3.5.3.1 Example 1

$x{\left(y^{\prime }\right)}^2-y{y}^{\prime }=-1$, is put in normal form (by replacing $y^{\prime }$ with $p$) and solving for $y$ gives

$$\begin{array}{rl}\text{(1)}& y& =xp+\frac{1}{p}& =xf\left(p\right)+g\left(p\right)\end{array}$$

Where $f\left(p\right)=p$ and $g\left(p\right)=\frac{1}{p}$. Since $f\left(p\right)=p$ then this is Clairaut ode. Taking derivative of the above w.r.t. $x$ gives

$$\begin{array}{rl}p& =\frac{d}{dx}(xp+g\left(p\right))\\ p& =p+(x+g^{\prime }\left(p\right))\frac{dp}{dx}\\ 0& =(x+g^{\prime }\left(p\right))\frac{dp}{dx}\end{array}$$

The general solution is given by

$$\begin{array}{rl}\frac{dp}{dx}& =0\\ p& =c_1\end{array}$$

Substituting this in (1) gives the general solution

$$y=c_{1}x+\frac{1}{c_1}$$

The term $(x+g^{\prime }\left(p\right))=0$ is used to find singular solutions.

$$\begin{array}{rl}x+g^{\prime }\left(p\right)& =x+\frac{d}{dp}\frac{1}{p}\\ & =x-\frac{1}{p^2}\end{array}$$

Hence $x-\frac{1}{p^2}=0$ or $p=\pm \frac{1}{\sqrt{x}}$. Substituting these back in (1) gives

$$\begin{array}{rl}{y}_1\left(x\right)& =xp+\frac{1}{p}\\ & =x\frac{1}{\sqrt{x}}+\sqrt{x}\\ \text{(3)}& & =2\sqrt{x}{y}_2\left(x\right)& =-x\sqrt{\frac{1}{x}}-\sqrt{x}\\ \text{(4)}& & =-2\sqrt{x}\end{array}$$

Eq. (2) is the general solution and (3,4) are the singular solutions.

Another method to find the singular solutions if it exists is called the p-discriminant. This is used only for first order ode with nonlinear in $y^{\prime }$. We set up the following two equations

$$\begin{array}{rl}F(x,y,y^{\prime })& =0\\ \frac{\partial F(x,y,y^{\prime })}{\partial y^{\prime }}& =0\end{array}$$

We eliminate $y^{\prime }$ and obtain $G(x,y)=0$ equation. This is the singular solution. But we still have to check if it satisfies the ode and also if it is true singular solution curve. More on this later. Let us now just find the singular solution found above but using the p-discriminant method. The above two equations are

$$\begin{array}{rl}y-x{y}^{\prime }-\frac{1}{y^{\prime }}& =0\\ -x+\frac{1}{{\left(y^{\prime }\right)}^2}& =0\end{array}$$

Second equation gives ${\left(y^{\prime }\right)}^2=\frac{1}{x}$. Hence $y^{\prime }=\pm \sqrt{\frac{1}{x}}$. Hence the first equation now gives (starting with positive root)

$$\begin{array}{rl}y-x\sqrt{\frac{1}{x}}-\frac{1}{\sqrt{\frac{1}{x}}}& =0\\ y& =x\sqrt{\frac{1}{x}}+\frac{1}{\sqrt{\frac{1}{x}}}\\ & =\frac{x\sqrt{\frac{1}{x}}\sqrt{\frac{1}{x}}+1}{\sqrt{\frac{1}{x}}}\\ & =2\sqrt{x}\end{array}$$

And for the second root $y^{\prime }=-\sqrt{\frac{1}{x}}$ we obtain $y=-2\sqrt{x}$. We see these are the same singular solutions obtained earlier.