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question about Paul's Math website on linear diff. eq.

I'm looking at Paul's math page (Paul Dawkins of Lamar U) and on this page:
http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
about 1/2 down in example 1 he says:
Question:
dv/dt = 9.8 - 0.196v
Solution:
dv/dt + 0.196v = 9.8
u(t) = e^(integral(0.196dt)) = e^0.196t
OK, I get it so far - but how do they get from this:
e^(0.196t) dv/dt + 0.196e^(0.196t)v = 9.8e^(0.196t)
to this:
(e^(0.196t)v)' = 9.8e^(0.196t)

0
zliminator
Asked:
zliminator
1 Solution
 
Peter KwanCommented:
Just follow the logic from equation 4 to equation 5, by substituting "u(t)" with e^0.196t and "y" with "v".

This is because d(e^ct)/dt = c * e^ct => d(e^0.196t)/dt = 0.196 * (e^0.196t)
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