Page 72 - Handbook Of Integral Equations

x

2 2
45. A sin (λx)+ B sin (λt) y(t) dt = f(x).
a
2
For B = –A, see equation 1.5.44. This is a special case of equation 1.9.4 with g(x) = sin (λx).
Solution:
2A x 2B
1 d – –

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

2 2
46. A sin (λx)+ B sin (µt)+ C y(t) dt = f(x).
a
2
2
This is a special case of equation 1.9.6 with g(x)= A sin (λx) and h(t)= B sin (µt)+ C.
x

47. sin[λ(x – t)] sin[λ(x + t)]y(t) dt = f(x), f(a)= f (a)=0.
x
a
Using the trigonometric formula

sin(α – β) sin(α + β)= 1 cos(2β) – cos(2α) , α = λx, β = λt,
2
we reduce the original equation to an equation of the form 1.5.5 with A = B =1:
x

cos(2λx) – cos(2λt) y(t) dt = –2f(x).
a
1 d f (x)

x
Solution: y(x)= .
λ dx sin(2λx)
x

48. sin(λx) sin(µt) + sin(βx) sin(γt) y(t) dt = f(x).
a
This is a special case of equation 1.9.15 with g 1 (x) = sin(λx), h 1 (t) = sin(µt), g 2 (x) = sin(βx),
and h 2 (t) = sin(γt).
x

3
49. sin [λ(x – t)]y(t) dt = f(x).
a
It is assumed that f(a)= f (a)= f (a)= f (a)=0.

x xx xxx
3
1
3
Using the formula sin β = – sin 3β + sin β, we arrive at an equation of the form 1.5.41:
4 4
x

1 3
– sin[3λ(x – t)] + sin[λ(x – t)] y(t) dt = f(x).
4 4
a
x
3 3
50. sin (λx) – sin (λt) y(t) dt = f(x), f(a)= f (a)=0.

x
a
3
This is a special case of equation 1.9.2 with g(x) = sin (λx).
x
3 3
51. A sin (λx)+ B sin (λt) y(t) dt = f(x).
a
3
This is a special case of equation 1.9.4 with g(x) = sin (λx).
Solution:
sign sin(λx) d – 3A x – 3B
y(x)= sin(λx) A+B sin(λt) A+B f (t) dt .

t
A + B dx a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 50