Question

Sketch the solution to the initial value problem and determine its maximum value. see post below for equation

Answer 1

\[dy/dt=2y-2yt \]

Answer 2

where y(0)=3

Answer 3

First, you have to find y(t)

Answer 4

\[dy/dt = 2y-2yt\]
Now, you separate the y's and t's
\[dy/dt=2y(1-t)\]
Next, divide both sides by 2y and multiply both sides by dt
\[dy/2y=(1-t)dt\]
Now we need to integrate both sides.
\[\int\limits_{}^{}dy/2y=\int\limits_{}^{}(1-t)dt\]
\[\ln(y)/2=t-t^2/2 +c\]
\[\ln(y)=2t-t^2+2c\]
e^(lny)=e^(2t-t^2 +2c)
[e^lny = y]
So,
y= e^(2t-t^2+2c)
----- By using laws of exponent---
y = (e^2t/e^t^2) * e^2c

As y(0)=3 or f(0)=3

3 = (e^0/e^0)*e^2c
3=e^2c
c=3 -----------------------> as e^2c is just a constant.
Thus, y = e^(2t-t^2+6)

Answer 5

In order to find the maximum first you find the critical points[points where the first derivative is 0.]

Answer 6

Hint :
-find the first derivative
-solve for t as y(prime) equals 0. And that will be your critical points.
-plug in the critical points in the y function, then you will get the maximum.(the maximum value obtained is going to be the max.)