y(0) = y 0 (0) = y 00 (0) = y 000 (0) = 0
8. (D2 + n2 )y = a sin(nt + α); y(0) = y 0 (0) = 0
9. ty 00 + y 0 + 4ty = 0; y(0) = 3, y 0 (0) = 0
10. ty 00 + 2y 0 + ty = 0; y(0+) = 1, y(π) = 0
11. (D − 2)x − (D + 1)y = 6e3t , (2D − 3) + (D − 3)y = 6e3t ; x(0) = 3, y(0) = 0
12. Solve Initial Value problem:
y 00 + 2y 0 + 5y = h(t);
where h(t) =
1,
0,
y(0) = 0, y 0 (0) = 0
0 < t < π,
t > π,
13. In an electrical circuit with e.m.f. E(t), resistance R and inductance L,
di
+ Ri = E(t). If the switch is
the current i builds up at the rate given by L dt
connected at t = 0 and disconnected at t = a, find the current i at any instant.
14. Obtain the deflection of a weightless beam of length l and freely supported
at ends, when a concentrated load W acts at x = a. The differential equation
d4 y
for deflection being EI dx
4 = W δ(x − a).
15. The coordinate (x, y) of a particle moving along a plane curve at any time
dx
t, given by dy
dt + 2x = sin 2t, dt − 2y = cos t(t > 0). If at t = 0, x = 1
and y = 0, show by Laplace transform that the particle moves along the curve
4x2 + 4xy + 5y 2 = 4.
ANSWERS
1. y= −7et + 4e2t + 4te2t
2. y = 13 e−t (sin t + sin 2t)
2 t
5 t
3. y = et − tet − t 2e + t60e
4
4
4. y = 5 cos 3t + 5 sin 3t + 15 cos 2t
5. y = 1 + 2t
6. x = 30 cos t − 15 sin 3t + 3e−t + 2 cos 2t; y = 30 cos 3t − 60 sin t − 3e−t + sin 2t
7. y = −1 + 12 (cosh t + cos t)
8. y = 2na 2 (cos α sin nt − nt cos(α + nt))
4
9. y = 3(1 − t2 + t4 − ...)
10. y = sint t
11. x = et + (
2tet + 2e3t , y = et − tet − e3t
1
1 − et cos 2t − 10
sin 2t,
0<t<π
12. y(t) =
e−t cos 2t(eπ − 1) + 21 e−t sin 2t(eπ − 1), t > π
13.
i=
E
−Rt/L
),
R (1 − e
E −Rt/L Ra/L
e
(e
−
1),
R
14. y(a) =
1
3
·
W
EI
·
a2 b2
l
0<t<a
;
t>a