Page 118 - Handbook Of Integral Equations

x

18. K(x – t)y(t) dt = Ae λx .
–∞
Solution:
A λx ∞ –λz
y(x)= e , B = K(z)e dz = L{K(z), λ}.
B 0
x
n λx
19. K(x – t)y(t) dt = Ax e , n =1, 2, ...
–∞
1 . Solution with n =1:
◦
A λx AC λx
y 1 (x)= xe + e ,
B B 2
∞ ∞

B = K(z)e –λz dz, C = zK(z)e –λz dz.
0 0
It is convenient to calculate the coefficients B and C using tables of Laplace transforms
according to the formulas B = L{K(z), λ} and C = L{zK(z), λ}.
2 . Solution with n =2:
◦
A 2 λx AC λx AC 2 AD λx
y 2 (x)= x e +2 xe + 2 – e ,
B B 2 B 3 B 2
∞ ∞ ∞

2
B = K(z)e –λz dz, C = zK(z)e –λz dz, D = z K(z)e –λz dz.
0 0 0
3 . Solution with n =3, 4, ... is given by:
◦
∂ ∂ n e λx ∞ –λz
y n (x)= y n–1 (x)= A , B(λ)= K(z)e dz.
∂λ ∂λ n B(λ) 0
x
20. K(x – t)y(t) dt = A cosh(λx).
–∞
Solution:
A λx A –λx 1 A A
1 A A
y(x)= e + e = + cosh(λx)+ – sinh(λx),
2B – 2B + 2 B – B + 2 B – B +
∞ ∞

B – = K(z)e –λz dz, B + = K(z)e λz dz.
0 0
x
21. K(x – t)y(t) dt = A sinh(λx).
–∞
Solution:
A λx A –λx 1 A A
1 A A
y(x)= e – e = – cosh(λx)+ + sinh(λx),
2B – 2B + 2 B – B + 2 B – B +
∞ ∞

B – = K(z)e –λz dz, B + = K(z)e λz dz.
0 0
x
22. K(x – t)y(t) dt = A cos(λx).
–∞
Solution:
A
y(x)= B c cos(λx) – B s sin(λx) ,
2
B + B 2 s
c
∞ ∞

B c = K(z) cos(λz) dz, B s = K(z) sin(λz) dz.
0 0

© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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