Solve the initial value problem. dz dt 1² +3tx=t5 In(t) +4, 2(1)=0 97 z (t) = ½ tª la(t) - ¹+² - 07/0 4913 z(t) = ½ tª ln(t) — z(t) = ²ln(t) - -t4. + 1 ²+2+³ - 97 —— 49 13 z(t) = tª ln(t) — 2/t¹ + ²/3 - 72500 7t³
Solve the initial value problem. dz dt 1² +3tx=t5 In(t) +4, 2(1)=0 97 z (t) = ½ tª la(t) - ¹+² - 07/0 4913 z(t) = ½ tª ln(t) — z(t) = ²ln(t) - -t4. + 1 ²+2+³ - 97 —— 49 13 z(t) = tª ln(t) — 2/t¹ + ²/3 - 72500 7t³
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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Transcribed Image Text:Solve the initial value problem.
1² +3tx=t5 In(t) +4, 2(1)=0
dz
dt
97
z (t) = ½ tª la(t) - ¹+² - 07/0
4913
(t)- m(t)-¹+
=
EM
In(
z(t) = ²ln(t) - ²+2+³.
1
49
97
49
——
13
z(t) = tª ln(t) — 2/t¹ + ²/3 - 72500
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