3.5.3.24 Example 24

$$x{y}^{\prime }-y=\sqrt{x^2-y^2}$$

Let $y^{\prime }=p$ and rearranging gives

$$xp-y=\sqrt{x^2-y^2}$$

Solving for $y$ gives two solutions

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

We will here solve the first one above. The second one will have similar solution. Comparing the above to

$$\begin{array}{}\text{(2)}& y=xf\left(p\right)+g\left(p\right)\end{array}$$

shows that

$$\begin{array}{rl}f& =\frac{p}{2}+\frac{1}{2}\sqrt{2-p^2}\\ g& =0\end{array}$$

Since $f\left(p\right)\ne p$ then this is d’Almbert ode. Taking derivative of (2) w.r.t. $x$ gives

$$\begin{array}{rl}p& =\frac{d}{dx}\left(xf\left(p\right)\right)\\ & =f\left(p\right)+x{f}^{\prime }\left(p\right)\frac{dp}{dx}\\ & =(\frac{p}{2}+\frac{1}{2}\sqrt{2-p^2})+x(\frac{1}{2}-\frac{p}{2\sqrt{2-p^2}})\frac{dp}{dx}\\ \text{(3)}& p-(\frac{p}{2}+\frac{1}{2}\sqrt{2-p^2})& =x(\frac{1}{2}-\frac{p}{2\sqrt{2-p^2}})\frac{dp}{dx}\end{array}$$

Singular solution is when $\frac{dp}{dx}=0$ which results in

$$\begin{array}{rl}p-(\frac{p}{2}+\frac{1}{2}\sqrt{2-p^2})& =0\\ \frac{p}{2}-\frac{1}{2}\sqrt{2-p^2}& =0\end{array}$$

Hence $p=1$.  Substituting this in (2) gives singular solution

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

To find general solution, we need to solve (3) for $p$. EQ (3) becomes

$$\begin{array}{rl}\frac{dp}{dx}& =\frac{\frac{p}{2}-\frac{1}{2}\sqrt{2-p^2}}{\frac{x}{2}-\frac{xp}{2\sqrt{2-p^2}}}\\ & =-\frac{1}{x}\sqrt{2-p^2}\end{array}$$

This is separable ode.

$$\begin{array}{rl}\frac{-dp}{\sqrt{2-p^2}}& =\frac{1}{x}dx\\ -\arcsin \left(\frac{\sqrt{2}}{2}p\right)& =\ln x+c_1\end{array}$$

Substituting this into (1) gives

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