How do you solve the differential equation given #f'(s)=6s-8s^3#, f(2)=3?

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Answer 1 · 2016-11-11T23:10:47

# f(s) = 3s^2 - 2s^4 + 23 #

Let # y = f(s) => dy/(ds)=6s-8s^3 #

This is a first order separable DE, so we can separate the variables as follows:

# intdy = int 6s-8s^3 ds #

Integrating gives:

# y = 3s^2 - 2s^4 + C #

We know # f(2)=3 => 3=(3)(4)-(2)(16) + C#
# :. 3=12-32 + C =>C=23#

Hence, the solution is:

# y = 3s^2 - 2s^4 + 23 #