How do I solve the equation #dy/dt = 2y - 10#?

Question

27225 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-01-31T20:07:21

You can use a technique known as Separation of Variables.
Take all the #y# to one side and the #t# on the other...
You get:

#dy/(2y-10)=dt#

Now you can integrate both sides with respect to the correspondent variables:

#int1/(2y-10)dy=intdt#
#int1/(2(y-5))dy=intdt#

And finally
#1/2ln(y-5)=t+c#

Now you can express #y# as:
#ln(y-5)=2t+c#
#y-5=c_1e^(2t)# where #c_1=e^c#
#y=c_1e^(2t)+5#

You can substitute back to check your result (calculating #dy/dt#) remembering that now it is: #y=c_1e^(2t)+5#