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January 19, 2017
Posted by **dyingincalc** on Monday, February 28, 2011 at 11:21pm.

dy/dt=k(y-a)

i) y=1-ae^(kt)

ii) y=a+Ce^(kt)

- Calculus AB -
**MathMate**, Tuesday, March 1, 2011 at 10:35amdy/dt=k(y-a) ... (1)

i) y=1-ae^(kt) (particular solution)

differentiate y w.r.t. t:

d(1-ae^(kt))/dt

= k(e^(kt)+a)-ka

= ky-ka

= k(y-a)

= right-hand side, QED

ii) y=a+Ce^(kt)

differentiate with respect to t:

dy/dt

=Cke^(kt)

=k(a+Ce^(kt))-ka

=ky-ka

=k(y-a)

=right-hand side QED