Example 10

\(y=xy^{\prime }+ax\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives

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

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

\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) \\ p-f & =xf^{\prime }\frac {dp}{dx}\end{align*}

\begin{equation} -a\sqrt {1+p^{2}}=x\left ( 1+\frac {ap}{\sqrt {1+p^{2}}}\right ) \frac {dp}{dx} \tag {2A}\end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(-a\sqrt {1+p^{2}}=0\). This gives no real solution for \(p.\) Hence no singular solution exists.

The general solution is when \(\frac {dp}{dx}\neq 0\) in (2A). Since (2A) is nonlinear, inversion is needed.

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

\[ \ln x\left ( p\right ) =-\frac {1}{2}\ln \left ( p^{2}+1\right ) -\frac {1}{a}\operatorname {arcsinh}\left ( p\right ) \]

\begin{equation} x=c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( p\right ) \right ) }}{\sqrt {p^{2}+1}} \tag {3}\end{equation}

There are now two choices to take. The ﬁrst is by solving for \(p\) from the above in terms of \(x\) and substituting the result in (1) to obtain explicit solution for \(y\left ( x\right ) \), and the second choice is by solving for \(p\) algebraically from (1) and substituting the result in (3). The second choice is easier in this case but gives an implicit solution. Solving for \(p\) from (1) gives

\begin{align*} p_{1} & =-\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\\ p_{2} & =\frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\end{align*}

Substituting each one of these solutions back in (3) gives two implicit solutions

\begin{align*} x & =c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( -\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt {\left ( -\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}}\\ x & =c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( \frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt {\left ( \frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}}\end{align*}

3.5.3.10 Example 10