Question

Sketch the solution to the initial value problem and determine its maximum value. dy/dt=2y-2yt y(0)=3

Answer 1

Arrange the given equation.
(dy/2y) = 1-t
now integrate both side .
\[\int\limits_{y0}^{y}dy/2y =\int\limits_{0}^{t}1-t\]
(1/2)* log(y/y0) = t-(t^2/2)
y0 is nothing but value of y at initial point of time at t=0
so y0=3 ; (from initial boundary condition)
log(y/3) = 2*{t-(t^2/2)}
y = 3*e^2*{t-(t^2/2)} (1)

y will max. when 2*{t-(t^2/2)} will be max. because base 'e' is >1.
so next task is to maximize t-(t^2/2)
t-(t^2/2) = t*(1-(t/2)) max at t=1, that is 0.5
put in eqn.(1)
Ymax= 3*e^2