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b

33. y(x)+ cos(λx + µt)f t, y(t) dt = h(x).
a
Using the formula cos(λx + µt) = cos(λx) cos(µt) – sin(µt) sin(λx), we arrive at an equation
of the form 6.8.39:
b

y(x)+ cos(λx)f 1 t, y(t) + sin(λx)f 2 t, y(t) dt = h(x),
a

f 1 t, y(t) = cos(µt)f t, y(t) , f 2 t, y(t) = – sin(µt)f t, y(t) .
b

34. y(x)+ sin(λx + µt)f t, y(t) dt = h(x).
a
Using the formula sin(λx + µt) = cos(λx) sin(µt) + cos(µt) sin(λx), we arrive at an equation
of the form 6.8.39:
b

y(x)+ cos(λx)f 1 t, y(t) + sin(λx)f 2 t, y(t) dt = h(x),
a

f 1 t, y(t) = sin(µt)f t, y(t) , f 2 t, y(t) = cos(µt)f t, y(t) .
b

35. y(x)+ |x – t|f t, y(t) dt = g(x), a ≤ x ≤ b.
a
1 . Let us remove the modulus in the integrand:
◦
x b

y(x)+ (x – t)f t, y(t) dt + (t – x)f t, y(t) dt = g(x). (1)
a x
Differentiating (1) with respect to x yields
x b

y (x)+ f t, y(t) dt – f t, y(t) dt = g (x). (2)
x
x
a x
Differentiating (2), we arrive at a second-order ordinary differential equation for y = y(x):

y xx +2f(x, y)= g (x). (3)
xx
◦
2 . Let us derive the boundary conditions for equation (3). We assume that –∞ < a < b < ∞.
By setting x = a and x = b in (1), we obtain the relations
b

y(a)+ (t – a)f t, y(t) dt = g(a),
a (4)
b

y(b)+ (b – t)f t, y(t) dt = g(b).
a
Let us solve equation (3) for f(x, y) and substitute the result into (4). Integrating by parts
yields the desired boundary conditions for y(x):

y(a)+ y(b)+(b – a) g (b) – y (b) = g(a)+ g(b),
x x
(5)

y(a)+ y(b)+(a – b) g (a) – y (a) = g(a)+ g(b).

x x
Let us point out a useful consequence of (5):

y (a)+ y (b)= g (a)+ g (b), (6)
x
x
x
x
which can be used together with one of conditions (5).
Equation (3) under the boundary conditions (5) determines the solution of the original
integral equation (there may be several solutions). Conditions (5) make it possible to calculate
the constants of integration that occur in solving the differential equation (3).
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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