How do you solve #x'(t)+x3 =0#?

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Answer 1 · 2015-08-13T02:24:39

#x = ke^(-3t)# (for constant #k#)

#x'(t)+x3=0# is equivalent to:

#dx/dt +3x=0# and, in turn:

#dx/x=-3dt#

#int dx/x=-int3dt#

#lnx = -3t +C#

#x = e^(-3t+C)#

#x = e^(-3t)e^C#

#x = ke^(-3t)# (for constant #k#)

(Check the answer by differentiating.)