Find solutions of the Riccati equation y′ + y + y2 = 2

$y' + y + y^2 = 2.$

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

If $-2 \leq b < 1$ , find a solution on the interval $(-\infty, +\infty)$ with $y= b$ when $x = 0$ .
If $b \geq 1$ or $b<-2$ , find a solution on the interval $(-\infty, +\infty)$ with $y = b$ when $x = 0$ .

First, we find the two constant solutions. If $u$ is constant then $u' = 0$ so

$u' + u + u^2 = 2 \quad \implies \quad u^2+u-2 = 0 \quad \implies \quad u= 1 \text{ or } -2.$

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

$y = u + \frac{1}{v}$

where $v$ is a solution of

$v' - v (P(x) + 2Q(x)u) = Q(x).$

In the present case we have $P(x) = Q(x) = 1$ , so $v$ is the solution of either

$v' -3v = 1 \qquad \text{or} \qquad v' + 3v = 1,$

for the cases $u = 1$ and $u=-2$ , 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 $P(x) = 3$ , $Q(x) = 1$ and let $a = 0$ , and $c = v(0)$ . Then we have

$\begin{align*} v &= ce^{3x} + e^{3x} \int_0^x e^{-3t} \, dt \\ &= ce^{3x} + e^{3x} \frac{1}{3}( -e^{-3x} + 1) \\ &= ce^{3x} - \frac{1}{3} + \frac{1}{3}e^{3x} \\ &= \frac{e^{3x}(3c+1) -1}{3}. \end{align*}$

This gives us the first solution

$y_1 = u + \frac{1}{v} = \frac{e^{3x}(3c+1) + 2}{e^{3x}(3c+1) - 1}.$

Evaluating the second differential equation, this time with $P(x) = -3$ , $Q(x) = 1$ , $a = 0$ and $c = v(0)$ we have,

$\begin{align*} v &= ce^{-3x} + e^{-3x} \int_0^x e^{3t} \, dt \\ &= ce^{-3x} + \frac{1}{3} - \frac{1}{3}e^{-3x} \\ &= \frac{3c-1+e^{3x}}{3e^{3x}}. \end{align*}$

Therefore, the second solution is

$y_2 = u + \frac{1}{v} = \frac{-2(3c-1) + e^{3x}}{(3c-1)+e^{3x}}.$

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

We want $-2 \leq b < 1$ , so we choose $y_2$ . Then $y_2(0) = b \implies c = \frac{1}{b+2}$ . (This follows since $y(0) = -2 + \frac{1}{v(0)}$ .) Therefore, $(3c-1) = -\frac{b-1}{b+2}$ , so,
$\begin{align*} y_2 &= \frac{2 \left( \frac{b-1}{b+2} \right) + e^{3x}}{-\left( \frac{b-1}{b+2} \right) +e^{3x}} \\[9pt] &= \frac{ \left( \frac{b+2}{b-1} e^{3x} + 2}{\left( \frac{b+2}{b-1} \right) e^{3x} -1} \\[9pt] &= \frac{Ce^{3x} +2}{Ce^{3x} -1} \end{align*}$

where $C = \frac{b-1}{b+2}$ .
In this case we want $b \geq 1$ or $b <-2$ , so we choose $y_1$ . Since $y_1(0) = b \implies c =\frac{1}{b-1}$ we have $(3c+1) = \frac{b+2}{b-1}$ . Therefore,
$\begin{align*} y_1 &= \frac{e^{3x} \left( \frac{b+2}{b-1} \right) + 2}{e^{3x} \left( \frac{b+2}{b-1} \right) -1} \\[9pt] &= \frac{e^{3x} + 2 \left(\frac{b-1}{b+2}\right)}{e^{3x} - \left( \frac{b-1}{b+2} \right)} \\[9pt] &= \frac{e^{3x} + 2C}{e^{3x} - C} \end{align*}$

where $C = \frac{b-1}{b+2}$ .

3 comments

October 22, 2019 at 9:45 am

hteica says:

(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.

- June 23, 2022 at 8:35 am
  
  S says:
  
  I agree.
  
June 17, 2016 at 8:59 pm

tom says:

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.

Stumbling Robot

Find solutions of the Riccati equation y′ + y + y² = 2

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