1 (1) u(x) = x + f * u(t)dt, u(x) = x+ 24 2 (2) u(x) = ² = x + √² xtu(t)dt, u(x) = x (3) u(x) = x + [xtu²(t)dt, u(x) = 2x

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
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Chapter6: Vector Spaces
Section6.7: Applications
Problem 2EQ
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In exercises 1-10, verify that the given function is a solution of the corresponding
integral or integro-differential equation:
·[* u(t)dt, u(x) = x+
/0
(2) u(z) = 3/² + √²
(3) u(x) = x +
(1) u(x) = x +
(4) u(x) = x -
-So
24
xtu(t)dt, u(x) = x
xtu²(t)dt, u(x) = 2x
(x-t)u(t)dt, u(x) = sinx
(5) u(2)=2eoshæ– sinh – 1+
(6) u(x) = x +
(7) u'(x) = 2x − x² +
1
-[* tu(t)dt, u(x) = cosh x
- [² tu³ (t)dt, u(x) =
(8) u" (x) = x cos x - 2 sin x +
4tu(t)dt, u(0) = 0, u(x) = x²
- [* tu(t)dt, u(0) = 0, u'(0) = 1, u(x) = sin x
(x-t)²u(t)dt = x³, u(x) = 3
=X
(10) f (x - 1)¹/2u(t)dt = 2³/², u(x) :
=
3
19
2
Transcribed Image Text:In exercises 1-10, verify that the given function is a solution of the corresponding integral or integro-differential equation: ·[* u(t)dt, u(x) = x+ /0 (2) u(z) = 3/² + √² (3) u(x) = x + (1) u(x) = x + (4) u(x) = x - -So 24 xtu(t)dt, u(x) = x xtu²(t)dt, u(x) = 2x (x-t)u(t)dt, u(x) = sinx (5) u(2)=2eoshæ– sinh – 1+ (6) u(x) = x + (7) u'(x) = 2x − x² + 1 -[* tu(t)dt, u(x) = cosh x - [² tu³ (t)dt, u(x) = (8) u" (x) = x cos x - 2 sin x + 4tu(t)dt, u(0) = 0, u(x) = x² - [* tu(t)dt, u(0) = 0, u'(0) = 1, u(x) = sin x (x-t)²u(t)dt = x³, u(x) = 3 =X (10) f (x - 1)¹/2u(t)dt = 2³/², u(x) : = 3 19 2
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