3.5.3.24 Example 24

\[ xy^{\prime }-y=\sqrt {x^{2}-y^{2}}\]

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

\[ xp-y=\sqrt {x^{2}-y^{2}}\]

Solving for \(y\) gives two solutions

\begin{align} y & =x\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) \tag {1}\\ y & =x\left ( \frac {p}{2}-\frac {1}{2}\sqrt {2-p^{2}}\right ) \nonumber \end{align}

We will here solve the first one above. The second one will have similar solution. Comparing the above to \(y=xf\left ( p\right ) +g\left ( p\right ) \) shows that

\begin{align} f & =\frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\tag {2}\\ g & =0\nonumber \end{align}

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

\begin{align} p & =\frac {d}{dx}\left ( xf\left ( p\right ) \right ) \nonumber \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}\nonumber \\ & =\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) +x\left ( \frac {1}{2}-\frac {p}{2\sqrt {2-p^{2}}}\right ) \frac {dp}{dx}\nonumber \\ p-\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) & =x\left ( \frac {1}{2}-\frac {p}{2\sqrt {2-p^{2}}}\right ) \frac {dp}{dx}\tag {3}\end{align}

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

\begin{align*} p-\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) & =0\\ \frac {p}{2}-\frac {1}{2}\sqrt {2-p^{2}} & =0 \end{align*}

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

\begin{align*} y & =x\left ( \frac {1}{2}+\frac {1}{2}\sqrt {2-1}\right ) \\ & =x \end{align*}

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

\begin{align*} \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{align*}

This is separable ode.

\begin{align*} \frac {-dp}{\sqrt {2-p^{2}}} & =\frac {1}{x}dx\\ -\arcsin \left ( \frac {\sqrt {2}}{2}p\right ) & =\ln x+c_{1}\end{align*}

Substituting this into (1) gives

\begin{align*} y & =x\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) \\ & =x\left ( \frac {-\frac {2}{\sqrt {2}}\sin \left ( \ln x+c_{1}\right ) }{2}+\frac {1}{2}\sqrt {2-\left ( -\frac {2}{\sqrt {2}}\sin \left ( \ln x+c_{1}\right ) \right ) ^{2}}\right ) \\ & =x\left ( \frac {-\sin \left ( \ln x+c_{1}\right ) }{\sqrt {2}}+\frac {1}{2}\sqrt {2-2\sin ^{2}\left ( \ln x+c_{1}\right ) }\right ) \end{align*}