we know, Now, Senndu = enxsen - ((San. In lon) du 0 0 = x ln x - = [using product Rule] - Sandn nhan [en (x+1) + In (y+2)] dudy I en (x+1) andy + 0 0 0 1 In (3+2) Ludy = 0 + = [[(2012-2)+1]dy = (2402-1) d + + j'en (+3) day [(1+2) en (4+2) = (4+2)]' 1 = (2 n2-1) ["] + [(3 en3-3)-(21m2-2)) = (2ln2-1) (1-0) + 3 en 3-3-24m2+2 = 2ln 2-1 + 3 en 3-3-2 en 2 + 2 = 3 ln 3-2 = in 3-2 = ln 27-2 = −2+ en 27

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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How to get the part that I circled??? Please explain.
we know,
Now,
Senndu = enxsen - ((San. In lon) du
0
0
=
x ln x -
=
[using product Rule]
- Sandn
nhan
[en (x+1) + In (y+2)] dudy
I en (x+1) andy +
0
0
0
1
In (3+2) Ludy
=
0
+
=
[[(2012-2)+1]dy
= (2402-1) d
+
+
j'en (+3) day
[(1+2) en (4+2) = (4+2)]'
1
= (2 n2-1) ["] + [(3 en3-3)-(21m2-2))
=
(2ln2-1) (1-0) + 3 en 3-3-24m2+2
= 2ln 2-1 + 3 en 3-3-2 en 2 + 2
= 3 ln 3-2
=
in 3-2
= ln 27-2
= −2+ en 27
Transcribed Image Text:we know, Now, Senndu = enxsen - ((San. In lon) du 0 0 = x ln x - = [using product Rule] - Sandn nhan [en (x+1) + In (y+2)] dudy I en (x+1) andy + 0 0 0 1 In (3+2) Ludy = 0 + = [[(2012-2)+1]dy = (2402-1) d + + j'en (+3) day [(1+2) en (4+2) = (4+2)]' 1 = (2 n2-1) ["] + [(3 en3-3)-(21m2-2)) = (2ln2-1) (1-0) + 3 en 3-3-24m2+2 = 2ln 2-1 + 3 en 3-3-2 en 2 + 2 = 3 ln 3-2 = in 3-2 = ln 27-2 = −2+ en 27
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