Page 339 - Schaum's Outline of Theory and Problems of Advanced Calculus

330 IMPROPER INTEGRALS [CHAP. 12

12.36. Solve the differential equation Y ðxÞþ YðxÞ¼ x; Yð0Þ¼ 0; Y ð0Þ¼ 2.
00
0
Take the Laplace transform of both sides of the given differential equation. Then by Problem 12.35,
lfY ðxÞþ YðxÞg ¼ lfxg; lfY ðxÞg þ lfYðxÞg ¼ 1=s 2
00
00
2
and so s lfYðxÞg sYð0Þ Y ð0Þþ lfYðxÞg ¼ 1=s 2
0
Solving for lfYðxÞg using the given conditions, we find
2s 2 1 1
2 2 s 2 s þ 1
2
lfYðxÞg ¼ ¼ þ ð1Þ
s ðs þ 1Þ
by methods of partial fractions.
1 1 1 1
Since ¼ lfxg and ¼ lfsin xg; it follows that þ ¼ lfx þ sin xg:
2
2
s 2 s þ 1 s 2 s þ 1
Hence, from (1), lfYðxÞg ¼ lfx þ sin xg, from which we can conclude that YðxÞ¼ x þ sin x which is,
in fact, found to be a solution.
Another method:
1
If lfFðxÞg ¼ f ðsÞ,we call f ðsÞ the inverse Laplace transform of FðxÞ and write f ðsÞ¼ l fFðxÞg.
1
1
1
By Problem 12.78, l f f ðsÞþ gðsÞg ¼ l f f ðsÞg þ l fgðsÞg. Then from (1),

1 1 1 1 1 1 1

YðxÞ¼ l þ ¼ l þ l ¼ x þ sin x
2
2
s 2 s þ 1 s 2 s þ 1
Inverse Laplace transforms can be read from the table on Page 315.
Supplementary Problems

IMPROPER INTEGRALS OF THE FIRST KIND
12.37. Test for convergence:
ð 2 ð ð 2
1 x þ 1 1 dx 1 x dx
dx
0 x þ 1 1 x þ 4 1 ðx þ x þ 1Þ
ðaÞ 4 ðdÞ 4 ðgÞ 2 5=2
ð ð ð
1 xdx 1 2 þ sin x 1 ln xdx
dx
3
ðbÞ p ffiffiffiffiffiffiffiffiffiffiffiffiffi ðeÞ 2 ðhÞ x
2 x 1 1 x þ 1 1 x þ e
ð ð ð 2
1 dx 1 xdx 1 sin x
dx
ðcÞ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð f Þ 3 ðiÞ 2
1 x 3x þ 2 2 ðln xÞ 0 x
Ans.(a)conv., (b)div., (c)conv., (d)conv., (e)conv., ( f )div., (g)conv., (h)div., (i)conv.
ð
1 dx
12.38. Prove that ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi if b > jaj.
2
1 x þ 2ax þ b 2 b a 2
2
ð ð ð
1 1 1
2
x
12.39. Test for convergence: (a) e x ln xdx; ðbÞ e x lnð1 þ e Þ dx; ðcÞ e x cosh x dx.
1 0 0
Ans. (a)conv., (b)conv., (c)div.