Consider the Riccati equation

This equation has two constant solutions. Starting with these and the previous exercise (linked above) find further solutions in the cases:

- If , find a solution on the interval with when .
- If or , find a solution on the interval with when .

First, we find the two constant solutions. If is constant then so

From Exercise 19 (linked above) we know we can obtain additional solutions to the Riccati equation by

where is a solution of

In the present case we have , so is the solution of either

for the cases and , respectively. Each of these is an first-order linear differential equation which we can solve using Theorem 8.3 (page 310 of Apostol). For the first one we have , and let , and . Then we have

This gives us the first solution

Evaluating the second differential equation, this time with , , and we have,

Therefore, the second solution is

Finally, for the specific cases in (a) and (b).

- We want , so we choose . Then . (This follows since .) Therefore, , so,
where .

- In this case we want or , so we choose . Since we have . Therefore,
where .

(a) I don’t think it is correct to choose y1, instead we should choose y1. Since b might be minus 2 and c = 1/0. Similarly in (b), we should choose y2 instead.

I agree.

Something doesn’t look right to me. The first line for y1 (involving b) is the same as the second for y2, meaning they must be identical solutions? Your work looks correct: is this a chance coincidence? Confusing because the constants weren’t picked arbitrarily.

Also, since y= u +1/c in both cases, u constant, isn’t the choice of y1 or y2 arbitrary since c can be anything? Now if b had to fulfill both conditions for ONE c, then the choice of solutions would be relevant.