given function then derivative Now dz dr dz d४ sing rcass-38³ He Have to find it's dz dr = dz 5 = 5 Z = dz + 5 = 5 (Cre スニ d using + (*). sin r (reasr-3r³ sinr r cosr-38³ sin'r ·Cos838³3 siner rCast-3r3 C 5 (sint) (rear-383)6 sinr rcasr-3√3 4 5 d dr vul-uv! V2 first sinr rcosr-3r³ (rear-3r³) casr-sinx (cur-rsinr-gr²) (rcer-3√³)² E [res²³r-³r³ car - sincer + rsin³²r + 38's int (rear-3r³¹) 2 grsinr sincer3rcont 28-35²3r+r) []+= d √²+24 du de = √Frau [ 2007) - 2 (0-3) ] + (42) * (u+1) ૨(૫) (4+1) du (4+1)³ √4²+24 لال dw du = √4²+24 d = (u-1) (u²³ +5u²+74-1) (u+1)³(√²+24)
given function then derivative Now dz dr dz d४ sing rcass-38³ He Have to find it's dz dr = dz 5 = 5 Z = dz + 5 = 5 (Cre スニ d using + (*). sin r (reasr-3r³ sinr r cosr-38³ sin'r ·Cos838³3 siner rCast-3r3 C 5 (sint) (rear-383)6 sinr rcasr-3√3 4 5 d dr vul-uv! V2 first sinr rcosr-3r³ (rear-3r³) casr-sinx (cur-rsinr-gr²) (rcer-3√³)² E [res²³r-³r³ car - sincer + rsin³²r + 38's int (rear-3r³¹) 2 grsinr sincer3rcont 28-35²3r+r) []+= d √²+24 du de = √Frau [ 2007) - 2 (0-3) ] + (42) * (u+1) ૨(૫) (4+1) du (4+1)³ √4²+24 لال dw du = √4²+24 d = (u-1) (u²³ +5u²+74-1) (u+1)³(√²+24)
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
![given function
Now
dz
dr
then file have to find it's
derivative
dz
dr
dz
dr
dz
dr
idr
=
dz
= 5
Z
5
Z=
seny
rcass-323
d
dr...
sinar
Cass-383
-373
using d (u)
sin r
(rcasr-3r³
rcos - 38³
sing
sin'r
5 (sint)
(reast-3√³)²
r cosr-3√3
[³]
Sinr
=
rcas - 3r
5
(rcas
d
dr
val-4v1
V2
-38
first
sinr
rcosr-3r³
casr-sinr (ar-rsinr_9r²)
(rear-38³)
·ces²³r-3r²³² cer - sincer + rsin³²r +9y²sin
(rear-3√³)2
gr*sinx-sinr cest-35%(3r+r)
dw
du
dw =
du
dw
du
= [(+1)] + [4=;)² de virtual
√4²+24
du
= √²+24 d
√ur²rau [2 (14-71) - 2 (4-13²] + [4=)²2 (4+1)
4²+24
(u+1)³
4²+24
= (u-1) (u²³+5u²+74-1)
(u+1)²³ (√²+24)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4be50db6-19e4-41fe-8b0c-6d619944736f%2F9443deae-8a41-4ef8-8a64-7e24b43a220b%2F50ekd0i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:given function
Now
dz
dr
then file have to find it's
derivative
dz
dr
dz
dr
dz
dr
idr
=
dz
= 5
Z
5
Z=
seny
rcass-323
d
dr...
sinar
Cass-383
-373
using d (u)
sin r
(rcasr-3r³
rcos - 38³
sing
sin'r
5 (sint)
(reast-3√³)²
r cosr-3√3
[³]
Sinr
=
rcas - 3r
5
(rcas
d
dr
val-4v1
V2
-38
first
sinr
rcosr-3r³
casr-sinr (ar-rsinr_9r²)
(rear-38³)
·ces²³r-3r²³² cer - sincer + rsin³²r +9y²sin
(rear-3√³)2
gr*sinx-sinr cest-35%(3r+r)
dw
du
dw =
du
dw
du
= [(+1)] + [4=;)² de virtual
√4²+24
du
= √²+24 d
√ur²rau [2 (14-71) - 2 (4-13²] + [4=)²2 (4+1)
4²+24
(u+1)³
4²+24
= (u-1) (u²³+5u²+74-1)
(u+1)²³ (√²+24)
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