The Laplace Trans! orm Method 227
We now prove the Laplace transform is a linear operator.
LINEARITY PROPERTY OF THE LAPLACE TRANSFORM
The Laplace transform is a linear operator. That is, if Ji ( x) and h ( x) are
functions which have Laplace transforms for s > s 1 ands> s 2 , respectively,
and if c 1 and c2 are constants, then
Proof: Let s > max(s 1 , s 2 ). Then, by definition,
.C[cif1(x) + c2h(x)] = 1= {cif1(x) + c2h(x)}e-sx dx
= C1 1= Ji (x )e-sx dx + C21= h (x )e-sx dx
= c1.C[fi(x) ] + c2.C[h(x)].
EXAMPLE 4 Using the Linearity Property to Calculate a
Laplace Transform
Calculate .C[4x^2 + 3].
SOLUTION
Using the linearity property of the Laplace transform, we find
2 1 8 3
.C[4x^2 + 3] = 4.C[x^2 ] + 3.C[l] = 43 + 3-= 3 + - for s > 0.
s s s s
EXAMPLE 5 Using an Identity and Linearity
Calculate .C[sinh bx].
SOLUTION
1 1
Since sinh bx = -ebx - -e-bx, we find using the linearity property of the
2 2
Laplace transform,
l [
.C[sinhbx = .C - e^1 bx - -^1 e-bx] =-.Ce^1 [ bx] - -J._,^1 r [ e-bx]
2 2 2 2
1 1 1 1 b
- ---- ---- -- for s > lbl.
- 2 s - b 2 s + b - s^2 - b2