Can a first order homogeneous ODE have singular solution(s) (free of a constant)

Question

Can $$y'=\frac{px+qy}{rx+sy}$$ have singular (essential) solution(s) free of a constant of integration? Here, $p,q,r,s$ are independent of $x$ and $y$.

Answer 1

If $y=f(x)$ is solution of an ODE then $f(x)+C$ or $Cf(x/C)$ may or may not be a solution of the ODE For example $y'+y=0$, it's solution is $e^{-x}$ but $y= e^{-x}+3$ does not satisfy it. Similarly, for $y'+y^2=0$, $y=1/x$ is a solution but $y=3/(x/3)$ does not satisfy it. Here, the second ODE is non-linear and $y=1/x$ is its singular (fixed or essential) solution which is free of any arbitrary constant.

We want to bring to attention that the commonly encountered first order homogeneous ODE: $$y'=\frac{px+qy}{rx+sy},\quad p,q,r,s \in \Re ~~~(1)$$ may have singular solutions which are often missed out. To get them one should take $y=vx$ (take $v$ as constant for change!). We get $$sv^2+(r-q)v-p=0$$. So $$v_1,v_2=\frac{(q-r)\pm\sqrt{(r-q)^2+4ps}}{2s}.$$ If $(r-q)^2 \ge -4ps$, there will be two real one real root giving us the singular solutions of this (1) ODE as two lines $y=v_1x$ and $y=v_2x$.

One usually takes $y=vx$ and $y'=xv'+x$ in (1) to get the general solution having one constant but misses out the possible real singular solutions: $y=v_1x$ and $y=v_2x$ as brought out here.

Also, if (1) is written as $\dot X=M X$ as below $$ \frac{d}{dt} \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} r & s\\ p & q \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ then $v_1$ and $v_2$ are nothing but eigenvalues of the matrix $M$.