Page 69 - Handbook Of Integral Equations

√
x
2
cos λ x – t 2
29. √ y(t) dt = f(x).
2
0 x – t 2
Solution: √
2
2 d x cosh λ x – t 2
y(x)= t √ f(t) dt.
π dx 0 x – t 2
2
√
∞ cos λ t – x
2 2
30. √ y(t) dt = f(x).
2
x t – x 2
Solution:
√
2
2 d ∞ cosh λ t – x 2
y(x)= – t √ f(t) dt.
π dx 2 2
x t – x
x

β γ
31. Ax + B cos (λt)+ C]y(t) dt = f(x).
a
γ
β
This is a special case of equation 1.9.6 with g(x)= Ax and h(t)= B cos (λt)+ C.
x

γ β
32. A cos (λx)+ Bt + C]y(t) dt = f(x).
a
β
γ
This is a special case of equation 1.9.6 with g(x)= A cos (λx) and h(t)= Bt + C.
x
λ µ β γ
33. Ax cos t + Bt cos x y(t) dt = f(x).
a
λ
µ
γ
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t) = cos t, g 2 (x)= B cos x,
β
and h 2 (t)= t .
1.5-2. Kernels Containing Sine
x
34. sin[λ(x – t)]y(t) dt = f(x), f(a)= f (a)=0.

x
a
1

Solution: y(x)= f (x)+ λf(x).
xx
λ
x

35. sin[λ(x – t)] + b y(t) dt = f(x).
a
Differentiating the equation with respect to x yields an equation of the form 2.5.3:
x

λ 1

y(x)+ cos[λ(x – t)]y(t) dt = f (x).
x
b b
a
x

36. sin(λx + βt)y(t) dt = f(x).
a
For β = –λ, see equation 1.5.34. Assume that β ≠ –λ.
Differentiating the equation with respect to x twice yields
x

sin[(λ + β)x]y(x)+ λ cos(λx + βt)y(t) dt = f (x), (1)
x
a
x
2
sin[(λ + β)x]y(x) + λ cos[(λ + β)x]y(x) – λ sin(λx + βt)y(t) dt = f (x). (2)
xx
x
a
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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