How to make a smart guess for this ODE
I am dealing with a strange problem currently,
we have a differential equation $$y(x)^2 = \pm \sqrt{-A \cos(x) - B \cos^2(x)+y'(x)-C},$$ where $C, A$ and $B $ are parameters. (The case that either $A$ or $B$ is zero is not interesting to me as these solutions are well-studied in the literature). So the idea of $C$ is to be some kind of adjustment variable, which means that I don't care about its value and it's only purpose is to make it easier to find solutions.
Now my question is: Is there a smart way to do this so that I get a wide range of solutions and is there even a solution for all different kinds of $A $ and $B$, because I am particularly interested in having a free choice in these two paramters? But even if you are able to suggest a better guess to me that gives me more solutions, this would be tremendously helpful!
Somehow I don't have much experience with differential equation and making this guess the way I did it, is probably not very sophisticated so I would be highly interested in hearing about better methods to do this. (Probably, if anybody here would tell me that he is able to solve this would remedy all my troubles, but I doubt that it is possible ;-))
Solution 1:
Let me note that this is a Riccati equation. Therefore, given a solution for some initial conditions we can solve the equation (with the same parameters) for arbitrary initial conditions. For example, consider $$y^2-y'-\beta^2=-\beta\cos x-\beta^2\cos^2 x \tag{1}$$ which is solved by $y(x)=\beta\sin x$. Substituting $\displaystyle y(x)=\beta\sin x+\frac{1}{v(x)}$ into (1) transforms the equation into a linear 1st order ODE $$v'+2\beta\sin x \,v+1=0,$$ with the general solution $$v(x)=\int_x^ae^{2\beta(\cos x-\cos t)}dt,$$ where $a$ is an arbitrary integration constant.
Riccati property also implies that the change of variables $\displaystyle y=-\frac{u'}{u}$ in the equation $y^2-y'+f=0$ transforms it into a linear 2nd order ODE $$u''+fu=0, \tag{2}$$ where in our case $f(x)=A\cos x+B\cos^2 x+C$. Obviously, for $A=0$ or $B=0$ (studied cases) this can be transformed into the Mathieu's equation. This provides some insight: let us make the change of variables $t=\cos\frac{x}{2}$. It will transform (2) into $$(1-t^2)u_{tt}-tu_t+4\underbrace{\left[A(2t^2-1)+B(2t^2-1)^2+C\right]}_{Q(t^2)}u=0\tag{3}$$ This ODE has 3 singularities on $\mathbb{P}^1$: two regular singular points at $\pm1$ and one irregular at $\infty$. The solutions listed in the question have the following form in terms of $u(t)$: \begin{align} u_1(t)=&e^{2\beta t^2},\\ u_2(t)=&te^{2\beta t^2},\\ u_3(t)=&\sqrt{1-t^2}e^{2\beta t^2},\\ u_4(t)=&t\sqrt{1-t^2}e^{2\beta t^2}. \end{align} Here $u_{2,3}$ correspond to $\pm$ sign in the second family.
Apparently one can construct an infinite number of solutions of each of the four types. Let me first give one more example: $$u_5(t)=\left(t^2+\sigma\right)e^{2\beta t^2},$$ which solves (3) for the following parameter values: \begin{align} A=3\beta,\qquad B=\beta^2,\qquad C=1-2\beta-\beta^2-4\beta\sigma,\qquad \beta=\frac{1+2\sigma}{8\sigma(1+\sigma)}. \end{align} In terms of the initial function $y(x)$, this one-parameter family of solutions is written as $$y_5(x)=\frac{\sin x}{8}\left[\frac{1}{\sigma}+\frac{1}{1+\sigma}+ \frac{8}{1+2\sigma+\cos x}\right].\tag{4}$$
Next let us generalize this result and show that for any positive integer $n$ there is a one-parameter family of solutions of the form $$u(t)=P_{n}(t^2)\,e^{2\beta t^2},\tag{5}$$ where $P_{n}(z)$ is a $n$-th degree polynomial in $z$ with coefficients depending on $\beta$. Indeed, write $$P_n(z)=z^n+\sum_{k=0}^{n-1}a_kz^k,\tag{6}$$ then (3) gives the following equation for $P_n(z)$: \begin{align}z(1-z)P_n''(z)+\left(\frac12+(4\beta-1)z-4\beta z^2\right)P_n'(z)+\\ +\biggl(Q(z)-\beta\left[4\beta z^2+2(1-2\beta)z-1\right]\biggr)P_n(z)=0. \tag{7} \end{align} Recall that $Q(z)$ is a 2nd degree polynomial, hence the left side of (7) is a $(n+2)$th degree polynomial. Thus we have to satisfy $n+3$ identities with $n+4$ parameters $A,B,C,\beta,a_0,\ldots,a_{n-1}$, and the statement follows.
In particular, substituting the expansion (6) into (7) and setting the coefficient of $z^{n+2}$ to $0$, we find the constraint $B=\beta^2$. Similarly equating the coefficient of $z^{n+1}$ to $0$, we get $A=(2n+1)\beta$. Hence the one-parameter family (5) is characterized by $$A=(2n+1)\beta,\qquad B=\beta^2.$$
Similar reasoning allows to establish the following
Theorem. For any $n\in\mathbb{Z}_{\ge 0}$ there are four one-parameter families of solutions of (3) of the form
\begin{align} u^{\mathrm{(I)}}(t)\;\;=&P_n^{\mathrm{(I)}}(t^2)\,e^{2\beta t^2},\\ u^{\mathrm{(II)}}(t)\;=&tP_n^{\mathrm{(II)}}(t^2)\,e^{2\beta t^2},\\ u^{\mathrm{(III)}}(t)=&\sqrt{1-t^2}P_n^{\mathrm{(III)}}(t^2)\,e^{2\beta t^2},\\ u^{\mathrm{(IV)}}(t)\,=&t\sqrt{1-t^2}P_n^{\mathrm{(IV)}}(t^2)\,e^{2\beta t^2}. \end{align} Here $P_n^{\mathrm{(k)}}(z)$ with $k=\mathrm{I,II,III,IV}$ are $n$th degree polynomials of the form (6) whose coefficients can be determined from (3). The appropriate parameter values are \begin{align} (A,B)^{\mathrm{(I)}}\;\;=&\Bigl((2n+1)\beta,\beta^2\Bigr),\\ (A,B)^{\mathrm{(II)}}\;=&\Bigl((2n+2)\beta,\beta^2\Bigr),\\ (A,B)^{\mathrm{(III)}}=&\Bigl((2n+2)\beta,\beta^2\Bigr),\\ (A,B)^{\mathrm{(IV)}}\,=&\Bigl((2n+3)\beta,\beta^2\Bigr). \end{align} The solutions mentioned in the question correspond to setting $n=0$ in the above.