fx₁y (x, y) = (a) To find 1 = → 1 = .. √ftcore cé {ere the ce e -J 1 = √ (c=³² [e¹ J² Jáx 2 = & [ - ce²² + = ² e ²² 1 = ¢/ Constant c dy dx C=2 and a 1 (cex- ce²²₂) dx O [ - 6 +0 - (- ₁ + ² ) ] Therefore, fx(x)= OLY≤x≤00 we write x fx (x) = √²³²ce²²³²e-³ dy = free ed free ³dy 2 ] 0² = ₂0²Fe²¹] ² e-y :: |fx(x) = ² 2 €²²-20-22 Similarly, y fy(y) = √26²6²³dx Į = 26²³-26²4 ปี 20 20 Alsewhere. 2 e $100 fy (4) = $.20²2 ₂0²3 e ง for xt [0,00] For YE[o,. -24 *-2 0-²*, 0² ~ ²00 ( a elsewhere 021200 elsewher
fx₁y (x, y) = (a) To find 1 = → 1 = .. √ftcore cé {ere the ce e -J 1 = √ (c=³² [e¹ J² Jáx 2 = & [ - ce²² + = ² e ²² 1 = ¢/ Constant c dy dx C=2 and a 1 (cex- ce²²₂) dx O [ - 6 +0 - (- ₁ + ² ) ] Therefore, fx(x)= OLY≤x≤00 we write x fx (x) = √²³²ce²²³²e-³ dy = free ed free ³dy 2 ] 0² = ₂0²Fe²¹] ² e-y :: |fx(x) = ² 2 €²²-20-22 Similarly, y fy(y) = √26²6²³dx Į = 26²³-26²4 ปี 20 20 Alsewhere. 2 e $100 fy (4) = $.20²2 ₂0²3 e ง for xt [0,00] For YE[o,. -24 *-2 0-²*, 0² ~ ²00 ( a elsewhere 021200 elsewher
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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Question
100%
TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT PLEASE
![fx,y (x, y) =
(a) To find
∞0x
1 =
[Sce
- 1 =
..
=
and
the
се-хе-у
ce-xe-J dy dx
c=2
сечему
e
{
fecé² [erja Jáx
dx
a
-22
- [ - ce² + — ē ²² ]
√ (ce ² - ce²²²) dx
10
=
[ - 0 +0 - (- ₁ + ≤ ) ]
S2
fx (x)
посухую
Constant c, we write
(6)
fx (x) = √²ce³e³ ds = [ ²2 e ² dy
J
x
26²² [²e²1²
Therefore,
fxlx)=
Similarly,
y
fy (y) = √2€²e²³ da
fy (y)
x) = 2 ē²²-20-22
elsewhere.
zey-2e²1
26
O
= {26³ 26²3
For
xt [0,00]
-X
2 é* - 2 €²²*, 0²*²00
-29
e
e
For YE [0,0]
elsewhere
027200
elsewhere](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4cdec65-7837-40d3-b239-1cf845d9f9e2%2F51fc09dc-1686-4014-bf64-37b939999e32%2F5jhz7l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:fx,y (x, y) =
(a) To find
∞0x
1 =
[Sce
- 1 =
..
=
and
the
се-хе-у
ce-xe-J dy dx
c=2
сечему
e
{
fecé² [erja Jáx
dx
a
-22
- [ - ce² + — ē ²² ]
√ (ce ² - ce²²²) dx
10
=
[ - 0 +0 - (- ₁ + ≤ ) ]
S2
fx (x)
посухую
Constant c, we write
(6)
fx (x) = √²ce³e³ ds = [ ²2 e ² dy
J
x
26²² [²e²1²
Therefore,
fxlx)=
Similarly,
y
fy (y) = √2€²e²³ da
fy (y)
x) = 2 ē²²-20-22
elsewhere.
zey-2e²1
26
O
= {26³ 26²3
For
xt [0,00]
-X
2 é* - 2 €²²*, 0²*²00
-29
e
e
For YE [0,0]
elsewhere
027200
elsewhere
![© E[x] = √n fx(x) dx √x (26²=-26²33 dx
C
20
= [(n=²* - 210²9 ) an
27524
[-2(x+1)=²*² + (2x+1)
€ [x] = 3/1/2
Similarly E(Y) =
Vor (x) = √(x) = {(x²) - [E(X)]²
V(X)
.: V(X) =
[(x²) = (x² (2 ex-zeza jdx
х2
2e
61
(22+1) €29
e²a Jo
7/22
-27
2
= [resses gren]]
[x²
е
тае
42
बात
√y (20²2 ₂0²³] dy
ง
¾/1/2
SIA
+
е
7/21 - ( ²/2 ) ² = 7
클
2
4e-2
oft
4
Smilarly,
vry) = E(7²) - (EN))²
E (y² = √² y² (2€²³²_₂e²²4) do
.: € (Y²) = 7/₂2
...
N(Y) =
17
Imit
3
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4cdec65-7837-40d3-b239-1cf845d9f9e2%2F51fc09dc-1686-4014-bf64-37b939999e32%2Fc39cia_processed.jpeg&w=3840&q=75)
Transcribed Image Text:© E[x] = √n fx(x) dx √x (26²=-26²33 dx
C
20
= [(n=²* - 210²9 ) an
27524
[-2(x+1)=²*² + (2x+1)
€ [x] = 3/1/2
Similarly E(Y) =
Vor (x) = √(x) = {(x²) - [E(X)]²
V(X)
.: V(X) =
[(x²) = (x² (2 ex-zeza jdx
х2
2e
61
(22+1) €29
e²a Jo
7/22
-27
2
= [resses gren]]
[x²
е
тае
42
बात
√y (20²2 ₂0²³] dy
ง
¾/1/2
SIA
+
е
7/21 - ( ²/2 ) ² = 7
클
2
4e-2
oft
4
Smilarly,
vry) = E(7²) - (EN))²
E (y² = √² y² (2€²³²_₂e²²4) do
.: € (Y²) = 7/₂2
...
N(Y) =
17
Imit
3
4
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