5.3.2.17 Example 17

$$x{\left(y^{\prime }\right)}^2-2y{y}^{\prime }+4x=0$$

Solving for $y$ gives

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

where $f=\frac{1}{2}p+2\frac{1}{p},g=0$. Since $f\left(p\right)\ne p$ then this is d’Almbert ode. Taking derivative and simplifying gives

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

Using values for $f,g$ the above simplifies to

$$\begin{array}{rl}p-\frac{1}{2}p-2\frac{1}{p}& =x(\frac{1}{2}-\frac{2}{p^2})\frac{dp}{dx}\\ \text{(2A)}& \frac{1}{2}p-\frac{2}{p}& =x(\frac{1}{2}-\frac{2}{p^2})\frac{dp}{dx}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ which results in $\frac{1}{2}p-\frac{2}{p}=0$ or $\frac{1}{2}{p}^2-2=0$ or $p^2=4$ or $p=\pm 2$. Hence $y=\pm 2x$ are the singular solutions.

The general solution is when $\frac{dp}{dx}\ne 0$ in (2A). Since (2A) is nonlinear, inversion is needed. General solution can be shown to be

$$\begin{array}{}\text{(3)}& y=-\frac{1}{2}(-\frac{x^2}{c_1^2}-4)c_1\end{array}$$

Will now show a more general method to find singular solution that works for any first order ode. This requires finding the general solution above first. Let the general solution be

$$\begin{array}{rl}\mathrm{\Phi }(x,y,c)& =0\\ & =y+\frac{1}{2}(-\frac{x^2}{c_1^2}-4)c_1\end{array}$$

The ode is

$$\begin{array}{rl}F(x,y,y^{\prime })& =0\\ & =x{\left(y^{\prime }\right)}^2-2y{y}^{\prime }+4x\end{array}$$

First we find the p-discriminant curve. This is found by eliminating $y^{\prime }$ from

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

Or

$$\begin{array}{rl}x{\left(y^{\prime }\right)}^2-2y{y}^{\prime }+4x& =0\\ 2x{y}^{\prime }-2y& =0\end{array}$$

Second equation gives $y^{\prime }=\frac{y}{x}$. Substituting into first equation gives $x{\left(\frac{y}{x}\right)}^2-2y\left(\frac{y}{x}\right)+4x=0$ or $\frac{y^2}{x}-2\frac{y^2}{x}+4x=0$ or $y=\pm 2x$. These are the candidate singular solutions

$$y_s=\pm 2x$$

Next, we verify these satisfy the ode itself. We see both do. Next we have to check that for an arbitrary point $x_0$ the following two equations are satisfied

$$\begin{array}{rl}{y}_g\left(x_0\right)& =y_s\left(x_0\right)\\ y_g^{\prime }\left(x_0\right)& =y_s^{\prime }\left(x_0\right)\end{array}$$

Where $y_g\left(x\right)$ is the general solution obtained above in (3). Starting with $y_s=2x$ the above two equations now become

$$\begin{array}{rl}-\frac{1}{2}(-\frac{x_0^2}{c_1^2}-4)c_1& =2x_0\\ -\frac{1}{2}(-\frac{2x_0}{c_1^2})c_1& =2\end{array}$$

Or

$$\begin{array}{rl}\frac{x_0^2}{2c_1}+2c_1& =2x_0\\ \frac{x_0}{c_1}& =2\end{array}$$

Second equation gives $c_1=\frac{x_0}{2}$. Using this in first equation gives

$$\begin{array}{rl}\frac{x_0^2}{2\frac{x_0}{2}}+2\left(\frac{x_0}{2}\right)& =2x_0\\ x_0+x_0& =2x_0\\ 2x_0& =2x_0\end{array}$$

Which shows it is satisfied. Hence this shows that $y_s=2x$ is indeed a singular solution. Now we have to do the same for second $y_s=-2x$. Hence the steps of this method are the following

1. Find $y_s$ using p-discriminant method by eliminating $y^{\prime }$ from $F=0$ and $\frac{\partial F}{\partial y^{\prime }}=0$.
2. Verify that each $y_s$ found satisfies the ode.
3. Find general solution to the ode $y_g\left(x\right)$.
4. Verify that the two equations $y_g\left(x_0\right)=y_s\left(x_0\right)$ and $y_g^{\prime }\left(x_0\right)=y_s^{\prime }\left(x_0\right)$ are satisfied at an arbitrary point $x_0$. If so, then $y_s$ is singular solution. (envelope of the family of curves of the general solution).