now, I know you can do this ...
dy/y = 2 dt
ln y = 2t + ln c
y = c*e^(2t)
...
The particular solution of the differential equation dy/dt= 2y for which y(0) = 60 is
a) y= 60e^2t
b) y= 60e^0.5t
c) y= 59 + e^t
d) y= 30e^t
3 answers
I don't understand how to get the solution to this problem
I assume you get to my last step ok.
y = c*e^(2x)
y(0) = 60, s0
c*e^0 = 60
c = 60
y = 60e^(2x)
Looks like you need to do a bunch of DE problems to get the technique down. Lots of online example, and I'm sure your text has some worked out as well.
y = c*e^(2x)
y(0) = 60, s0
c*e^0 = 60
c = 60
y = 60e^(2x)
Looks like you need to do a bunch of DE problems to get the technique down. Lots of online example, and I'm sure your text has some worked out as well.
0 answers