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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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TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT

27.
we
міх, у)
position vector
in R².
have
1 μ =
Tr
=
√x² + y²
तारे
to Show
7(4²) = 24
1
" =
is
Now, if f(x,y) be
у
the
x i + y j
that
when (x, y) + (0,0)
length of
for each point (x,y)
and
of(x,y) = (i3/2 + $ 24 ) f(x,y)
тых, у
id
fixe
х,у)
әх
бу
the
a function of R²,
of R², then
First
TM=
ㅋ
we
Fr
since, r(x,y)=√x² + y²
és Jr =
Tr=
Hence
show
1 r
μ
Now,
H
( iz + j) n = [ ² 2 + j 2 ) √x ² + y ²
i z
ау
dæ
- (22//W) + 12 (²)
i s
=/i
туя
dx
бу
=/ix
= [ix + √2² +4²
X_1
=
=(²
that
i x
1
√x² + y²
1 r
we
√x² + y²
روندان
the proof.
=
(x² + y j)
x 2x + jx 1
Ir
+ in the stre)
j y
+ y
show that
2 Jetty: x*y) ; (x₁y) + (09)
{ x=
; (x, y) = (0,0)
; (x, y) = (0,0)
• • r = √x²³² + y²
μ = x² + y j
xi
and (x, y) = (0,0)
Since,
→
介
r = √√x ²² + y²
=> μ²³ = (√x² + y²
(√x² + y²) ²
⇒ x² = x² + y²
☺☺.
موہو لا
رمندان
ů
= (₁²) = (iz + $ $ ) x² = ( i 2/2 + 1 + ) (x ² + y ² )
is
j
ter
d
j
dy
dy
⇒ O (re?).
=
روہواں
7[r²) = (i (2x) + j(ay
z x i + 2 Y j
obe ²) = 2(x²+ y j)
(1
11
i 2 ( x ³² + y ² ) + j } (x² + y ² )
2
dxc
бу
= 2 μ
Hence, the proof
poo
x² + y²
11
=
A
r = x² + y j }
xi
ہو
2(xity ý)
Transcribed Image Text:27. we міх, у) position vector in R². have 1 μ = Tr = √x² + y² तारे to Show 7(4²) = 24 1 " = is Now, if f(x,y) be у the x i + y j that when (x, y) + (0,0) length of for each point (x,y) and of(x,y) = (i3/2 + $ 24 ) f(x,y) тых, у id fixe х,у) әх бу the a function of R², of R², then First TM= ㅋ we Fr since, r(x,y)=√x² + y² és Jr = Tr= Hence show 1 r μ Now, H ( iz + j) n = [ ² 2 + j 2 ) √x ² + y ² i z ау dæ - (22//W) + 12 (²) i s =/i туя dx бу =/ix = [ix + √2² +4² X_1 = =(² that i x 1 √x² + y² 1 r we √x² + y² روندان the proof. = (x² + y j) x 2x + jx 1 Ir + in the stre) j y + y show that 2 Jetty: x*y) ; (x₁y) + (09) { x= ; (x, y) = (0,0) ; (x, y) = (0,0) • • r = √x²³² + y² μ = x² + y j xi and (x, y) = (0,0) Since, → 介 r = √√x ²² + y² => μ²³ = (√x² + y² (√x² + y²) ² ⇒ x² = x² + y² ☺☺. موہو لا رمندان ů = (₁²) = (iz + $ $ ) x² = ( i 2/2 + 1 + ) (x ² + y ² ) is j ter d j dy dy ⇒ O (re?). = روہواں 7[r²) = (i (2x) + j(ay z x i + 2 Y j obe ²) = 2(x²+ y j) (1 11 i 2 ( x ³² + y ² ) + j } (x² + y ² ) 2 dxc бу = 2 μ Hence, the proof poo x² + y² 11 = A r = x² + y j } xi ہو 2(xity ý)
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