264 Ordinary Differential Equations
SOLUTION
Taking the Laplace transform of the differential equation in (2) and using
the linearity property of the Laplace transform, we get
£[y"(x)] + £[y(x)] = 2£[u(x - 1)] - £[u(x - 2)]
and
2 -s -2s
-y'(O)-sy(O) + s^2 £[y(x) ] + £[y(x)] = _e_ - _e -.
s sImposing the initial conditions of (2) and solving for £[y(x )], we find
(4)
2e-s e-2s
--- - +1
£[y(x)] = s s2 + sl2e-s
s(s^2 +1)e -2s 1
-~-+-s(s2+1) s^2 +1·
By partial fraction expansion
(5)
1 1 ss(s^2 +1) -;-s^2 +1·
From Table 5.1, we see that
£ [1] = ~)
ss
£[cos x] = - 2 --,s + 1
. 1
and £[smx] = - 2 -- for s > 0.
s + 1
Consequently, from equation (5) the Laplace transform of the function
1 1
f(x) = 1 - cosx is s(s 2 + l). Substituting £[1 - cosx] for -
8
(-
8- 2 -+-l-) and
1
£[sin x] for - 2 --in equation (4), we obtain
s + 1
(6) £[y(x)] = 2e-s £[1-cosx] - e-^2 s £[1 - cosx] + £[sinx].
Applying the second translation property to equation (6), we get
£[y(x)] = £[2u(x - 1)(1 - cos(x - 1))] - £[u(x - 2)(1 - cos(x - 2))]
- £[sinx]
= £[2u(x - 1)(1 - cos(x - 1)) - u(x - 2)(1 - cos(x - 2)) + sinx].
Hence,
(7) y(x) = 2u(x - 1)(1 - cos(x - 1)) - u(x - 2)(1 - cos(x - 2)) + sin x
is t he so lution of the initial value problem (2). Or, writing y(x) in a more
conventional way, we have