Example 2. Convert the triple integral to a single integral. Using the formula (70) yields subject to the initial conditions -fff u(t) dtdtdt, I(x) = = 6 (x-t)²u(t) dt, [(x) = or equivalently the equivalent single integral. Returning to the main goal of this section, we discuss the technique that will be used to convert an initial value problem to an equivalent Volterra integral equation. Without loss of generality, and for simplicity reasons, we apply this technique to a third order initial value problem given by y" (x) + p(x)y" (x) + g(x)y'(x) +r(x)y(x) = g(x) y(0) = a,y (0) = 8,y" (0)=%, a, 3 andy are constants. (81) The coefficient functions p(x), q(x) and (x) are analytic functions by assuming that these functions have Taylor expansions about the origin. Besides, we assume that g(x) is continuous through the interval of discussion. To transform (80) into an equivalent Volterra integral equation, we first set y" (x) = u(x), (82) where u(x) is a continuous function on the interval of discussion. Based on (82), it remains to find other relations for y and its derivatives as single integrals involving u(x). This can be simply performed by integrating both sides of (82) from 0 to x where we find y" (x) - y" (0) = = √² ₁² u(t)dt, y" (x) = 7+ +S u(t)dt, y (x)=B+ya+ fu(t)dt dt. y(x) = a + 3x + (78) Similarly we integrate both sides of (85) from 0 to x to obtain +/=/= 7²2² + [₁²] [6²] [6²² (79) (80) obtained upon using the initial condition y" (0) = y. To obtain y'(x) we integrate both sides of (84) from 0 to x to find that u(t)dtdtdt. y(x) = 3+72+ + √² (2₂- (r-t)u(t)dt, (83) (84) (85) Using the conversion formulas (71) and (72), to reduce the double and triple integrals in (85) and (86) respectively to single integrals, we obtain (86) (87)
Example 2. Convert the triple integral to a single integral. Using the formula (70) yields subject to the initial conditions -fff u(t) dtdtdt, I(x) = = 6 (x-t)²u(t) dt, [(x) = or equivalently the equivalent single integral. Returning to the main goal of this section, we discuss the technique that will be used to convert an initial value problem to an equivalent Volterra integral equation. Without loss of generality, and for simplicity reasons, we apply this technique to a third order initial value problem given by y" (x) + p(x)y" (x) + g(x)y'(x) +r(x)y(x) = g(x) y(0) = a,y (0) = 8,y" (0)=%, a, 3 andy are constants. (81) The coefficient functions p(x), q(x) and (x) are analytic functions by assuming that these functions have Taylor expansions about the origin. Besides, we assume that g(x) is continuous through the interval of discussion. To transform (80) into an equivalent Volterra integral equation, we first set y" (x) = u(x), (82) where u(x) is a continuous function on the interval of discussion. Based on (82), it remains to find other relations for y and its derivatives as single integrals involving u(x). This can be simply performed by integrating both sides of (82) from 0 to x where we find y" (x) - y" (0) = = √² ₁² u(t)dt, y" (x) = 7+ +S u(t)dt, y (x)=B+ya+ fu(t)dt dt. y(x) = a + 3x + (78) Similarly we integrate both sides of (85) from 0 to x to obtain +/=/= 7²2² + [₁²] [6²] [6²² (79) (80) obtained upon using the initial condition y" (0) = y. To obtain y'(x) we integrate both sides of (84) from 0 to x to find that u(t)dtdtdt. y(x) = 3+72+ + √² (2₂- (r-t)u(t)dt, (83) (84) (85) Using the conversion formulas (71) and (72), to reduce the double and triple integrals in (85) and (86) respectively to single integrals, we obtain (86) (87)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Example 2. Convert the triple integral
to a single integral.
Using the formula (70) yields
subject to the initial conditions
- 5 fő fu(t) dtdtdt,
I(x) =
=16 (x-t)²u(t) dt,
[(x) =
or equivalently
the equivalent single integral.
Returning to the main goal of this section, we discuss the technique that will be used
to convert an initial value problem to an equivalent Volterra integral equation. Without
loss of generality, and for simplicity reasons, we apply this technique to a third order
initial value problem given by
y" (x) + p(x)y" (x) + q(r)y'(x) +r(x)y(x) = g(x)
y(0) = a,y (0) = 8.y" (0)=, a, 3 andy are constants.
(81)
The coefficient functions p(x), q(x) and (x) are analytic functions by assuming that
these functions have Taylor expansions about the origin. Besides, we assume that g(x) is
continuous through the interval of discussion. To transform (80) into an equivalent
Volterra integral equation, we first set
y" (x) = u(x).
(82)
where u(x) is a continuous function on the interval of discussion. Based on (82), it
y" (x) - y" (0) = = √² ₁²
remains to find other relations for y and its derivatives as single integrals involving
u(x). This can be simply performed by integrating both sides of (82) from 0 to x where
we find
y" (x) = 7+
+S
y (x)=B+ya+
y(x) = a + 3x +
u(t)dt,
u(t)dt,
f*f* u(t)dt dt.
(78)
Similarly we integrate both sides of (85) from 0 to x to obtain
+/=/= 7²2² + [₁²] [²] [²²
(79)
obtained upon using the initial condition y" (0) = y. To obtain y'(x) we integrate both
sides of (84) from 0 to x to find that
y(x) = 3+72+
(80)
u(t)dtdtdt.
+ √² (2²-
(x-t)u(t)dt,
(83)
(84)
(85)
Using the conversion formulas (71) and (72), to reduce the double and triple integrals in
(85) and (86) respectively to single integrals, we obtain
(86)
(87)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6344b2d-8fba-4ae4-a76b-36914c334404%2F63b2b73b-d8c0-4949-8af7-85dbae26ba66%2Fw4xpbjl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 2. Convert the triple integral
to a single integral.
Using the formula (70) yields
subject to the initial conditions
- 5 fő fu(t) dtdtdt,
I(x) =
=16 (x-t)²u(t) dt,
[(x) =
or equivalently
the equivalent single integral.
Returning to the main goal of this section, we discuss the technique that will be used
to convert an initial value problem to an equivalent Volterra integral equation. Without
loss of generality, and for simplicity reasons, we apply this technique to a third order
initial value problem given by
y" (x) + p(x)y" (x) + q(r)y'(x) +r(x)y(x) = g(x)
y(0) = a,y (0) = 8.y" (0)=, a, 3 andy are constants.
(81)
The coefficient functions p(x), q(x) and (x) are analytic functions by assuming that
these functions have Taylor expansions about the origin. Besides, we assume that g(x) is
continuous through the interval of discussion. To transform (80) into an equivalent
Volterra integral equation, we first set
y" (x) = u(x).
(82)
where u(x) is a continuous function on the interval of discussion. Based on (82), it
y" (x) - y" (0) = = √² ₁²
remains to find other relations for y and its derivatives as single integrals involving
u(x). This can be simply performed by integrating both sides of (82) from 0 to x where
we find
y" (x) = 7+
+S
y (x)=B+ya+
y(x) = a + 3x +
u(t)dt,
u(t)dt,
f*f* u(t)dt dt.
(78)
Similarly we integrate both sides of (85) from 0 to x to obtain
+/=/= 7²2² + [₁²] [²] [²²
(79)
obtained upon using the initial condition y" (0) = y. To obtain y'(x) we integrate both
sides of (84) from 0 to x to find that
y(x) = 3+72+
(80)
u(t)dtdtdt.
+ √² (2²-
(x-t)u(t)dt,
(83)
(84)
(85)
Using the conversion formulas (71) and (72), to reduce the double and triple integrals in
(85) and (86) respectively to single integrals, we obtain
(86)
(87)
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