6.3.2.7 Example 7

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

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

Hence $f=\frac{1}{p},g\left(p\right)=p$. Taking derivative w.r.t. $x$ gives

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

Using $f=\frac{1}{p},g=p$. Since $f\left(p\right)\ne p$ then this is d’Almbert ode. the above simplifies to

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

The singular solution is found by setting $\frac{dp}{dx}=0$ in (2) which results in $Q\left(p\right)=0$ or $p-1=0$ or $p=1$. Substituting these values in (1) gives the solutions

$$\begin{array}{}\text{(3)}& y_1\left(x\right)=x+1\end{array}$$

The general solution is found by finding $p$ from (2A). Since (2A) is not linear and not separable in $p$, then inversion is needed. Writing (2) as

$$\begin{array}{rl}\frac{dx}{dp}& =\frac{1-\frac{x}{p^2}}{p-\frac{1}{p}}\\ & =\frac{1}{p-p^3}(x-p^2)\end{array}$$

Hence

$$\frac{dx}{dp}+\frac{x}{p(p^2-1)}=\frac{p^2}{p(p^2-1)}$$

This is now linear ODE in $x\left(p\right)$. The solution is

$$\begin{array}{rl}x& =\frac{p\sqrt{(p-1)(1+p)}\ln (p+\sqrt{p^2-1})}{(1+p)(p-1)}+c_1\frac{p}{\sqrt{(1+p)(p-1)}}\\ \text{(4)}& & =\frac{p\sqrt{p^2-1}\ln (p+\sqrt{p^2-1})}{p^2-1}+c_1\frac{p}{\sqrt{p^2-1}}\end{array}$$

Now we need to eliminate $p$ from (1,4). From (1) since $y=\frac{1}{p}x+p$ then solving for $p$ gives

$$\begin{array}{rl}{p}_1& =\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x}\\ p_2& =\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x}\end{array}$$

Substituting each $p_i$ in (4) gives the general solution (implicit) in $y\left(x\right)$. First solution is

$$x=\frac{(\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x})\sqrt{{(\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x})}^2-1}\ln (\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x}+\sqrt{{(\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x})}^2-1})}{{(\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x})}^2-1}+c_1\frac{\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x}}{\sqrt{{(\frac{y}{2}+\frac{1}{2}\sqrt{y^2-4x})}^2-1}}$$

And second solution is

$$x=\frac{(\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x})\sqrt{{(\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x})}^2-1}\ln (\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x}+\sqrt{{(\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x})}^2-1})}{{(\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x})}^2-1}+c_1\frac{\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x}}{\sqrt{{(\frac{y}{2}-\frac{1}{2}\sqrt{y^2-4x})}^2-1}}$$