# Solving a Riccati ODE Twice

$$y' - \frac{1}{t}y = y^2 - \frac{3}{t^2}, y_p = \frac{1}{t}$$

Method 1

To obtain a Bernoulli ODE, we plug $$y = \frac{1}{t} + u$$ into the Riccati ODE, yielding $$u' - \frac{3}{t}u = u^2$$. To obtain a linear ODE, we plug $$u = \frac{1}{v}$$ into the Bernoulli ODE, yielding $$v' + \frac{3}{t}v = -1$$.

Thus, $$v = \frac{C - t^4}{4t^3}$$, which entails that $$u = \frac{4t^3}{C - t^4}$$, which entails that $$y = \frac{3t^4 + C}{Ct - t^5}$$.

Method 2

To obtain a second-order linear ODE, we plug $$y = -\frac{w'}{w}$$ into the Riccati ODE, yielding $$w'' - \frac{1}{t}w' - \frac{3}{t^2}w = 0$$. Since our particular solution $$\frac{1}{t}$$ also solves this homogeneous equation, we can use reduction of order to find that $$w = \frac{At^4 +B}{t}$$. It is interesting to note that the first-order linear ODE found through reduction of order is $$x' - \frac{3}{t}x = 0$$, which is similar but not identical to the first-order linear ODE in $$v$$ found through Method 1. Plugging $$w = \frac{At^4 +B}{t}$$ into $$y = -\frac{w'}{w}$$ yields the solution $$y = \frac{B - 3At^4}{At^5 + Bt}$$.

These two solutions match iff $$A = -1$$, but we wouldn't have known that if we'd only used Method 2 to solve the Riccati ODE. I tried plugging $$y = \frac{B - 3At^4}{At^5 + Bt}$$ into the Riccati ODE to fish for the fact that $$A = -1$$, but instead I got $$B = 0$$, which isn't even generally true. Why does $$B = 0$$ come out of this process, and if we solved the ODE with Method 2, how could we have discovered that $$A = -1$$ without doing the problem a second time using Method 1.

• You could also just cancel and set $C=-B/A$. So if you got $B=0$ you made some error while inserting. – LutzL Oct 9 at 17:17

Both of your approaches are correctly carried out. You get more generally the first form from the second for $$A\ne 0$$ by canceling $$-A$$ in the quotient and setting $$C=-\frac BA$$.
To the second approach one could remark that the equation you get is of the Euler-Cauchy type so that you do not even need the provided solution, you get the basis solutions $$t^m$$ from the characteristic polynomial $$0=m(m-1)-m-3=(m+1)(m-3)$$, so that $$w=At^3+Bt^{-1}$$.
• Is there any way I could have known to divide by $-A$ instead of, say, $B$ or $-B$ (and then set $C = \frac{A}{B}$ or $C = -\frac{B}{A}$ without going through Method 1? – user10478 Oct 11 at 3:11
• No. These other variants, and even with the two constants, are all valid general solution formulas. Note that if you reduce to one constant, you will lose a valid solution, $t^{-1}$ or $-3t^{-1}$. – LutzL Oct 11 at 6:01