In Exercises 23-34, (a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. d P d t = 0.024 P , where P = 32 when t = 0
In Exercises 23-34, (a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. d P d t = 0.024 P , where P = 32 when t = 0
Solution Summary: The author explains that the particular solution of the provided differential equation is P=32e0.024t.
In Exercises 23-34, (a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation.
d
P
d
t
=
0.024
P
,
where
P
=
32
when
t
=
0
With integration, one of the major concepts of calculus. Differentiation is the derivative or rate of change of a function with respect to the independent variable.
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