Linear Equations of the First Kind With Variable Limit of Integration
1.3. Equations Whose Kernels Contain Hyperbolic Functions
49.
x a
Aexp(λx2) +Bexp(λt2) +C
y(t)dt=f(x).
This is a special case of equation 1.9.5 withg(x) = exp(λx2).
50.
x
a
Aexp(λx2) +Bexp(µt2)
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Aexp(λx2) andh(t) =Bexp(µt2).
51.
x a
√x–t exp[λ(x2–t2)]y(t)dt=f(x).
Solution:
y(x) = 2
πexp(λx2) d2 dx2
x
a
exp(–λt2)
√x–t f(t)dt.
52.
x a
exp[λ(x2–t2)]
√x–t y(t)dt=f(x).
Solution:
y(x) = 1
πexp(λx2) d dx
x a
exp(–λt√ 2) x–t f(t)dt.
53.
x a
(x–t)λexp[µ(x2–t2)]y(t)dt=f(x), 0 <λ< 1.
Solution:
y(x) =kexp(µx2) d2 dx2
x a
exp(–µt2)
(x–t)λ f(t)dt, k= sin(πλ) πλ . 54.
x
a
exp[λ(xβ–tβ)]y(t)dt =f(x).
Solution: y(x) =fx(x)–λβxβ–1f(x).
1.3. Equations Whose Kernels Contain Hyperbolic
3.
x a
cosh[λ(x–t)] +b
y(t)dt=f(x).
Forb= 0, see equation 1.3.1. Forb=–1, see equation 1.3.2. Forλ= 0, see equation 1.1.1.
Differentiating the equation with respect tox, we arrive at an equation of the form 2.3.16:
y(x) + λ b+ 1
x
a
sinh[λ(x–t)]y(t)dt= fx(x) b+ 1. 1◦. Solution withb(b+ 1) < 0:
y(x) = fx(x)
b+ 1 – λ2 k(b+ 1)2
x
a
sin[k(x–t)]ft(t)dt, where k=λ –b
b+ 1. 2◦. Solution withb(b+ 1) > 0:
y(x) = fx(x)
b+ 1 – λ2 k(b+ 1)2
x a
sinh[k(x–t)]ft(t)dt, where k=λ b
b+ 1. 4.
x a
cosh(λx+βt)y(t)dt=f(x).
Forβ =–λ, see equation 1.3.1.
Differentiating the equation with respect toxtwice, we obtain cosh[(λ+β)x]y(x) +λ
x
a
sinh(λx+βt)y(t)dt=fx(x), (1)
cosh[(λ+β)x]y(x)
x+λsinh[(λ+β)x]y(x) +λ2 x
a
cosh(λx+βt)y(t)dt=fxx (x). (2) Eliminating the integral term from (2) with the aid of the original equation, we arrive at thefirst-order linear ordinary differential equation
wx+λtanh[(λ+β)x]w=fxx(x)–λ2f(x), w= cosh[(λ+β)x]y(x). (3) Settingx=ain (1) yields the initial conditionw(a) =fx(a). On solving equation (3) with this condition, after some manipulations we obtain the solution of the original integral equation in the form
y(x) = 1
cosh[(λ+β)x]fx(x)– λsinh[(λ+β)x]
cosh2[(λ+β)x]f(x)
+ λβ
coshk+1[(λ+β)x]
x
a
f(t) coshk–2[(λ+β)t]dt, k= λ λ+β. 5.
x a
[cosh(λx)–cosh(λt)]y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = cosh(λx).
Solution: y(x) = 1 λ
d dx
fx(x) sinh(λx)
.
6.
x a
[Acosh(λx) +Bcosh(λt)]y(t)dt=f(x).
ForB=–A, see equation 1.3.5. This is a special case of equation 1.9.4 withg(x) = cosh(λx).
Solution: y(x) = 1 A+B
d dx
cosh(λx)– A A+B
x a
cosh(λt)– B
A+Bft(t)dt
.
7.
x a
Acosh(λx) +Bcosh(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acosh(λx) andh(t) =Bcosh(µt) +C.
8.
x a
A1cosh[λ1(x–t)] +A2cosh[λ2(x–t)]
y(t)dt=f(x).
The equation is equivalent to the equation x
a
B1sinh[λ1(x–t)] +B2sinh[λ2(x–t)]
y(t)dt=F(x), B1 = A1
λ1, B2= A2
λ2, F(x) = x
a
f(t)dt,
of the form 1.3.41. (Differentiating this equation yields the original equation.) 9.
x
a
cosh2[λ(x–t)]y(t)dt=f(x).
Differentiation yields an equation of the form 2.3.16:
y(x) +λ x
a
sinh[2λ(x–t)]y(t)dt=fx(x).
Solution:
y(x) =fx(x)– 2λ2 k
x
a
sinh[k(x–t)]ft(t)dt, where k=λ√ 2.
10.
x
a
cosh2(λx)–cosh2(λt)
y(t)dt =f(x), f(a) =fx(a) = 0.
Solution: y(x) = 1 λ
d dx
fx(x) sinh(2λx)
. 11.
x a
Acosh2(λx) +Bcosh2(λt)
y(t)dt=f(x).
ForB=–A, see equation 1.3.10. This is a special case of equation 1.9.4 withg(x) = cosh2(λx).
Solution:
y(x) = 1 A+B
d dx
cosh(λx)–A+B2A x
a
cosh(λt)–A+B2B ft(t)dt
.
12.
x a
Acosh2(λx) +Bcosh2(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acosh2(λx), andh(t) =Bcosh2(µt) +C.
13.
x a
cosh[λ(x–t)] cosh[λ(x+t)]y(t)dt=f(x).
Using the formula
cosh(α–β) cosh(α+β) = 12[cos(2α) + cos(2β)], α=λx, β =λt, we transform the original equation to an equation of the form 1.4.6 withA=B= 1:
x a
[cosh(2λx) + cosh(2λt)]y(t)dt= 2f(x).
Solution:
y(x) = d dx
1
√cosh(2λx) x
a
ft(t)dt
√cosh(2λt)
.
14.
x a
[cosh(λx) cosh(µt) + cosh(βx) cosh(γt)]y(t)dt=f(x).
This is a special case of equation 1.9.15 with g1(x) = cosh(λx),h1(t) = cosh(µt), g2(x) = cosh(βx), andh2(t) = cosh(γt).
15.
x a
cosh3[λ(x–t)]y(t)dt=f(x).
Using the formula cosh3β= 14cosh 3β+34 coshβ, we arrive at an equation of the form 1.3.8:
x
a
1
4cosh[3λ(x–t)] +34cosh[λ(x–t)]
y(t)dt=f(x).
16.
x a
cosh3(λx)–cosh3(λt)
y(t)dt =f(x), f(a) =fx(a) = 0.
Solution: y(x) = 1 3λ
d dx
fx(x) sinh(λx) cosh2(λx)
.
17.
x
a
Acosh3(λx) +Bcosh3(λt)
y(t)dt=f(x).
ForB=–A, see equation 1.3.16. This is a special case of equation 1.9.4 withg(x) = cosh3(λx).
Solution:
y(x) = 1 A+B
d dx
cosh(λx)–A+B3A x a
cosh(λt)–A+B3B ft(t)dt
.
18.
x a
Acosh2(λx) cosh(µt) +Bcosh(βx) cosh2(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Acosh2(λx),h1(t) = cosh(µt),g2(x) = Bcosh(βx), andh2(t) = cosh2(γt).
19.
x a
cosh4[λ(x–t)]y(t)dt=f(x).
Let us transform the kernel of the integral equation using the formula cosh4β = 18cosh 4β+ 12cosh 2β+ 38, where β=λ(x–t),
and differentiate the resulting equation with respect tox. Then we obtain an equation of the form 2.3.18:
y(x) +λ x
a
1
2sinh[4λ(x–t)] + sinh[2λ(x–t)]
y(t)dt=fx(x).
20.
x
a
[cosh(λx)–cosh(λt)]ny(t)dt=f(x), n= 1, 2,. . .
The right-hand side of the equation is assumed to satisfy the conditionsf(a) =fx(a) =· · ·= fx(n)(a) = 0.
Solution: y(x) = sinh(λx) λnn!
1 sinh(λx)
d dx
n+1
f(x).
21.
x a
√coshx–cosht y(t)dt=f(x).
Solution:
y(x) = 2
πsinhx 1 sinhx
d dx
2 x a
sinht f(t)dt
√coshx–cosht.
22.
x
a
y(t)dt
√coshx–cosht =f(x).
Solution:
y(x) = 1 π
d dx
x a
sinht f(t)dt
√coshx–cosht.
23.
x a
(coshx–cosht)λy(t)dt =f(x), 0 <λ< 1.
Solution:
y(x) =ksinhx 1 sinhx
d dx
2 x
a
sinht f(t)dt
(coshx–cosht)λ, k= sin(πλ) πλ . 24.
x a
(coshµx–coshµt)y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = coshµx.
Solution: y(x) = 1 µ
d dx
fx(x) sinhxcoshµ–1x
.
25.
x
a
Acoshµx+Bcoshµt
y(t)dt=f(x).
ForB=–A, see equation 1.3.24. This is a special case of equation 1.9.4 withg(x) = coshµx.
Solution:
y(x) = 1 A+B
d dx
cosh(λx)–A+BAµ x
a
cosh(λt)–A+BBµ ft(t)dt
.
26.
x a
y(t)dt
(coshx–cosht)λ =f(x), 0 <λ< 1.
Solution:
y(x) = sin(πλ) π
d dx
x
a
sinht f(t)dt (coshx–cosht)1–λ. 27.
x a
(x–t) cosh[λ(x–t)]y(t)dt=f(x), f(a) =fx(a) = 0.
Differentiating the equation twice yields y(x) + 2λ
x
a
sinh[λ(x–t)]y(t)dt+λ2 x
a
(x–t) cosh[λ(x–t)]y(t)dt=fxx(x).
Eliminating the third term on the right-hand side with the aid of the original equation, we arrive at an equation of the form 2.3.16:
y(x) + 2λ x
a
sinh[λ(x–t)]y(t)dt=fxx (x)–λ2f(x).
28.
x a
cosh λ√
x–t
√x–t y(t)dt=f(x).
Solution:
y(x) = 1 π
d dx
x a
cos λ√
x–t
√x–t f(t)dt.
29.
x 0
cosh λ√
x2–t2
√x2–t2 y(t)dt=f(x).
Solution:
y(x) = 2 π
d dx
x
0
tcos λ√
x2–t2
√x2–t2 f(t)dt.
30.
∞
x
cosh λ√
t2–x2
√t2–x2 y(t)dt=f(x).
Solution:
y(x) =–2 π
d dx
∞
x
tcos λ√
t2–x2
√t2–x2 f(t)dt.
31.
x a
Axβ+Bcoshγ(λt) +C]y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Axβandh(t) =Bcoshγ(λt) +C.
32.
x a
Acoshγ(λx) +Btβ+C]y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acoshγ(λx) andh(t) =Btβ+C.
33.
x
a
Axλcoshµt+Btβcoshγx
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = coshµt,g2(x) =Bcoshγx, andh2(t) =tβ.
1.3-2. Kernels Containing Hyperbolic Sine 34.
x a
sinh[λ(x–t)]y(t)dt=f(x), f(a) =fx(a) = 0.
Solution: y(x) = 1
λfxx (x)–λf(x).
35.
x a
sinh[λ(x–t)] +b
y(t)dt =f(x).
Differentiating the equation with respect tox, we arrive at an equation of the form 2.3.3:
y(x) +λ b
x a
cosh[λ(x–t)]y(t)dt= 1 bfx(x).
Solution:
y(x) = 1 bfx(x) +
x a
R(x–t)ft(t)dt, R(x) = λ
b2 exp
–λx 2b
λ
2bk sinh(kx)–cosh(kx)
, k= λ√ 1 + 4b2
2b .
36.
x a
sinh(λx+βt)y(t)dt =f(x).
Forβ =–λ, see equation 1.3.34. Assume thatβ≠–λ.
Differentiating the equation with respect toxtwice yields sinh[(λ+β)x]y(x) +λ
x
a
cosh(λx+βt)y(t)dt=fx(x), (1)
sinh[(λ+β)x]y(x)
x+λcosh[(λ+β)x]y(x) +λ2 x
a
sinh(λx+βt)y(t)dt=fxx (x). (2) Eliminating the integral term from (2) with the aid of the original equation, we arrive at the first-order linear ordinary differential equation
wx+λcoth[(λ+β)x]w=fxx(x)–λ2f(x), w= sinh[(λ+β)x]y(x). (3) Settingx=ain (1) yields the initial conditionw(a) =fx(a). On solving equation (3) with this condition, after some manipulations we obtain the solution of the original integral equation in the form
y(x) = 1
sinh[(λ+β)x]fx(x)– λcosh[(λ+β)x]
sinh2[(λ+β)x] f(x)
– λβ
sinhk+1[(λ+β)x]
x a
f(t) sinhk–2[(λ+β)t]dt, k= λ λ+β. 37.
x a
[sinh(λx)–sinh(λt)]y(t)dt=f(x), f(a) =fx(a) = 0.
This is a special case of equation 1.9.2 withg(x) = sinh(λx).
Solution: y(x) = 1 λ
d dx
fx(x) cosh(λx)
.
38.
x a
[Asinh(λx) +Bsinh(λt)]y(t)dt=f(x).
ForB=–A, see equation 1.3.37. This is a special case of equation 1.9.4 withg(x) = sinh(λx).
Solution: y(x) = 1 A+B
d dx
sinh(λx)– A A+B
x a
sinh(λt)– B
A+Bft(t)dt
.
39.
x
a
[Asinh(λx) +Bsinh(µt)]y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Asinh(λx), andh(t) =Bsinh(µt).
40.
x a
µsinh[λ(x–t)]–λsinh[µ(x–t)]
y(t)dt=f(x).
It is assumed thatf(a) =fx(a) =fxx (a) =fxxx (a) = 0.
Solution:
y(x) = fxxxx –(λ2+µ2)fxx +λ2µ2f
µλ3–λµ3 , f =f(x).
41.
x a
A1sinh[λ1(x–t)] +A2sinh[λ2(x–t)]
y(t)dt=f(x), f(a) =fx(a) = 0.
1◦. Introduce the notation I1=
x
a
sinh[λ1(x–t)]y(t)dt, I2= x
a
sinh[λ2(x–t)]y(t)dt, J1=
x
a
cosh[λ1(x–t)]y(t)dt, J2= x
a
cosh[λ2(x–t)]y(t)dt.
Let us successively differentiate the integral equation four times. As a result, we have (the first line is the original equation):
A1I1+A2I2=f, f =f(x), (1)
A1λ1J1+A2λ2J2 =fx, (2)
(A1λ1+A2λ2)y+A1λ21I1+A2λ22I2=fxx , (3) (A1λ1+A2λ2)yx +A1λ31J1+A2λ32J2=fxxx , (4) (A1λ1+A2λ2)yxx + (A1λ31+A2λ32)y+A1λ41I1+A2λ42I2=fxxxx . (5) EliminatingI1 andI2 from (1), (3), and (5), we arrive at the following second-order linear ordinary differential equation with constant coefficients:
(A1λ1+A2λ2)yxx–λ1λ2(A1λ2+A2λ1)y=fxxxx –(λ21+λ22)fxx +λ21λ22f. (6) The initial conditions can be obtained by substitutingx=ainto (3) and (4):
(A1λ1+A2λ2)y(a) =fxx(a), (A1λ1+A2λ2)yx(a) =fxxx (a). (7) Solving the differential equation (6) under conditions (7) allows us tofind the solution of the integral equation.
2◦. Denote
∆=λ1λ2
A1λ2+A2λ1
A1λ1+A2λ2. 2.1. Solution for∆> 0:
(A1λ1+A2λ2)y(x) =fxx (x) +Bf(x) +C x
a
sinh[k(x–t)]f(t)dt, k=√
∆, B=∆–λ21–λ22, C= 1
√∆
∆2–(λ21+λ22)∆+λ21λ22 . 2.2. Solution for∆< 0:
(A1λ1+A2λ2)y(x) =fxx (x) +Bf(x) +C x
a
sin[k(x–t)]f(t)dt, k=√
–∆, B=∆–λ21–λ22, C= √1 –∆
∆2–(λ21+λ22)∆+λ21λ22 . 2.3. Solution for∆= 0:
(A1λ1+A2λ2)y(x) =fxx(x)–(λ21+λ22)f(x) +λ21λ22 x
a
(x–t)f(t)dt.
2.4. Solution for∆=∞:
y(x) = fxxxx –(λ21+λ22)fxx +λ21λ22f
A1λ31+A2λ32 , f =f(x).
In the last case, the relation A1λ1+A2λ2 = 0 is valid, and the right-hand side of the integral equation is assumed to satisfy the conditionsf(a) =fx(a) =fxx (a) =fxxx (a) = 0.
42.
x a
Asinh[λ(x–t)] +Bsinh[µ(x–t)] +Csinh[β(x–t)]
y(t)dt=f(x).
It assumed thatf(a) =fx(a) = 0. Differentiating the integral equation twice yields (Aλ+Bµ+Cβ)y(x) +
x
a
Aλ2sinh[λ(x–t)] +Bµ2sinh[µ(x–t)]
y(t)dt +Cβ2
x a
sinh[β(x–t)]y(t)dt=fxx (x).
Eliminating the last integral with the aid of the original equation, we arrive at an equation of the form 2.3.18:
(Aλ+Bµ+Cβ)y(x) +
x a
A(λ2–β2) sinh[λ(x–t)] +B(µ2–β2) sinh[µ(x–t)]
y(t)dt=fxx (x)–β2f(x).
In the special caseAλ+Bµ+Cβ= 0, this is an equation of the form 1.3.41.
43.
x a
sinh2[λ(x–t)]y(t)dt=f(x), f(a) =fx(a) =fxx (a) = 0.
Differentiating yields an equation of the form 1.3.34:
x
a
sinh[2λ(x–t)]y(t)dt= 1 λfx(x).
Solution: y(x) = 12λ–2fxxx (x)–2fx(x).
44.
x a
sinh2(λx)–sinh2(λt)
y(t)dt=f(x), f(a) =fx(a) = 0.
Solution: y(x) = 1 λ
d dx
fx(x) sinh(2λx)
.
45.
x
a
Asinh2(λx) +Bsinh2(λt)
y(t)dt=f(x).
ForB=–A, see equation 1.3.44. This is a special case of equation 1.9.4 withg(x) = sinh2(λx).
Solution:
y(x) = 1 A+B
d dx
sinh(λx)–A+B2A x a
sinh(λt)–A+B2B ft(t)dt
.
46.
x a
Asinh2(λx) +Bsinh2(µt)
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Asinh2(λx) andh(t) =Bsinh2(µt).
47.
x
a
sinh[λ(x–t)] sinh[λ(x+t)]y(t)dt=f(x).
Using the formula
sinh(α–β) sinh(α+β) = 12[cosh(2α)–cosh(2β)], α=λx, β=λt, we reduce the original equation to an equation of the form 1.3.5:
x a
[cosh(2λx)–cosh(2λt)]y(t)dt= 2f(x).
Solution: y(x) = 1 λ
d dx
fx(x) sinh(2λx)
.
48.
x a
Asinh(λx) sinh(µt) +Bsinh(βx) sinh(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Asinh(λx),h1(t) = sinh(µt),g2(x) = Bsinh(βx), andh2(t) = sinh(γt).
49.
x a
sinh3[λ(x–t)]y(t)dt=f(x), f(a) =fx(a) =fxx (a) =fxxx (a) = 0.
Using the formula sinh3β= 14 sinh 3β– 34sinhβ, we arrive at an equation of the form 1.3.41:
x a
1
4sinh[3λ(x–t)]– 34sinh[λ(x–t)]
y(t)dt=f(x).
50.
x a
sinh3(λx)–sinh3(λt)
y(t)dt=f(x), f(a) =fx(a) = 0.
This is a special case of equation 1.9.2 withg(x) = sinh3(λx).
51.
x
a
Asinh3(λx) +Bsinh3(λt)
y(t)dt=f(x).
This is a special case of equation 1.9.4 withg(x) = sinh3(λx).
Solution:
y(x) = 1 A+B
d dx
sinh(λx)–A+B3A x
a
sinh(λt)–A+B3B ft(t)dt
.
52.
x
a
Asinh2(λx) sinh(µt) +Bsinh(βx) sinh2(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Asinh2(λx),h1(t) = sinh(µt),g2(x) = Bsinh(βx), andh2(t) = sinh2(γt).
53.
x a
sinh4[λ(x–t)]y(t)dt=f(x).
It is assumed thatf(a) =fx(a) =· · ·=fxxxx (a) = 0.
Let us transform the kernel of the integral equation using the formula sinh4β= 18cosh 4β– 12cosh 2β+ 38, where β =λ(x–t),
and differentiate the resulting equation with respect tox. Then we arrive at an equation of the form 1.3.41:
λ x
a
1
2sinh[4λ(x–t)]–sinh[2λ(x–t)]
y(t)dt=fx(x).
54.
x
a
sinhn[λ(x–t)]y(t)dt=f(x), n= 2, 3,. . . It is assumed thatf(a) =fx(a) =· · ·=fx(n)(a) = 0.
Let us differentiate the equation with respect toxtwice and transform the kernel of the resulting integral equation using the formula cosh2β= 1 + sinh2β, whereβ=λ(x–t). Then we have
λ2n2 x
a
sinhn[λ(x–t)]y(t)dt+λ2n(n–1) x
a
sinhn–2[λ(x–t)]y(t)dt=fxx(x).
Eliminating thefirst term on the left-hand side with the aid of the original equation, we obtain x
a
sinhn–2[λ(x–t)]y(t)dt= 1 λ2n(n–1)
fxx (x)–λ2n2f(x) .
This equation has the same form as the original equation, but the exponent of the kernel has been reduced by two.
By applying this technique sufficiently many times, wefinally arrive at simple integral equations of the form 1.1.1 (for evenn) or 1.3.34 (for oddn).
55.
x a
sinh λ√
x–t
y(t)dt=f(x).
Solution:
y(x) = 2 πλ
d2 dx2
x a
cos λ√
x–t
√x–t f(t)dt.
56.
x a
√sinhx–sinht y(t)dt=f(x).
Solution:
y(x) = 2
πcoshx 1 coshx
d dx
2 x a
cosht f(t)dt
√sinhx–sinht.
57.
x
a
y(t)dt
√sinhx–sinht =f(x).
Solution:
y(x) = 1 π
d dx
x a
cosht f(t)dt
√sinhx–sinht.
58.
x a
(sinhx–sinht)λy(t)dt=f(x), 0 <λ< 1.
Solution:
y(x) =kcoshx 1 coshx
d dx
2 x a
cosht f(t)dt
(sinhx–sinht)λ, k= sin(πλ) πλ . 59.
x
a
(sinhµx–sinhµt)y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = sinhµx.
Solution: y(x) = 1 µ
d dx
fx(x) coshxsinhµ–1x
.
60.
x a
Asinhµ(λx) +Bsinhµ(λt)
y(t)dt=f(x).
This is a special case of equation 1.9.4 withg(x) = sinhµ(λx).
Solution withB≠–A:
y(x) = 1 A+B
d dx
sinh(λx)–A+BAµ x a
sinh(λt)–A+BBµ ft(t)dt
.
61.
x
a
y(t)dt
(sinhx–sinht)λ =f(x), 0 <λ< 1.
Solution:
y(x) = sin(πλ) π
d dx
x a
cosht f(t)dt (sinhx–sinht)1–λ. 62.
x
a
(x–t) sinh[λ(x–t)]y(t)dt=f(x), f(a) = fx(a) =fxx (a) = 0.
Double differentiation yields 2λ
x
a
cosh[λ(x–t)]y(t)dt+λ2 x
a
(x–t) sinh[λ(x–t)]y(t)dt=fxx (x).
Eliminating the second term on the left-hand side with the aid of the original equation, we arrive at an equation of the form 1.3.1:
x a
cosh[λ(x–t)]y(t)dt= 1 2λ
fxx (x)–λ2f(x) . Solution:
y(x) = 1
2λfxxx (x)–λfx(x) +12λ3 x
a
f(t)dt.
63.
x
a
Axβ+Bsinhγ(λt) +C]y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Axβandh(t) =Bsinhγ(λt) +C.
64.
x a
Asinhγ(λx) +Btβ+C]y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Asinhγ(λx) andh(t) =Btβ+C.
65.
x
a
Axλsinhµt+Btβsinhγx
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = sinhµt,g2(x) =Bsinhγx, andh2(t) =tβ.
1.3-3. Kernels Containing Hyperbolic Tangent
66.
x a
tanh(λx)–tanh(λt)
y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = tanh(λx).
Solution: y(x) = 1 λ
cosh2(λx)fx(x)
x. 67.
x a
Atanh(λx) +Btanh(λt)
y(t)dt=f(x).
ForB=–A, see equation 1.3.66. This is a special case of equation 1.9.4 withg(x) = tanh(λx).
Solution: y(x) = 1 A+B
d dx
tanh(λx)– A A+B
x a
tanh(λt)– B
A+Bft(t)dt
.
68.
x a
Atanh(λx) +Btanh(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Atanh(λx) andh(t) =Btanh(µt) +C.
69.
x a
tanh2(λx)–tanh2(λt)
y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = tanh2(λx).
Solution: y(x) = d dx
cosh3(λx)fx(x) 2λsinh(λx)
.
70.
x a
Atanh2(λx) +Btanh2(λt)
y(t)dt=f(x).
ForB=–A, see equation 1.3.69. This is a special case of equation 1.9.4 withg(x) = tanh2(λx).
Solution: y(x) = 1 A+B
d dx
tanh(λx)– 2A A+B
x
a
tanh(λt)– 2B
A+Bft(t)dt
.
71.
x a
Atanh2(λx) +Btanh2(µt) +C
y(t)dt =f(x).
This is a special case of equation 1.9.6 withg(x) =Atanh2(λx) andh(t) =Btanh2(µt) +C.
72.
x a
tanh(λx)–tanh(λt)n
y(t)dt =f(x), n= 1, 2,. . .
The right-hand side of the equation is assumed to satisfy the conditionsf(a) =fx(a) =· · ·= fx(n)(a) = 0.
Solution: y(x) = 1 λnn! cosh2(λx)
cosh2(λx) d dx
n+1
f(x).
73.
x a
√tanhx–tanht y(t)dt =f(x).
Solution:
y(x) = 2
πcosh2x cosh2x d dx
2 x a
f(t)dt cosh2t√
tanhx–tanht. 74.
x
a
y(t)dt
√tanhx–tanht =f(x).
Solution:
y(x) = 1 π
d dx
x a
f(t)dt cosh2t√
tanhx–tanht. 75.
x a
(tanhx–tanht)λy(t)dt=f(x), 0 <λ< 1.
Solution:
y(x) = sin(πλ)
πλcosh2x cosh2x d dx
2 x a
f(t)dt
cosh2t(tanhx–tanht)λ.
76.
x a
(tanhµx–tanhµt)y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = tanhµx.
Solution: y(x) = 1 µ
d dx
coshµ+1xfx(x) sinhµ–1x
.
77.
x
a
Atanhµx+Btanhµt
y(t)dt=f(x).
ForB=–A, see equation 1.3.76. This is a special case of equation 1.9.4 withg(x) = tanhµx.
Solution:
y(x) = 1 A+B
d dx
tanh(λx)–A+BAµ x a
tanh(λt)–A+BBµ ft(t)dt
.
78.
x a
y(t)dt
[tanh(λx)–tanh(λt)]µ =f(x), 0 <µ< 1.
This is a special case of equation 1.9.42 withg(x) = tanh(λx) andh(x)≡1.
Solution:
y(x) = λsin(πµ) π
d dx
x
a
f(t)dt
cosh2(λt)[tanh(λx)–tanh(λt)]1–µ. 79.
x
a
Axβ+Btanhγ(λt) +C]y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Axβandh(t) =Btanhγ(λt) +C.
80.
x
a
Atanhγ(λx) +Btβ+C]y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Atanhγ(λx) andh(t) =Btβ+C.
81.
x
a
Axλtanhµt+Btβtanhγx
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = tanhµt,g2(x) =Btanhγx, andh2(t) =tβ.
1.3-4. Kernels Containing Hyperbolic Cotangent
82.
x
a
coth(λx)–coth(λt)
y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = coth(λx).
Solution: y(x) =–1 λ
d dx
sinh2(λx)fx(x) .
83.
x
a
Acoth(λx) +Bcoth(λt)
y(t)dt=f(x).
ForB=–A, see equation 1.3.82. This is a special case of equation 1.9.4 withg(x) = coth(λx).
Solution: y(x) = 1 A+B
d dx
tanh(λx) A
A+B
x a
tanh(λt) B
A+Bft(t)dt
.
84.
x a
Acoth(λx) +Bcoth(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acoth(λx) andh(t) =Bcoth(µt) +C.
85.
x a
coth2(λx)–coth2(λt)
y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = coth2(λx).
Solution: y(x) =– d dx
sinh3(λx)fx(x) 2λcosh(λx)
.
86.
x a
Acoth2(λx) +Bcoth2(λt)
y(t)dt=f(x).
ForB=–A, see equation 1.3.85. This is a special case of equation 1.9.4 withg(x) = coth2(λx).
Solution: y(x) = 1 A+B
d dx
tanh(λx) 2A
A+B
x a
tanh(λt) 2B
A+Bft(t)dt
.
87.
x a
Acoth2(λx) +Bcoth2(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acoth2(λx) andh(t) =Bcoth2(µt) +C.
88.
x a
coth(λx)–coth(λt)n
y(t)dt=f(x), n= 1, 2,. . .
The right-hand side of the equation is assumed to satisfy the conditionsf(a) =fx(a) =· · ·= fx(n)(a) = 0.
Solution: y(x) = (–1)n λnn! sinh2(λx)
sinh2(λx) d dx
n+1
f(x).
89.
x
a
(cothµx–cothµt)y(t)dt=f(x).
This is a special case of equation 1.9.2 withg(x) = cothµx.
Solution: y(x) =–1 µ
d dx
sinhµ+1xfx(x) coshµ–1x
.
90.
x
a
Acothµx+Bcothµt
y(t)dt =f(x).
ForB=–A, see equation 1.3.89. This is a special case of equation 1.9.4 withg(x) = cothµx.
Solution:
y(x) = 1 A+B
d dx
tanhxA+BAµ x a
tanhtA+BBµ ft(t)dt
.
91.
x a
Axβ+Bcothγ(λt) +C]y(t)dt =f(x).
This is a special case of equation 1.9.6 withg(x) =Axβandh(t) =Bcothγ(λt) +C.
92.
x
a
Acothγ(λx) +Btβ+C]y(t)dt =f(x).
This is a special case of equation 1.9.6 withg(x) =Acothγ(λx) andh(t) =Btβ+C.
93.
x a
Axλcothµt+Btβcothγx
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = cothµt,g2(x) =Bcothγx, andh2(t) =tβ.
1.3-5. Kernels Containing Combinations of Hyperbolic Functions
94.
x
a
cosh[λ(x–t)] +Asinh[µ(x–t)]
y(t)dt =f(x).
Let us differentiate the equation with respect toxand then eliminate the integral with the hyperbolic cosine. As a result, we arrive at an equation of the form 2.3.16:
y(x) + (λ–A2µ) x
a
sinh[µ(x–t)]y(t)dt=fx(x)–Aµf(x).
95.
x a
Acosh(λx) +Bsinh(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acosh(λx) andh(t) =Bsinh(µt) +C.
96.
x a
Acosh2(λx) +Bsinh2(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) = cosh2(λx) andh(t) =Bsinh2(µt) +C.
97.
x
a
sinh[λ(x–t)] cosh[λ(x+t)]y(t)dt=f(x).
Using the formula
sinh(α–β) cosh(α+β) = 12
sinh(2α)–sinh(2β)
, α=λx, β=λt, we reduce the original equation to an equation of the form 1.3.37:
x a
sinh(2λx)–sinh(2λt)
y(t)dt= 2f(x).
Solution: y(x) = 1 λ
d dx
fx(x) cosh(2λx)
.
98.
x
a
cosh[λ(x–t)] sinh[λ(x+t)]y(t)dt=f(x).
Using the formula
cosh(α–β) sinh(α+β) = 12
sinh(2α) + sinh(2β)
, α=λx, β =λt, we reduce the original equation to an equation of the form 1.3.38 withA=B= 1:
x
a
sinh(2λx) + sinh(2λt)
y(t)dt= 2f(x).
99.
x a
Acosh(λx) sinh(µt) +Bcosh(βx) sinh(γt)
y(t)dt =f(x).
This is a special case of equation 1.9.15 withg1(x) =Acosh(λx),h1(t) = sinh(µt),g2(x) = Bcosh(βx), andh2(t) = sinh(γt).
100.
x
a
sinh(λx) cosh(µt) + sinh(βx) cosh(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) = sinh(λx),h1(t) = cosh(µt),g2(x) = sinh(βx), andh2(t) = cosh(γt).
101.
x a
cosh(λx) cosh(µt) + sinh(βx) sinh(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 with g1(x) = cosh(λx),h1(t) = cosh(µt), g2(x) = sinh(βx), andh2(t) = sinh(γt).
102.
x
a
Acoshβ(λx) +Bsinhγ(µt)
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acoshβ(λx) andh(t) =Bsinhγ(µt).
103.
x
a
Asinhβ(λx) +Bcoshγ(µt)
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Asinhβ(λx) andh(t) =Bcoshγ(µt).
104.
x
a
Axλcoshµt+Btβsinhγx
y(t)dt =f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = coshµt,g2(x) =Bsinhγx, andh2(t) =tβ.
105.
x a
(x–t) sinh[λ(x–t)]–λ(x–t)2 cosh[λ(x–t)]
y(t)dt=f(x).
Solution:
y(x) = x
a
g(t)dt, where
g(t) = π
2λ 1 64λ5
d2 dt2 –λ2
6 t a
(t–τ)52 I5
2
[λ(t–τ)]f(τ)dτ.
106.
x
a
sinh[λ(x–t)]
x–t –λcosh[λ(x–t)]
y(t)dt=f(x).
Solution:
y(x) = 1 2λ4
d2 dx2 –λ2
3 x a
sinh[λ(x–t)]f(t)dt.
107.
x a
sinh λ√
x–t –λ√
x–tcosh λ√
x–t
y(t)dt=f(x), f(a) =fx(a) = 0.
Solution:
y(x) =– 4 πλ3
d3 dx3
x
a
cos λ√
x–t
√x–t f(t)dt.
108.
x a
Axλsinhµt+Btβcoshγx
y(t)dt =f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = sinhµt,g2(x) =Bcoshγx, andh2(t) =tβ.
109.
x
a
Atanh(λx) +Bcoth(µt) +C
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Atanh(λx) andh(t) =Bcoth(µt) +C.
110.
x a
Atanh2(λx) +Bcoth2(µt)
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) = tanh2(λx) andh(t) =Bcoth2(µt).
111.
x a
tanh(λx) coth(µt) + tanh(βx) coth(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 with g1(x) = tanh(λx), h1(t) = coth(µt),g2(x) = tanh(βx), andh2(t) = coth(γt).
112.
x a
coth(λx) tanh(µt) + coth(βx) tanh(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 with g1(x) = coth(λx), h1(t) = tanh(µt),g2(x) = coth(βx), andh2(t) = tanh(γt).
113.
x
a
tanh(λx) tanh(µt) + coth(βx) coth(γt)
y(t)dt=f(x).
This is a special case of equation 1.9.15 with g1(x) = tanh(λx), h1(t) = tanh(µt),g2(x) = coth(βx), andh2(t) = coth(γt).
114.
x a
Atanhβ(λx) +Bcothγ(µt)
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Atanhβ(λx) andh(t) =Bcothγ(µt).
115.
x a
Acothβ(λx) +Btanhγ(µt)
y(t)dt=f(x).
This is a special case of equation 1.9.6 withg(x) =Acothβ(λx) andh(t) =Btanhγ(µt).
116.
x a
Axλtanhµt+Btβcothγx
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = tanhµt,g2(x) =Bcothγx, andh2(t) =tβ.
117.
x a
Axλcothµt+Btβtanhγx
y(t)dt=f(x).
This is a special case of equation 1.9.15 withg1(x) =Axλ,h1(t) = cothµt,g2(x) =Btanhγx, andh2(t) =tβ.