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x
µ
11. y(x)+ k sin[λ(x – t)]y (t) dt = A sin(λx)+ B cos(λx)+ C.
a
µ
This is a special case of equation 5.8.17 with f(y)= ky .
Solution in an implicit form:
y
2 2 2 2 –1/2
λ D – λ u +2λ Cu – 2kλF(u) du = ±(x – a),
y 0
2
2
2
y 0 = A sin(λa)+ B cos(λa)+ C, D = A + B – C , F(u)= 1 u µ+1 – y µ+1 .
0
µ +1
5.3-2. Equations Containing Arbitrary Functions
x
λ
12. K(x, t)y (t) dt = f(x).
a
λ
The substitution w(x)= y (x) leads to the linear equation
x
K(x, t)w(t) dt = f(x).
a
x
k
13. y(x)+ f(t)y (t) dt = A.
a
Solution:
1
x 1–k
y(x)= A 1–k +(k – 1) f(t) dt .
a
x
k
14. y(x) – f(x)g(t)y (t) dt =0.
a
◦
1 . Differentiating the equation with respect to x and eliminating the integral term (using the
original equation), we obtain the Bernoulli ordinary differential equation
f (x)
x
k
y – f(x)g(x)y – y =0, y(a)=0.
x
f(x)
◦
2 . Solution with k <1:
1
x 1–k
k
y(x)= f(x) (1 – k) f (t)g(t) dt .
a
Additionally, for k > 0, there is the trivial solution y(x) ≡ 0.
x t
λ
k
15. y(x)+ x λ–kλ–1 f y (t) dt = Ax .
0 x
λ
A solution: y(x)= βx , where β is a root of the algebraic equation
1
k kλ
Iβ + β – A =0, I = f(z)z dz.
0
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 349