Linear Equations of the Second Kind With Variable Limit of Integration
2.1. Equations Whose Kernels Contain Power-Law Functions
2.1-1. Kernels Linear in the Arguments xandt
1. y(x) –λ x
a
y(t)dt=f(x).
Solution:
y(x) =f(x) +λ x
a
eλ(x–t)f(t)dt.
2. y(x) +λx x
a
y(t)dt =f(x).
Solution:
y(x) =f(x) –λ x
a
xexp1
2λ(t2–x2) f(t)dt.
3. y(x) +λ x
a
ty(t)dt=f(x).
Solution:
y(x) =f(x) –λ x
a
texp1
2λ(t2–x2) f(t)dt.
4. y(x) +λ x
a
(x–t)y(t)dt=f(x).
This is a special case of equation 2.1.34 withn= 1.
1◦. Solution withλ> 0:
y(x) =f(x) –k x
a
sin[k(x–t)]f(t)dt, k=√ λ.
2◦. Solution withλ< 0:
y(x) =f(x) +k x
a
sinh[k(x–t)]f(t)dt, k=√ –λ.
5. y(x) + x
a
A+B(x–t)
y(t)dt=f(x).
1◦. Solution withA2> 4B:
y(x) =f(x)– x
a
R(x–t)f(t)dt, R(x) = exp
–12Ax
Acosh(βx) +2B–A2
2β sinh(βx)
, β= 14A2–B.
2◦. Solution withA2< 4B:
y(x) =f(x)– x
a
R(x–t)f(t)dt, R(x) = exp
–12Ax
Acos(βx) +2B–A2
2β sin(βx)
, β = B– 14A2. 3◦. Solution withA2= 4B:
y(x) =f(x)– x
a
R(x–t)f(t)dt, R(x) = exp
–12Ax
A– 14A2x .
6. y(x)– x
a
Ax+Bt+C
y(t)dt=f(x).
ForB =–Asee equation 2.1.5. This is a special case of equation 2.9.6 withg(x) =–Axand h(t) =–Bt–C.
By differentiation followed by the substitutionY(x) = x
a
y(t)dt, the original equation can be reduced to the second-order linear ordinary differential equation
Yxx –
(A+B)x+C
Yx–AY =fx(x) (1) under the initial conditions
Y(a) = 0, Yx(a) =f(a). (2)
A fundamental system of solutions of the homogeneous equation (1) withf ≡0 has the form
Y1(x) =Φ
α, 12;kz2
, Y2(x) =Ψ
α, 12;kz2 , α= A
2(A+B), k= A+B
2 , z=x+ C A+B, whereΦ
α,β;x andΨ
α,β;x
are degenerate hypergeometric functions.
Solving the homogeneous equation (1) under conditions (2) for an arbitrary function f =f(x) and taking into account the relationy(x) =Yx(x), we thus obtain the solution of the integral equation in the form
y(x) =f(x)– x
a
R(x,t)f(t)dt, R(x,t) = ∂2
∂x∂t
Y1(x)Y2(t)–Y2(x)Y1(t) W(t)
, W(t) = 2√ πk Γ(α) exp
k
t+ C
A+B 2
.
2.1-2. Kernels Quadratic in the Argumentsxandt 7. y(x) +A
x a
x2y(t)dt =f(x).
This is a special case of equation 2.1.50 withλ= 2 andµ= 0.
Solution:
y(x) =f(x)–A x
a
x2exp1
3A(t3–x3) f(t)dt.
8. y(x) +A x
a
xty(t)dt=f(x).
This is a special case of equation 2.1.50 withλ= 1 andµ= 1.
Solution:
y(x) =f(x)–A x
a
xtexp1
3A(t3–x3) f(t)dt.
9. y(x) +A x
a
t2y(t)dt=f(x).
This is a special case of equation 2.1.50 withλ= 0 andµ= 2.
Solution:
y(x) =f(x)–A x
a
t2exp1
3A(t3–x3) f(t)dt.
10. y(x) +λ x
a
(x–t)2y(t)dt=f(x).
This is a special case of equation 2.1.34 withn= 2.
Solution:
y(x) =f(x)– x
a
R(x–t)f(t)dt, R(x) = 23ke–2kx– 23kekx
cos√
3kx –√
3 sin√ 3kx
, k=1
4λ1/3
.
11. y(x) +A x
a
(x2–t2)y(t)dt=f(x).
This is a special case of equation 2.9.5 withg(x) =Ax2. Solution:
y(x) =f(x) + 1 W
x a
u1(x)u2(t)–u2(x)u1(t) f(t)dt,
where the primes denote differentiation with respect to the argument specified in the parenthe- ses;u1(x),u2(x) is a fundamental system of solutions of the second-order linear homogeneous ordinary differential equationuxx+ 2Axu = 0; and the functionsu1(x) andu2(x) are ex- pressed in terms of Bessel functions or modified Bessel functions, depending on the sign of the parameterA:
ForA> 0,
W = 3/π, u1(x) =√
x J1/3 89A x3/2
, u2(x) =√
x Y1/3 89A x3/2
. ForA< 0,
W = –32, u1(x) =√
x I1/3 89|A|x3/2
, u2(x) =√
x K1/3 89|A|x3/2
.
12. y(x) +A x
a
(xt–t2)y(t)dt=f(x).
This is a special case of equation 2.9.4 withg(t) =At. Solution:
y(x) =f(x) + A W
x a
t
y1(x)y2(t)–y2(x)y1(t) f(t)dt,
wherey1(x),y2(x) is a fundamental system of solutions of the second-order linear homo- geneous ordinary differential equation yxx +Axy = 0; the functions y1(x) and y2(x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of the parameterA:
ForA> 0,
W = 3/π, y1(x) =√
x J1/32
3
√A x3/2
, y2(x) =√
x Y1/32
3
√A x3/2 . ForA< 0,
W =–32, y1(x) =√ x I1/32
3
|A|x3/2
, y2(x) =√
x K1/32
3
|A|x3/2 . 13. y(x) +A
x a
(x2–xt)y(t)dt=f(x).
This is a special case of equation 2.9.3 withg(x) =Ax. Solution:
y(x) =f(x) + A W
x a
x
y1(x)y2(t)–y2(x)y1(t) f(t)dt,
wherey1(x),y2(x) is a fundamental system of solutions of the second-order linear homo- geneous ordinary differential equation yxx +Axy = 0; the functions y1(x) and y2(x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign of the parameterA:
ForA> 0,
W = 3/π, y1(x) =√
x J1/32
3
√A x3/2
, y2(x) =√
x Y1/32
3
√A x3/2 . ForA< 0,
W =–32, y1(x) =√ x I1/32
3
|A|x3/2
, y2(x) =√
x K1/32
3
|A|x3/2 . 14. y(x) +A
x a
(t2–3x2)y(t)dt=f(x).
This is a special case of equation 2.1.55 withλ= 1 andµ= 2.
15. y(x) +A x
a
(2xt–3x2)y(t)dt=f(x).
This is a special case of equation 2.1.55 withλ= 2 andµ= 1.
16. y(x)– x
a
(ABxt–ABx2+Ax+B)y(t)dt=f(x).
This is a special case of equation 2.9.16 withg(x) =Axandh(x) =B.
Solution:
y(x) =f(x) + x
a
R(x,t)f(t)dt, R(x,t) = (Ax+B) exp1
2A(x2–t2) +B2
x
t
exp1
2A(s2–t2) +B(x–s) ds.
17. y(x) + x
a
Ax2–At2+Bx–Ct+D
y(t)dt =f(x).
This is a special case of equation 2.9.6 withg(x) =Ax2+Bx+Dandh(t) =–At2–Ct.
Solution:
y(x) =f(x) + x
a
∂2
∂x∂t
Y1(x)Y2(t)–Y2(x)Y1(t) W(t)
f(t)dt.
Here Y1(x),Y2(x) is a fundamental system of solutions of the second-order homogeneous ordinary differential equationYxx +
(B–C)x+D
Yx+ (2Ax+B)Y = 0 (see A. D. Polyanin and V. F. Zaitsev (1996) for details about this equation):
Y1(x) = exp(–kx)Φ
α, 12; 12(C–B)z2
, Y2(x) = exp(–kx)Ψ
α, 12; 12(C–B)z2 , W(x) =–
√2π(C–B) Γ(α) exp1
2(C–B)z2–2kx
, k= 2A B–C, α=–4A2+ 2AD(C–B) +B(C–B)2
2(C–B)3 , z=x– 4A+ (C–B)D (C–B)2 , whereΦ
α,β;x andΨ
α,β;x
are degenerate hypergeometric functions andΓ(α) is the gamma function.
18. y(x)– x
a
Ax+B+ (Cx+D)(x–t)
y(t)dt=f(x).
This is a special case of equation 2.9.11 withg(x) =Ax+Bandh(x) =Cx+D.
Solution withA≠0:
y(x) =f(x) + x
a
Y2(x)Y1(t)–Y1(x)Y2(t) f(t) W(t)dt.
Here Y1(x),Y2(x) is a fundamental system of solutions of the second-order homogeneous ordinary differential equationYxx –(Ax+B)Yx–(Cx+D)Y = 0 (see A. D. Polyanin and V. F. Zaitsev (1996) for details about this equation):
Y1(x) = exp(–kx)Φ
α, 12; 12Az2
, Y2(x) = exp(–kx)Ψ
α, 12; 12Az2 , W(x) =–√
2πA Γ(α)–1
exp1
2Az2–2kx
, k=C/A, α= 12(A2D–ABC–C2)A–3, z=x+ (AB+ 2C)A–2, whereΦ
α,β;x andΨ
α,β;x
are degenerate hypergeometric functions,Γ(α) is the gamma function.
19. y(x) + x
a
At+B+ (Ct+D)(t–x)
y(t)dt=f(x).
This is a special case of equation 2.9.12 withg(t) =–At–Bandh(t) =–Ct–D.
Solution withA≠0:
y(x) =f(x)– x
a
Y1(x)Y2(t)–Y1(t)Y2(x) f(t) W(x)dt.
Here Y1(x),Y2(x) is a fundamental system of solutions of the second-order homogeneous ordinary differential equationYxx –(Ax+B)Yx–(Cx+D)Y = 0 (see A. D. Polyanin and V. F. Zaitsev (1996) for details about this equation):
Y1(x) = exp(–kx)Φ
α, 12; 12Az2
, Y2(x) = exp(–kx)Ψ
α, 12; 12Az2 , W(x) =–√
2πA Γ(α)–1
exp1
2Az2–2kx
, k=C/A, α= 12(A2D–ABC–C2)A–3, z=x+ (AB+ 2C)A–2, whereΦ
α,β;x andΨ
α,β;x
are degenerate hypergeometric functions andΓ(α) is the gamma function.
2.1-3. Kernels Cubic in the Arguments xandt
20. y(x) +A x
a
x3y(t)dt =f(x).
Solution:
y(x) =f(x)–A x
a
x3exp1
4A(t4–x4) f(t)dt.
21. y(x) +A x
a
x2ty(t)dt=f(x).
Solution:
y(x) =f(x)–A x
a
x2texp1
4A(t4–x4) f(t)dt.
22. y(x) +A x
a
xt2y(t)dt=f(x).
Solution:
y(x) =f(x)–A x
a
xt2exp1
4A(t4–x4) f(t)dt.
23. y(x) +A x
a
t3y(t)dt=f(x).
Solution:
y(x) =f(x)–A x
a
t3exp1
4A(t4–x4) f(t)dt.
24. y(x) +λ x
a
(x–t)3y(t)dt=f(x).
This is a special case of equation 2.1.34 withn= 3.
Solution:
y(x) =f(x)– x
a
R(x–t)f(t)dt, where
R(x) =
k
cosh(kx) sin(kx)–sinh(kx) cos(kx)
, k=3
2λ1/4
for λ> 0,
1 2s
sin(sx)–sinh(sx)
, s= (–6λ)1/4 for λ< 0.
25. y(x) +A x
a
(x3–t3)y(t)dt=f(x).
This is a special case of equation 2.1.52 withλ= 3.
26. y(x)–A x
a
4x3–t3
y(t)dt=f(x).
This is a special case of equation 2.1.55 withλ= 1 andµ= 3.
27. y(x) +A x
a
(xt2–t3)y(t)dt=f(x).
This is a special case of equation 2.1.49 withλ= 2.
28. y(x) +A x
a
x2t–t3
y(t)dt=f(x).
The transformationz=x2,τ =t2,y(x) =w(z) leads to an equation of the form 2.1.4:
w(z) + 12A z
a2
(z–τ)w(τ)dτ =F(z), F(z) =f(x).
29. y(x) + x
a
Ax2t+Bt3
y(t)dt =f(x).
The transformationz=x2,τ =t2,y(x) =w(z) leads to an equation of the form 2.1.6:
w(z) + z
a2
1
2Az+ 12Bτ
w(τ)dτ =F(z), F(z) =f(x).
30. y(x) +B x
a
2x3–xt2
y(t)dt=f(x).
This is a special case of equation 2.1.55 withλ= 2,µ= 2, andB =–2A.
31. y(x)–A x
a
4x3–3x2t
y(t)dt=f(x).
This is a special case of equation 2.1.55 withλ= 3 andµ= 1.
32. y(x) + x
a
ABx3–ABx2t–Ax2–B
y(t)dt =f(x).
This is a special case of equation 2.9.7 withg(x) =Ax2andλ=B.
Solution:
y(x) =f(x) + x
a
R(x–t)f(t)dt, R(x,t) = (Ax2+B) exp1
3A(x3–t3) +B2
x
t
exp1
3A(s3–t3) +B(x–s) ds.
33. y(x) + x
a
ABxt2–ABt3+At2+B
y(t)dt=f(x).
This is a special case of equation 2.9.8 withg(t) =At2andλ=B.
Solution:
y(x) =f(x) + x
a
R(x–t)f(t)dt, R(x,t) =–(At2+B) exp1
3A(t3–x3) +B2
x t
exp1
3A(s3–x3) +B(t–s) ds.
2.1-4. Kernels Containing Higher-Order Polynomials in xand t 34. y(x) +A
x
a
(x–t)ny(t)dt=f(x), n= 1, 2,. . .
1◦. Differentiating the equationn+ 1 times with respect toxyields an (n+ 1)st-order linear ordinary differential equation with constant coefficients fory=y(x):
yx(n+1)+An!y=fx(n+1)(x).
This equation under the initial conditionsy(a) =f(a), yx(a) =fx(a), . . ., y(n)x (a) =fx(n)(a) determines the solution of the original integral equation.
2◦. Solution:
y(x) =f(x) + x
a
R(x–t)f(t)dt, R(x) = 1
n+ 1 n
k=0
exp(σkx)
σkcos(βkx)–βksin(βkx) ,
where the coefficientsσkandβkare given by σk=|An!|n+11 cos
2πk n+ 1
, βk=|An!|n+11 sin 2πk
n+ 1
for A< 0, σk=|An!|n+11 cos
2πk+π n+ 1
, βk=|An!|n+11 sin
2πk+π n+ 1
for A> 0.
35. y(x) +A ∞
x
(t–x)ny(t)dt=f(x), n= 1, 2,. . .
The Picard–Goursat equation.This is a special case of equation 2.9.62 withK(z) =A(–z)n. 1◦. A solution of the homogeneous equation (f ≡0) is
y(x) =Ce–λx, λ=
–An! 1
n+1,
whereCis an arbitrary constant andA< 0. This is a unique solution forn= 0, 1, 2, 3.
The general solution of the homogeneous equation for any sign ofAhas the form y(x) =
s k=1
Ckexp(–λkx). (1)
HereCkare arbitrary constants andλkare the roots of the algebraic equationλn+1+An! = 0 that satisfy the condition Reλk> 0. The number of terms in (1) is determined by the inequality s≤2n
4
+ 1, where [a] stands for the integral part of a numbera. For more details about the solution of the homogeneous Picard–Goursat equation, see Subsection 9.11-1 (Example 1).
2◦. Forf(x) = m k=1
akexp(–βkx), whereβk > 0, a solution of the equation has the form
y(x) = m
k=1
akβkn+1
βkn+1+An!exp(–βkx), (2)
whereβkn+1+An!≠0. ForA> 0, this formula can also be used for arbitraryf(x) expandable into a convergent exponential series (which corresponds tom=∞).
3◦. Forf(x) =e–βx m k=1
akxk, whereβ> 0, a solution of the equation has the form
y(x) =e–βx m k=0
Bkxk, (3)
where the constantsBkare found by the method of undetermined coefficients. The solution can also be constructed using the formulas given in item 3◦, equation 2.9.55.
4◦. Forf(x) = cos(βx) m k=1
akexp(–µkx), a solution of the equation has the form
y(x) = cos(βx) m
k=1
Bkexp(–µkx) + sin(βx) m
k=1
Ckexp(–µkx), (4)
where the constantsBkandCk are found by the method of undetermined coefficients. The solution can also be constructed using the formulas given in 2.9.60.
5◦. Forf(x) = sin(βx) m k=1
akexp(–µkx), a solution of the equation has the form
y(x) = cos(βx) m
k=1
Bkexp(–µkx) + sin(βx) m
k=1
Ckexp(–µkx), (5)
where the constantsBkandCk are found by the method of undetermined coefficients. The solution can also be constructed using the formulas given in 2.9.61.
6◦. To obtain the general solution in item 2◦–5◦, the solution (1) of the homogeneous equation must be added to each right-hand side of (2)–(5).
36. y(x) +A x
a
(x–t)tny(t)dt =f(x), n= 1, 2,. . .
This is a special case of equation 2.1.49 withλ=n.
37. y(x) +A x
a
(xn–tn)y(t)dt=f(x), n= 1, 2,. . .
This is a special case of equation 2.1.52 withλ=n.
38. y(x) + x
a
ABxn+1–ABxnt–Axn–B
y(t)dt=f(x), n= 1, 2,. . . This is a special case of equation 2.9.7 withg(x) =Axnandλ=B.
Solution:
y(x) =f(x) + x
a
R(x–t)f(t)dt, R(x,t) = (Axn+B) exp
A n+ 1
xn+1–tn+1 +B2
x t
exp A
n+ 1
sn+1–tn+1
+B(x–s)
ds.
39. y(x) + x
a
ABxtn–ABtn+1+Atn+B
y(t)dt=f(x), n= 1, 2,. . . This is a special case of equation 2.9.8 withg(t) =Atnandλ=B.
Solution:
y(x) =f(x) + x
a
R(x–t)f(t)dt, R(x,t) =–(Atn+B) exp
A n+ 1
tn+1–xn+1 +B2
x t
exp A
n+ 1
sn+1–xn+1
+B(t–s)
ds.
2.1-5. Kernels Containing Rational Functions
40. y(x) +x–3 x
a
t
2Ax+ (1–A)t
y(t)dt=f(x).
This equation can be obtained by differentiating the equation x
a
Ax2t+ (1–A)xt2
y(t)dt=F(x), F(x) = x
a
t3f(t)dt, which has the form 1.1.17:
Solution:
y(x) = 1 x
d dx
x–A
x a
tA–1ϕt(t)dt
, ϕ(x) = 1 x
x a
t3f(t)dt.
41. y(x)–λ x
0
y(t)dt
x+t =f(x).
Dixon’s equation. This is a special case of equation 2.1.62 witha=b= 1 andµ= 0.
1◦. The solution of the homogeneous equation (f ≡0) is
y(x) =Cxβ (β >–1, λ> 0). (1) HereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation
λI(β) = 1, where I(β) = 1
0
zβdz
1 +z. (2)
2◦. For a polynomial right-hand side, f(x) =
N n=0
Anxn
the solution bounded at zero is given by
y(x) =
N n=0
An
1–(λ/λn)xn for λ<λ0, N
n=0
An
1–(λ/λn)xn+Cxβ for λ>λ0andλ≠λn, λn= 1
I(n), I(n) = (–1)n
ln 2 + n m=1
(–1)m m
,
whereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation (2).
For specialλ=λn(n= 1, 2,. . .), the solution differs in one term and has the form y(x) =
n–1 m=0
Am
1–(λn/λm)xm+ N m=n+1
Am
1–(λn/λm)xm–Anλ¯n λn
xnlnx+Cxn,
where λ¯n= (–1)n+1 π2
12 + n k=1
(–1)k k2
–1
.
Remark. For arbitraryf(x), expandable into power series, the formulas of item 2◦can be used, in which one should setN = ∞. In this case, the radius of convergence of the solutiony(x) is equal to the radius of convergence off(x).
3◦. For logarithmic-polynomial right-hand side, f(x) = lnx
N n=0
Anxn
, the solution with logarithmic singularity at zero is given by
y(x) =
lnx
N n=0
An
1–(λ/λn)xn+ N
n=0
AnDnλ
[1–(λ/λn)]2xn for λ<λ0, lnx
N n=0
An
1–(λ/λn)xn+ N
n=0
AnDnλ
[1–(λ/λn)]2xn+Cxβ for λ>λ0andλ≠λn, λn= 1
I(n), I(n) = (–1)n
ln 2 + n
k=1
(–1)k k
, Dn = (–1)n+1 π2
12 + n k=1
(–1)k k2
. 4◦. For arbitraryf(x), the transformation
x= 12e2z, t= 12e2τ, y(x) =e–zw(z), f(x) =e–zg(z) leads to an integral equation with difference kernel of the form 2.9.51:
w(z)–λ z
–∞
w(τ)dτ
cosh(z–τ) =g(z).
42. y(x)–λ x
a
x+b
t+b y(t)dt=f(x).
This is a special case of equation 2.9.1 withg(x) =x+b.
Solution:
y(x) =f(x) +λ x
a
x+b
t+beλ(x–t)f(t)dt.
43. y(x) = 2 (1–λ2)x2
x
λx
t
1 +ty(t)dt.
This equation is encountered in nuclear physics and describes deceleration of neutrons in matter.
1◦. Solution withλ= 0:
y(x) = C (1 +x)2, whereCis an arbitrary constant.
2◦. Forλ≠0, the solution can be found in the series form y(x) =
∞ n=0
Anxn. • Reference: I. Sneddon (1951).
2.1-6. Kernels Containing Square Roots and Fractional Powers
44. y(x) +A x
a
(x–t)√
t y(t)dt=f(x).
This is a special case of equation 2.1.49 withλ= 12. 45. y(x) +A
x a
√x–√ t
y(t)dt=f(x).
This is a special case of equation 2.1.52 withλ= 12. 46. y(x) +λ
x
a
y(t)dt
√x–t =f(x).
Abel’s equation of the second kind. This equation is encountered in problems of heat and mass transfer.
Solution:
y(x) =F(x) +πλ2 x
a
exp[πλ2(x–t)]F(t)dt, where
F(x) =f(x)–λ x
a
f(t)dt
√x–t.
• References: H. Brakhage, K. Nickel, and P. Rieder (1965), Yu. I. Babenko (1986).
47. y(x)–λ x
0
y(t)dt
√ax2+bt2 =f(x), a> 0, b> 0.
1◦. The solution of the homogeneous equation (f ≡0) is
y(x) =Cxβ (β >–1, λ> 0). (1) HereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation
λI(β) = 1, where I(β) = 1
0
zβdz
√a+bz2. (2) 2◦. For a polynomial right-hand side,
f(x) = N
n=0
Anxn
the solution bounded at zero is given by
y(x) =
N n=0
An
1–(λ/λn)xn for λ<λ0, N
n=0
An
1–(λ/λn)xn+Cxβ for λ>λ0andλ≠λn, λ0=
√b Arsinh
b/a, λn= 1
I(n), I(n) = 1
0
zndz
√a+bz2.
HereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation (2).
3◦. For specialλ=λn, (n= 1, 2,. . .), the solution differs in one term and has the form y(x) =
n–1 m=0
Am
1–(λn/λm)xm+ N m=n+1
Am
1–(λn/λm)xm–Anλ¯n λn
xnlnx+Cxn,
where λ¯n= 1
0
znlnz dz
√a+bz2 –1
.
4◦. For arbitraryf(x), expandable into power series, the formulas of item 2◦can be used, in which one should setN =∞. In this case, the radius of convergence of the solutiony(x) is equal to the radius of convergence off(x).
48. y(x) +λ x
a
y(t)dt
(x–t)3/4 =f(x).
This equation admits solution by quadratures (see equation 2.1.60 and Section 9.4-2).
2.1-7. Kernels Containing Arbitrary Powers 49. y(x) +A
x
a
(x–t)tλy(t)dt=f(x).
This is a special case of equation 2.9.4 withg(t) =Atλ. Solution:
y(x) =f(x) + A W
x a
y1(x)y2(t)–y2(x)y1(t)
tλf(t)dt,
wherey1(x),y2(x) is a fundamental system of solutions of the second-order linear homo- geneous ordinary differential equationyxx+Axλy = 0; the functionsy1(x) andy2(x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign ofA:
ForA> 0, W = 2q
π, y1(x) =√ x J1
2q
√ A q xq
, y2(x) =√ x Y1
2q
√ A q xq
, q= λ+ 2 2 , ForA< 0,
W =–q, y1(x) =√ x I1
2q
√
|A| q xq
, y2(x) =√ x K1
2q
√
|A| q xq
, q= λ+ 2 2 . 50. y(x) +A
x a
xλtµy(t)dt =f(x).
This is a special case of equation 2.9.2 withg(x) =–Axλandh(t) =tµ(λandµare arbitrary numbers).
Solution:
y(x) =f(x)– x
a
R(x,t)f(t)dt,
R(x,t) =
Axλtµexp A
λ+µ+ 1
tλ+µ+1–xλ+µ+1
for λ+µ+ 1≠0,
Axλ–Atµ+A for λ+µ+ 1 = 0.
51. y(x) +A x
a
(x–t)xλtµy(t)dt=f(x).
The substitutionu(x) =x–λy(x) leads to an equation of the form 2.1.49:
u(x) +A x
a
(x–t)tλ+µu(t)dt=f(x)x–λ.
52. y(x) +A x
a
(xλ–tλ)y(t)dt=f(x).
This is a special case of equation 2.9.5 withg(x) =Axλ. Solution:
y(x) =f(x) + 1 W
x a
u1(x)u2(t)–u2(x)u1(t) f(t)dt,
where the primes denote differentiation with respect to the argument specified in the paren- theses, andu1(x),u2(x) is a fundamental system of solutions of the second-order linear ho- mogeneous ordinary differential equationuxx+Aλxλ–1u= 0; the functionsu1(x) andu2(x) are expressed in terms of Bessel functions or modified Bessel functions, depending on the sign ofA:
ForAλ> 0, W = 2q
π, u1(x) =√ x J 1
2q
√ Aλ q xq
, u2(x) =√ x Y1
2q
√ Aλ q xq
, q= λ+ 1 2 , ForAλ< 0,
W =–q, u1(x) =√ x I1
2q
√
|Aλ|
q xq
, u2(x) =√ x λK 1
2q
√
|Aλ|
q xq
, q= λ+ 1 2 . 53. y(x)–
x
a
Axλtλ–1+Bt2λ–1
y(t)dt=f(x).
The transformation
z=xλ, τ =tλ, y(x) =Y(z) leads to an equation of the form 2.1.6:
Y(z)– z
b
A λz+ B
λτ
Y(τ)dτ =F(z), F(z) =f(x), b=aλ.
54. y(x)– x
a
Axλ+µtλ–µ–1+Bxµt2λ–µ–1
y(t)dt=f(x).
The substitutiony(x) =xµw(x) leads to an equation of the form 2.1.53:
w(x)– x
a
Axλtλ–1+Bt2λ–1
w(t)dt=x–µf(x).
55. y(x) +A x
a
λxλ–1tµ–(λ+µ)xλ+µ–1
y(t)dt=f(x).
This equation can be obtained by differentiating equation 1.1.51:
x a
1 +A(xλtµ–xλ+µ)
y(t)dt=F(x), F(x) = x
a
f(x)dx.
Solution:
y(x) = d dx
xλ Φ(x)
x a
t–λF(t)
tΦ(t)dt
, Φ(x) = exp
– Aµ µ+λxµ+λ
.
56. y(x) + x
a
ABxλ+1–ABxλt–Axλ–B
y(t)dt=f(x).
This is a special case of equation 2.9.7.
Solution:
y(x) =f(x) + x
a
R(x–t)f(t)dt, R(x,t) = (Axλ+B) exp
A λ+ 1
xλ+1–tλ+1 +B2
x
t
exp A
λ+ 1
sλ+1–tλ+1
+B(x–s)
ds.
57. y(x) + x
a
ABxtλ–ABtλ+1+Atλ+B
y(t)dt=f(x).
This is a special case of equation 2.9.8.
Solution:
y(x) =f(x) + x
a
R(x–t)f(t)dt, R(x,t) =–(Atλ+B) exp
A λ+ 1
tλ+1–xλ+1 +B2
x t
exp A
λ+ 1
sλ+1–xλ+1
+B(t–s)
ds.
58. y(x)–λ x
a
x+b t+b
µ
y(t)dt=f(x).
This is a special case of equation 2.9.1 withg(x) = (x+b)µ. Solution:
y(x) =f(x) +λ x
a
x+b t+b
µ
eλ(x–t)f(t)dt.
59. y(x)–λ x
a
xµ+b
tµ+b y(t)dt=f(x).
This is a special case of equation 2.9.1 withg(x) =xµ+b.
Solution:
y(x) =f(x) +λ x
a
xµ+b
tµ+beλ(x–t)f(t)dt.
60. y(x)–λ x
0
y(t)dt
(x–t)α =f(x), 0 <α< 1.
Generalized Abel equation of the second kind.
1◦. Assume that the numberαcan be represented in the form α= 1– m
n, where m= 1, 2,. . ., n= 2, 3,. . . (m<n).
In this case, the solution of the generalized Abel equation of the second kind can be written in closed form (in quadratures):
y(x) =f(x) + x
0
R(x–t)f(t)dt,
where R(x) =
n–1 ν=1
λνΓν(m/n)
Γ(νm/n) x(νm/n)–1+ b m
m–1
µ=0
εµexp εµbx + b
m n–1
ν=1
λνΓν(m/n) Γ(νm/n)
m–1 µ=0
εµexp εµbx
x
0
t(νm/n)–1exp –εµbt
dt
, b=λn/mΓn/m(m/n), εµ = exp
2πµi m
, i2=–1, µ= 0, 1,. . .,m–1.
2◦. Solution with anyαfrom 0 <α< 1:
y(x) =f(x) + x
0
R(x–t)f(t)dt, where R(x) = ∞
n=1
λΓ(1–α)x1–αn
xΓ
n(1–α) . • References: H. Brakhage, K. Nickel, and P. Rieder (1965), V. I. Smirnov (1974).
61. y(x)– λ xα
x 0
y(t)dt
(x–t)1–α =f(x), 0 <α≤1.
1◦. The solution of the homogeneous equation (f ≡0) is
y(x) =Cxβ (β >–1, λ> 0). (1) HereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation
λB(α,β+ 1) = 1, (2)
whereB(p,q) =01zp–1(1–z)q–1dzis the beta function.
2◦. For a polynomial right-hand side, f(x) =
N n=0
Anxn
the solution bounded at zero is given by
y(x) =
N
n=0
An
1–(λ/λn)xn for λ<α, N
n=0
An
1–(λ/λn)xn+Cxβ for λ>αandλ≠λn, λn= (α)n+1
n! , (α)n+1=α(α+ 1). . .(α+n).
HereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation (2).
For specialλ=λn(n= 1, 2,. . .), the solution differs in one term and has the form y(x) =
n–1 m=0
Am
1–(λn/λm)xm+ N m=n+1
Am
1–(λn/λm)xm–Anλ¯n
λnxnlnx+Cxn,
where λ¯n= 1
0
(1–z)α–1znlnz dz –1
.
3◦. For arbitraryf(x), expandable into power series, the formulas of item 2◦can be used, in which one should setN =∞. In this case, the radius of convergence of the solutiony(x) is equal to the radius of convergence off(x).
4◦. For
f(x) = ln(kx) N
n=0
Anxn, a solution has the form
y(x) = ln(kx) N
n=0
Bnxn+ N n=0
Dnxn,
where the constantsBnandDnare found by the method of undetermined coefficients. To obtain the general solution we must add the solution (1) of the homogeneous equation.
In Mikhailov (1966), solvability conditions for the integral equation in question were investigated for various classes off(x).
62. y(x)– λ xµ
x 0
y(t)dt
(ax+bt)1–µ =f(x).
Herea> 0,b> 0, andµis an arbitrary number.
1◦. The solution of the homogeneous equation (f ≡0) is
y(x) =Cxβ (β >–1, λ> 0). (1) HereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation
λI(β) = 1, where I(β) = 1
0
zβ(a+bz)µ–1dz. (2) 2◦. For a polynomial right-hand side,
f(x) = N
n=0
Anxn
the solution bounded at zero is given by
y(x) =
N n=0
An
1–(λ/λn)xn for λ<λ0, N
n=0
An
1–(λ/λn)xn+Cxβ for λ>λ0andλ≠λn, λn= 1
I(n), I(n) = 1
0
zn(a+bz)µ–1dz.
HereCis an arbitrary constant, andβ=β(λ) is determined by the transcendental equation (2).
3◦. For specialλ=λn (n= 1, 2,. . .), the solution differs in one term and has the form y(x) =
n–1 m=0
Am
1–(λn/λm)xm+ N m=n+1
Am
1–(λn/λm)xm–An
λ¯n
λnxnlnx+Cxn, where λ¯n=
1 0
zn(a+bz)µ–1lnz dz –1
.
4◦. For arbitraryf(x) expandable into power series, the formulas of item 2◦can be used, in which one should setN =∞. In this case, the radius of convergence of the solutiony(x) is equal to the radius of convergence off(x).