20. Suppose that y(x) denotes a solution of the first-order IVP \f$y'=x^{2}+y^{2},y(1)=-1\f$  and that y(x) possess at least a second derivative at x=1. In some neighborhood of x=1 use the DE to determine whether y(x) is increasing or decreasing and whether the graph y(x) is concave up or concave down.

The first thing I like to do when I solve math problems is to write down all the data that was given in the question. We know that y(x) would be a solution and also we know the following.

\f[\inline y'=x^{2}+y^{2}\f]

\f[\inline y(1)=-1\f]

So when x is equal to 1, it is given that y is equal to -1. From there, it gives us a point. We can use that point to find out that whether y(x) is increasing or decreasing. If y’ comes out positive at the point of (1,-1), y(x) is increasing and if it is negative, y(x) is decreasing. We can now find this out first and then figure out whether the graph of y(x) is concave up or down.

Let’s plug in the point (1,-1) into the DE.

\f[\inline y'=(1)^{2}+(-1)^{2} \f]

\f[\inline y'=1+1\f]

\f[\inline y'=2\f]

As I mentioned before, since y’ is positive 2 at the point (1,-1), y(x) is increasing within a neighbor of the point (1,-1).

Now, let’s think about the second part. If I want to find out the concavity of y(x), first I have to find the second derivative of the given DE. From there, I can most likely find out the concavity by the values that we already have. From there, if the second derivative comes out positive, its concave up and if it’s negative, then it is concave down.

Now, let’s start by calculating the second derivative.

\f[\inline \frac{d}{dx}(y')=\frac{d}{dx}(x^2 +y^2)\f]

\f[\inline y''=2x+2yy'\f]

We can now plug in the point (1,-1) and y’ value.

\f[\inline y''=2(1)+2(-1)(2)\f]

\f[\inline y''=2+-4\f]

\f[\inline y''=-2\f]

It looks like the second derivative is negative 2 at the point (1,-1). Therefore y(x) is concave down in the neighbor of the point (1,-1).

I hope that my explanation makes sense and it is correct. If you see any mistake, please point it out so that I can learn and correct it.