Question

Question:

First order differential equations:
Intro to differential equations:
Slope Fields:

Answer 1

Solve this:

Answer 2

In case you're wondering how to interpret the slope field:
\[\frac{\mathrm{d}y}{\mathrm{d}x}=y^2-4y+3=(y-3)(y-1)\]
For \(y>3\), you have \(\dfrac{\mathrm{d}y}{\mathrm{d}x}>0\), so the solution curve is increasing.
For \(1<y<3\), you have \(\dfrac{\mathrm{d}y}{\mathrm{d}x}<0\), so the solution is decreasing.
For \(y<1\), you have \(\dfrac{\mathrm{d}y}{\mathrm{d}x}>0\), so the solution is increasing.

Now, given the initial condition \((0,2)\), you know the particular solution curve is contained in the middle band of the slope field, where \(1<y<3\). The solution here is decreasing, so as \(x\to\infty\), you have the solution approaching the lower limit of this interval, i.e. \(\lim\limits_{x\to\infty}f(x)=1\), but note that this doesn't meant \(f(x)\) ever takes on the value of \(1\).

Because any solution in the interval \(1<y<3\) is decreasing, that must mean the starting point \(f(x)=2\) is the maximum value attained by the solution. Therefore the range of \(y=f(x)\) given that \(f(0)=2\) must be \((1,2]\).

See if you can apply the same reasoning to the second question.

Answer 3

So the least is -1 and 1

so [-1, 1)