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

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.

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  • $\begingroup$ You could also just cancel and set $C=-B/A$. So if you got $B=0$ you made some error while inserting. $\endgroup$ – LutzL Oct 9 at 17:17
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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 $w=At^3+Bt^{-1}$.

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  • $\begingroup$ 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? $\endgroup$ – user10478 Oct 11 at 3:11
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    $\begingroup$ 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 $-3t^{-1}$. $\endgroup$ – LutzL Oct 11 at 6:01

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