Part 2 of 5 Integrate both sides and use the power rule on the right side. 5+1 y = 48. + C 5+1 6 8 + C Part 3 of 5 Thus, the general solution of Y = 48t5 is y = 810 + C |C + 8t6 Part 4 of 5 d 8t6 + C dt 48t5. To check for the result, we have to prove that d fsrs + c = 8( = 6( %3D dt t II

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.4: Values Of The Trigonometric Functions
Problem 23E
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Part 2 of 5
Integrate both sides and use the power rule on the right side.
5+1
y = 48.
+ C
5+1
6
8
+ C
Part 3 of 5
Thus, the general solution of Y = 48t5 is y = 810 + C
|C + 8t6
Part 4 of 5
d
8t6 + C
dt
48t5.
To check for the result, we have to prove that
d fsrs + c = 8(
= 6(
%3D
dt
t
II
Transcribed Image Text:Part 2 of 5 Integrate both sides and use the power rule on the right side. 5+1 y = 48. + C 5+1 6 8 + C Part 3 of 5 Thus, the general solution of Y = 48t5 is y = 810 + C |C + 8t6 Part 4 of 5 d 8t6 + C dt 48t5. To check for the result, we have to prove that d fsrs + c = 8( = 6( %3D dt t II
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