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 N d t = 3 N , where N = 3.5 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 N d t = 3 N , where N = 3.5 when t = 0
Solution Summary: The author explains that the particular solution of the provided differential equation is N=3.5e3t.
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
N
d
t
=
3
N
,
where
N
=
3.5
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.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (4th Edition)
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