3.5.3.22 Example 22

\[ x\left ( y^{\prime }\right ) ^{3}=yy^{\prime }+1 \]

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

\begin{align} xp^{3} & =yp+1\nonumber \\ y & =\frac {xp^{3}-1}{p}\nonumber \\ & =xp^{2}-\frac {1}{p}\nonumber \\ & =xf+g \tag {1}\end{align}

Hence

\begin{align*} f & =p^{2}\\ g & =-\frac {1}{p}\end{align*}

Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative of (1) w.r.t. \(x\) gives

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

But \(f\left ( p\right ) =p^{2},f^{\prime }\left ( p\right ) =2p,g=-\frac {1}{p},g^{\prime }=\frac {1}{p^{2}}\) and the above becomes

\begin{align} p & =p^{2}+\left ( 2xp+\frac {1}{p^{2}}\right ) \frac {dp}{dx}\nonumber \\ p-p^{2} & =\left ( 2xp+\frac {1}{p^{2}}\right ) \frac {dp}{dx} \tag {2}\end{align}

The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p-p^{2}=0\). Hence \(p=0\) or \(p=1\). Substituting \(p=0\) in (1) gives 1/0 error. Hence this is not valid solution. Substituting \(p=1\) in (1) gives \(y=x-1\) which verifies the ode. Hence this is valid singular solution.

The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). This gives the ode

\[ \frac {dp}{dx}=\frac {p^{3}\left ( 1-p\right ) }{2xp^{3}+1}\]

But this is non-linear in \(p\). Hence inversion is needed. This becomes

\[ \frac {dx}{dp}=\frac {2xp^{3}+1}{p^{3}\left ( 1-p\right ) }\]

Which is now linear in \(x\left ( p\right ) \).  The solution is

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

We now need to eliminate \(p\). We have two equations to do that, (1) and (3). Here they are side by side

\begin{align} y & =xp^{2}-\frac {1}{p}\tag {1}\\ x & =\frac {2c_{1}p^{2}+2p-1}{2p^{2}\left ( p-1\right ) ^{2}} \tag {3}\end{align}

We can either solve for \(p\) from (1) and plugin in the value found into (3). Or we can solve for \(p\) from (3) and plugin the value found in (1). Using CAS we can just use the solve command. For an example, using Maple it gives

eq1:=y=x*p^2-1/p; 
eq2:=x= (2*_C1*p^2+2*p-1)/(2*p^2*(p-1)^2); 
solve({eq1,eq2},{y,p})
 

Whch gives

{p = RootOf(1 + 2*x*_Z^4 - 4*x*_Z^3 + (-2*c__1 + 2*x)*_Z^2 - 2*_Z), 
y = (x*RootOf(1 + 2*x*_Z^4 - 4*x*_Z^3 + (-2*c__1 + 2*x)*_Z^2 - 2*_Z)^3 - 1)/RootOf(1 + 2*x*_Z^4 - 4*x*_Z^3 + (-2*c__1 + 2*x)*_Z^2 - 2*_Z)}
                                                                                    
                                                                                    
 

Hence the general solution is

\[ y=\frac {x\operatorname {RootOf}\left ( 1+2xZ^{4}-4xZ^{3}+\left ( -2c_{1}+2x\right ) Z^{2}-2Z\right ) ^{3}-1}{\operatorname {RootOf}\left ( 1+2xZ^{4}-4xZ^{3}+\left ( -2c_{1}+2x\right ) Z^{2}-2Z\right ) }\]

And the singular solution is

\[ y=x-1 \]