DP-6P dt=0 P=5 when t=0 ?

Question

separation of variables.

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Answer 1 · 2018-04-25T17:45:46

#P(t) = 5 e^(6 t) #

#dP-6P dt=0 implies dP = 6 P dt#

# implies (dP)/P = 6 \ dt#

Integrate:

# ln P = 6 t + C#

# implies P = alpha e^(6 t) #

#P(0) = 5 implies 5 = alpha#

So:

# P(t) = 5 e^(6 t) #