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x

γ β
64. A sin (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 sin (λx) and h(t)= Bt + C.
x
λ µ β γ
65. Ax sin t + Bt sin x y(t) dt = f(x).
a
γ
µ
λ
This is a special case of equation 1.9.15 with g 1 (x)= Ax , h 1 (t) = sin t, g 2 (x)= B sin x,
β
and h 2 (t)= t .
1.5-3. Kernels Containing Tangent
x

66. tan(λx) – tan(λt) y(t) dt = f(x).
a
This is a special case of equation 1.9.2 with g(x) = tan(λx).
1 d 2
Solution: y(x)= cos (λx)f (x) .
x
λ dx
x

67. A tan(λx)+ B tan(λt) y(t) dt = f(x).
a
For B = –A, see equation 1.5.66. This is a special case of equation 1.9.4 with g(x) = tan(λx).
A x B
1 d – –

Solution: y(x)= tan(λx) A+B tan(λt) A+B f (t) dt .
t
A + B dx
a
x

68. A tan(λx)+ B tan(µt)+ C y(t) dt = f(x).
a
This is a special case of equation 1.9.6 with g(x)= A tan(λx) and h(t)= B tan(µt)+ C.
x

2 2
69. tan (λx) – tan (λt) y(t) dt = f(x).
a
2
This is a special case of equation 1.9.2 with g(x) = tan (λx).
3
d cos (λx)f (x)

x
Solution: y(x)= .
dx 2λ sin(λx)
x
2 2
70. A tan (λx)+ B tan (λt) y(t) dt = f(x).
a
2
For B = –A, see equation 1.5.69. This is a special case of equation 1.9.4 with g(x) = tan (λx).
2A x 2B
1 d – –

Solution: y(x)= tan(λx) A+B tan(λt) A+B f (t) dt .
t
A + B dx
a
x

2 2
71. A tan (λx)+ B tan (µt)+ C y(t) dt = f(x).
a
2
2
This is a special case of equation 1.9.6 with g(x)= A tan (λx) and h(t)= B tan (µt)+ C.
x
n
72. tan(λx) – tan(λt) y(t) dt = f(x), n =1, 2, ...
a
The right-hand side of the equation is assumed to satisfy the conditions f(a)= f (a)= ··· =

x
f (n) (a)=0.
x
n+1
1 2 d
Solution: y(x)= cos (λx) f(x).
2
n
λ n! cos (λx) dx
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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