What is a solution to the differential equation #dP-6Pdt=0# with P=5 when t=0?

Question

1346 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-27T17:27:41

#P = 5 e^{6t}#

#dP-6Pdt=0#

so separate it!!

#dP = 6Pdt#

#(dP)/P = 6dt#

#int (dP)/P = 6 int dt#

#ln P = 6t + alpha#

#P = e^{6t + alpha} = e^{6t} e^{ alpha} = beta e^{6t}#

NB: we have just replaced the constant #e^{ alpha}# by #beta#, a simpler form of constant, so #beta = e^{ alpha}#

with final form #P = beta e^{6t}#, we can plug in the initial value so

#5 = beta e^{6 times0} = beta#

so we have

#P = 5 e^{6t}#