3.5.3.4 Example 4

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

$$\begin{array}{rl}\text{(1)}& y& =x{p}^2& =xf\left(p\right)\end{array}$$

This is the case when $f\left(p\right)=p^2$ and $g\left(p\right)=0$. Since $f\left(p\right)\ne p$ then this is d’Almbert ode.  

Writing $f\equiv f\left(p\right)$ and $g\equiv g\left(p\right)$ to make notation simpler but remembering that $f$ is function of $p\left(x\right)$ which in turn is function of $x$. Same for $g\left(p\right)$.

$$y=xf$$

Taking derivative of the above w.r.t. $x$ gives

$$\begin{array}{rl}p& =\frac{d}{dx}\left(xf\right)\\ p& =f+x{f}^{\prime }\frac{dp}{dx}\\ p-f& =x{f}^{\prime }\frac{dp}{dx}\end{array}$$

Since $f=p^2$ then the above becomes

$$\begin{array}{}\text{(2)}& p-p^2=2xp\frac{dp}{dx}\end{array}$$

The singular solution is given when $\frac{dp}{dx}=0$ or $p-p^2=0$. This gives $p=0$ or $p=1$. Substituting these values of $p$ in (1) gives singular solutions

$$\begin{array}{}\text{(3)}& y_{s1}& =0\text{(4)}& y_{s2}& =x\end{array}$$

General solution is found when $\frac{dp}{dx}\ne 0$. Eq(2) is a first order ode in $p$. Now we could either solve ode (2) directly as it is for $p\left(x\right)$, or do an inversion and solve for $x\left(p\right)$.  If the ode is linear as is in $p$ then no need to do inversion. Since (2) is separable as is, no need to do an inversion. The solution to (2) is

$$\begin{array}{rl}{p}_1& =0\\ p_2& =1+\frac{c_1}{\sqrt{x}}\end{array}$$

For each $p$, there is a general solution. Substituting each of the above in (1) gives

$$\begin{array}{rl}{y}_1\left(x\right)& =0\\ y_2\left(x\right)& =x{(1+\frac{c_1}{\sqrt{x}})}^2\end{array}$$

Hence the final solutions are

$$\begin{array}{rl}y& =x\text{(singular)}\\ y& =0\\ y& =x{(1+\frac{c_1}{\sqrt{x}})}^2\end{array}$$

But $y=x$ can be obtained from the general solution when $c_1=0$. Hence it is removed. Therefore the final solutions are

$$\begin{array}{}\text{(6)}& y& =0\text{(7)}& y& =x{(1+\frac{c_1}{\sqrt{x}})}^2\end{array}$$

What will happen if we had done an inversion to $x\left(p\right)$? Let us find out. ode(5) now becomes

$$\begin{array}{rl}\frac{p-p^2}{p}\frac{dx}{dp}& =2x\\ \frac{dx}{2x}& =\frac{p}{p-p^2}dp\end{array}$$

This is also separable in $x$. Solving this for $x$ gives

$$x=\frac{c_1}{{(p-1)}^2}$$

Solving for $p$ from the above gives

$$\begin{array}{rl}{p}_1& =\frac{x+\sqrt{x{c}_1}}{x}\\ p_2& =\frac{x-\sqrt{x{c}_1}}{x}\end{array}$$

Substituting each of the above in (1) gives

$$\begin{array}{rl}{y}_1& =x{\left(\frac{x+\sqrt{x{c}_1}}{x}\right)}^2\\ & ={\frac{(x+\sqrt{x{c}_1})}{x}}^2\\ y_2& =x{\left(\frac{x-\sqrt{x{c}_1}}{x}\right)}^2\\ & ={\frac{(x-\sqrt{x{c}_1})}{x}}^2\end{array}$$

Now we see that singular solution $y=x$ can be obtained from the above general solutions from $c_1=0$. But $y=0$ can not. Hence the final solutions are

$$\begin{array}{}\text{(8)}& y& =0\text{(singular)}\text{(9)}& y& ={\frac{(x+\sqrt{x{c}_1})}{x}}^2\text{(10)}& y& ={\frac{(x-\sqrt{x{c}_1})}{x}}^2\end{array}$$

All solutions (6,7,8,9,10) are correct and verified. Maple gives the solutions given in (8,9,10) and not those in (6,7).

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

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

Second equation gives $y^{\prime }=0$. Hence the first equation now gives the singular solution as

$$y=0$$

Which is the same obtained earlier.