Example 12

\begin{aligned} {(y^{'})}^{2} & = e^{4 x - 2 y} (y^{'} - 1) \\ \ln {(y^{'})}^{2} & = (4 x - 2 y) + \ln (y^{'} - 1) \\ 4 x - 2 y & = \ln {(y^{'})}^{2} - \ln (y^{'} - 1) \\ 4 x - 2 y & = \ln \frac{{(y^{'})}^{2}}{y^{'} - 1} \\ 2 y & = 4 x - \ln \frac{{(y^{'})}^{2}}{y^{'} - 1} \\ y & = 2 x - \frac{1}{2} \ln (\frac{{(y^{'})}^{2}}{y^{'} - 1}) \\ = 2 x - \frac{1}{2} \ln (\frac{p^{2}}{p - 1}) \\ = x f + g \end{aligned}

Where $f = 2, g = - \frac{1}{2} \ln (\frac{p^{2}}{p - 1})$ . Since $f (p) \neq p$ then this is d’Almbert ode. Taking derivative w.r.t. $x$ gives

\begin{aligned} p & = (f + x f^{'} \frac{d p}{d x}) + (g^{'} \frac{d p}{d x}) \\ p & = f + (x f^{'} + g^{'}) \frac{d p}{d x} \\ p - f & = (x f^{'} + g^{'}) \frac{d p}{d x} \end{aligned}

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

\begin{matrix} (2A) & p - 2 = (\frac{2 - p}{2 p^{2} - 2 p}) \frac{d p}{d x} \end{matrix}

The singular solution is when $\frac{d p}{d x} = 0$ which gives $p = 2$ . From (1) this gives

y = 2 x - \frac{1}{2} \ln 4

The general solution is when $\frac{d p}{d x} \neq 0$ . Then (2) becomes

\begin{aligned} \frac{d p}{d x} & = (p - 2) (\frac{2 p^{2} - 2 p}{2 - p}) \\ = 2 p (1 - p) \end{aligned}

is now separable. Solving for $p$ gives

p = \frac{1}{1 + c e^{- 2 x}}

Substituting the above solutions of $p$ in (1) gives

\begin{aligned} y & = 2 x - \frac{1}{2} \ln (\frac{{(\frac{1}{1 + c e^{- 2 x}})}^{2}}{\frac{1}{1 + c e^{- 2 x}} - 1}) \\ = 2 x - \frac{1}{2} \ln (\frac{- e^{4 x}}{c (c + e^{2 x})}) \end{aligned}