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0.1 Solving \(F\left ( x,y,p\right ) =0\) for \(y\) (Clairaut and d’Alembert)

Given an ode \(F\left ( x,y,y^{\prime }\right ) =0\) where from now on \(p=y^{\prime }\), we first solve for \(y\). Writing the solution as\begin {equation} y=xf\left ( p\right ) +g\left ( p\right ) \tag {1} \end {equation} The following are the conditions to classify the ode as either Clairaut or d’Alembert or neither of these two type. We should always check these conditions before continuing.

0.1.1 Checking for Clairaut ode

If \(f\left ( p\right ) =p\) then if \(g\left ( p\right ) =0\) (example: \(y=xp)\) then this is separable ode. Neither Clairaut nor d’Alembert. If \(g\left ( p\right ) \neq 0\) but is linear in \(p\) (example \(y=xp+p\) ) then this is also separable ode. Neither Clairaut nor d’Alembert. If \(g\left ( p\right ) \neq 0\) and also nonlinear in \(p\) then this is Clairaut ode (example \(y=xp+p^{2}\))

0.1.2 Checking for d’Alembert ode

If \(f\left ( p\right ) \neq p\) then there are 2 possibilities: Either \(g\left ( p\right ) =0\) or \(g\left ( p\right ) \neq 0\). If \(g\left ( p\right ) =0\) then \(f\left ( p\right ) \) must be nonlinear in \(p\) but can not be the special case \(p^{\frac {1}{n}}\) where \(n\) integer. Because now it is separable. All other nonlinear forms of \(f\left ( p\right ) \) makes it d’Alembert even if \(g\left ( p\right ) =0\). For example \(y=xp^{\frac {1}{2}}\) is not d’Alembert (here \(f=p^{\frac {1}{2}}\)) but \(y=xp^{\frac {3}{2}}\)is. Similarly \(y=xp^{\frac {1}{3}}\) is not, but \(y=xp^{\frac {2}{3}}\) is. The reason is that if \(y=xp^{\frac {1}{n}}\)  then by raising both sides to \(n\) it becomes \(y^{n}=x^{n}p\) which is separable. Also \(y=xp^{n}\) for positive \(n\) is d’Alembert. For example \(y=xp^{2}\) is.

If however \(g\left ( p\right ) \neq 0\) then any form of \(f\left ( p\right ) \) is allowed, except of course \(f\left ( p\right ) =p\) which is the case we are in now. Example \(y=xp^{\frac {1}{2}}+p\). In the special case of \(xp^{\frac {1}{n}}\) for \(g\) \(\neq 0\), then \(g\) must be function of \(p\) and not constant. For example \(y=xp^{\frac {1}{2}}+p^{2}\) is d’Alembert but \(y=xp^{\frac {1}{2}}+1\) is not. Finally, if \(f\left ( p\right ) =p^{\frac {1}{n}}\) and \(g\left ( p\right ) =f\left ( p\right ) \) then it is not d’Alembert. For example \(y=xp^{\frac {1}{2}}+p^{\frac {1}{2}}\) is not, but \(y=xp^{\frac {1}{2}}+p^{\frac {1}{3}}\) is. However for all positive powers such as \(y=xp^{2}+p^{2}\) then it is d’Alembert. The checking for d’Alembert is much more complicated than for Clairaut.

Note In all the above cases \(f\left ( p\right ) \) and \(g\left ( p\right ) \) can not be function of \(x\) in any case.

0.1.3 Clairaut ode

ode internal name "clairaut"

This is the case \(f\left ( p\right ) =p\). Note as mentioned above, \(g\left ( p\right ) \) can not be zero in this case. Taking derivative of (1) w.r.t. \(x\) gives\[ y=xp+g\left ( p\right ) \] We write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(x\). Hence the above becomes\[ y=xp+g \] Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xp+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =p+\left ( x+g^{\prime }\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\right ) \frac {dp}{dx} \end {align*}

Where \(g^{\prime }\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p & =c_{1} \end {align*}

Substituting this in (1) gives the general solution \[ y=c_{1}x+g\left ( c_{1}\right ) \] The singular solution is found from solving \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) for \(p\) and substituting the result in (1). Let the solution of \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) be \(p_{i}\), therefore the singular solution is \[ y=xp_{i}+g\left ( p_{i}\right ) \] Examples below show will clarify this process.

0.1.4 d’Alembert ode

ode internal name "dAlembert"

This is the case when \(f\left ( p\right ) \neq p\). (1) becomes \[ y=xf\left ( p\right ) +g\left ( p\right ) \] Writing \(f\equiv f\left ( p\right ) \) and \(g\equiv g\left ( p\right ) \) to make notation simpler but remembering that \(f\) is function of \(p\left ( x\right ) \) which in turn is function of \(x\). Same for \(g\left ( p\right ) \).\[ y=xf+g \] Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xf+g\right ) \\ p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \end {align*}

Where \(f^{\prime }\) above means \(\frac {df}{dp}\) and \(g^{\prime }=\frac {dg}{dp}\). Hence\begin {align} p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \tag {2} \end {align}

The singular solution is given when \(\frac {dp}{dx}=0\) or \[ p-f=0 \] Solving the above for \(p_{i}\) and substituting the result back in (1) gives the singular solution(s).

The general solution is given by solving the ode in (2) for \(p\) and substituting the result in (1).  There are two cases. If the ode (2) is separable or linear in \(p\) as is, then this ode is solved for \(p\) directly. If (2) turns out to be none of these cases, then inverting it first gives\[ \frac {dx}{dp}=\frac {\left ( xf^{\prime }+g^{\prime }\right ) }{p-f}\] Which makes it linear ode in \(x\left ( p\right ) \). And this is solved for \(x\left ( p\right ) \) as function of \(p\).

Let \(x=h\left ( p\right ) +c_{1}\) be the solution. Now two possible cases exist. If we can isolate \(p\) from this, then substituting the result in (1) gives the general solution \(y\left ( x\right ) \) as explicit solution. If it is not possible to isolate \(p\) from this solution, another option is to solve for \(p\) from (1) instead and substitute the result into \(x=h\left ( p\right ) +c_{1}\). This gives an implicit solution for \(y\left ( x\right ) \). Examples below show how this is done.

0.2 Algorithm diagram

The above is summarized in the following flow chart.

0.3 Solved examples

0.3.1 Example 1

\(x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1\), is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives\begin {align} y & =xp+\frac {1}{p}\tag {1}\\ & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end {align}

Where \(f\left ( p\right ) =p\) and \(g\left ( p\right ) =\frac {1}{p}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx} \end {align*}

The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p & =c_{1} \end {align*}

Substituting this in (1) gives the general solution \[ y=c_{1}x+\frac {1}{c_{1}}\] The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular solutions. \begin {align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\frac {1}{p}\\ & =x-\frac {1}{p^{2}} \end {align*}

Hence \(x-\frac {1}{p^{2}}=0\) or \(p=\pm \frac {1}{\sqrt {x}}\). Substituting these back in (1) gives\begin {align} y_{1}\left ( x\right ) & =xp+\frac {1}{p}\nonumber \\ & =x\frac {1}{\sqrt {x}}+\sqrt {x}\nonumber \\ & =2\sqrt {x}\tag {3}\\ y_{2}\left ( x\right ) & =-x\sqrt {\frac {1}{x}}-\sqrt {x}\nonumber \\ & =-2\sqrt {x}\tag {4} \end {align}

Eq. (2) is the general solution and (3,4) are the singular solutions.

0.3.2 Example 2

\(y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives\begin {align} y & =xp-p^{2}\tag {1}\\ & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end {align}

Where \(f\left ( p\right ) =p\) and \(g\left ( p\right ) =-p^{2}\). Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx} \end {align*}

The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p & =c_{1} \end {align*}

Substituting this in (1) gives the general solution \[ y=c_{1}x-c_{1}^{2}\] The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular solutions. \begin {align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\left ( -p^{2}\right ) \\ & =x+2p \end {align*}

Hence \(x+2p=0\) or \(p=\frac {x}{2}\). Substituting this back in (1) gives\begin {align} y\left ( x\right ) & =\frac {x^{2}}{2}-\frac {x^{2}}{4}\nonumber \\ & =\frac {x^{2}}{4} \tag {3} \end {align}

Eq. (2) is the general solution and (3) is the singular solution.

0.3.3 Example 3

\(y=xy^{\prime }-\frac {1}{4}\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives\begin {align} y & =xp-\frac {1}{4}p^{2}\tag {1}\\ & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end {align}

Where \(f\left ( p\right ) =p\) and \(g\left ( p\right ) =-\frac {1}{4}p^{2}\). Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx} \end {align*}

The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p & =c_{1} \end {align*}

Substituting this in (1) gives the general solution \[ y=c_{1}x-\frac {1}{4}c_{1}^{2}\] The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular solutions. \begin {align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\left ( -\frac {1}{4}p^{2}\right ) \\ & =x-\frac {1}{2}p \end {align*}

Hence \(x-\frac {1}{2}p=0\) or \(p=2x\). Substituting this back in (1) gives\begin {align} y\left ( x\right ) & =2x^{2}-x^{2}\nonumber \\ & =x^{2} \tag {3} \end {align}

Eq. (2) is the general solution and (3) is the singular solution.

0.3.4 Example 4

\(y=x\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives\begin {align} y & =xp^{2}\tag {1}\\ & =xf\left ( p\right ) \nonumber \end {align}

This is the case when \(f\left ( p\right ) =p^{2}\) and \(g\left ( p\right ) =0\). Writing \(f\equiv f\left ( p\right ) \) and \(g\equiv g\left ( p\right ) \) to make notation simpler but remembering that \(f\) is function of \(p\left ( x\right ) \) which in turn is function of \(x\). Same for \(g\left ( p\right ) \).\[ y=xf \] Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xf\right ) \\ p & =f+xf^{\prime }\frac {dp}{dx}\\ p-f & =xf^{\prime }\frac {dp}{dx} \end {align*}

Since \(f=p^{2}\) then the above becomes\begin {equation} p-p^{2}=2xp\frac {dp}{dx} \tag {2} \end {equation} The singular solution is given when \(\frac {dp}{dx}=0\) or \(p-p^{2}=0\). This gives \(p=0\) or \(p=1\). Substituting these values of \(p\) in (1) gives singular solutions\begin {align} y_{s1} & =0\tag {3}\\ y_{s2} & =x \tag {4} \end {align}

General solution is found when \(\frac {dp}{dx}\neq 0\,\). Eq(2) is a first order ode in \(p\). Now we could either solve ode (2) directly as it is for \(p\left ( x\right ) \), or do an inversion and solve for \(x\left ( p\right ) \). We always try the first option first. Since (2) is separable as is, no need to do an inversion. Eq (2) is separable. The solution is\begin {align*} p_{1} & =0\\ p_{2} & =1+\frac {c_{1}}{\sqrt {x}} \end {align*}

For each \(p\), there is a general solution. Substituting each of the above in (1) gives\begin {align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2} \end {align*}

Hence the final solutions are\begin {align*} y & =x\qquad \text {(singular)}\\ y & =0\\ y & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2} \end {align*}

But \(y=x\) can be obtained from the general solution when \(c_{1}=0\). Hence it is removed. Therefore the final solutions are\begin {align} y & =0\tag {6}\\ y & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2} \tag {7} \end {align}

What will happen if we had done an inversion to \(x\left ( p\right ) \)? Let us find out. ode(5) now becomes\begin {align*} \frac {p-p^{2}}{p}\frac {dx}{dp} & =2x\\ \frac {dx}{2x} & =\frac {p}{p-p^{2}}dp \end {align*}

This is also separable in \(x\). Solving this for \(x\) gives\[ x=\frac {c_{1}}{\left ( p-1\right ) ^{2}}\] Solving for \(p\) from the above gives\begin {align*} p_{1} & =\frac {x+\sqrt {xc_{1}}}{x}\\ p_{2} & =\frac {x-\sqrt {xc_{1}}}{x} \end {align*}

Substituting each of the above in (1) gives\begin {align*} y_{1} & =x\left ( \frac {x+\sqrt {xc_{1}}}{x}\right ) ^{2}\\ & =\frac {\left ( x+\sqrt {xc_{1}}\right ) }{x}^{2}\\ y_{2} & =x\left ( \frac {x-\sqrt {xc_{1}}}{x}\right ) ^{2}\\ & =\frac {\left ( x-\sqrt {xc_{1}}\right ) }{x}^{2} \end {align*}

Now we see that singular solution \(y=x\) can be obtained from the above general solutions from \(c_{1}=0\). But \(y=0\) can not. Hence the final solutions are\begin {align} y & =0\qquad \text {(singular)}\tag {8}\\ y & =\frac {\left ( x+\sqrt {xc_{1}}\right ) }{x}^{2}\tag {9}\\ y & =\frac {\left ( x-\sqrt {xc_{1}}\right ) }{x}^{2} \tag {10} \end {align}

All solutions (6,7,8,9,10) are correct and verified. Maple gives the solutions given in (8,9,10) and not those in (6,7).

0.3.5 Example 5

\(y=x+\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =x+p^{2}\tag {1}\\ & =xf+g\nonumber \end {align}

Hence \(f\left ( p\right ) =1,g\left ( p\right ) =p^{2}\). Taking derivative w.r.t. \(x\) gives \begin {align} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \nonumber \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \tag {2} \end {align}

Using \(f=1,g=p^{2}\) the above simplifies to

\begin {equation} p-1=2p\frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) in (2) which results in \(p-f=0\) or \(p-1=0\). Hence \(p=1\).  Substituting these values of \(p\) in (1) gives singular solution as\begin {equation} y=x+1 \tag {3} \end {equation} General solution is found when \(\frac {dp}{dx}\neq 0\,\). Eq (2A) is a first order ode in \(p\). Now we could either solve ode (2) directly as it is for \(p\left ( x\right ) \), or do an inversion and solve for \(x\left ( p\right ) \). We always try the first option first. Since (2) is separable as is, no need to do an inversion. Solving (2) for \(p\) gives\[ p=\operatorname *{LambertW}\left ( c_{1}e^{\frac {x}{2}-1}\right ) +1 \] Substituting this in (1) gives the general solution\begin {equation} y\left ( x\right ) =x+\left ( \operatorname *{LambertW}\left ( c_{1}e^{\frac {x}{2}-1}\right ) +1\right ) ^{2} \tag {4} \end {equation} Note however that when \(c_{1}=0\) then the general solution becomes \(y\left ( x\right ) =x+1\). Hence (3) is a particular solution and not a singular solution. (4) is the only solution.

0.3.6 Example 6

\(\left ( y^{\prime }\right ) ^{2}-1-x-y=0\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =-x+\left ( p^{2}-1\right ) \tag {1}\\ & =xf+g\nonumber \end {align}

Hence \(f=-1,g\left ( p\right ) =\left ( p^{2}-1\right ) \). Taking derivative w.r.t. \(x\) gives \begin {align} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \nonumber \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \tag {2} \end {align}

Using \(f=-1,g=\left ( p^{2}-1\right ) \) the above simplifies to\begin {equation} p+1=2p\frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=-1\). Substituting this in (1) gives singular solution as\begin {equation} y\left ( x\right ) =-x \tag {3} \end {equation} The general solution is found by finding \(p\) from (2A). No need here to do the inversion as (2) is separable already. Solving (2) gives\begin {align*} p & =-\operatorname *{LambertW}\left ( -e^{-\frac {x}{2}-1+\frac {c_{2}}{2}}\right ) -1\\ & =-\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1 \end {align*}

Substituting the above in (1) gives the general solution\begin {align} y\left ( x\right ) & =-x+\left ( p^{2}-1\right ) \nonumber \\ y\left ( x\right ) & =-x+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1\right ) ^{2}-1 \tag {4} \end {align}

Note however that when \(c_{1}=0\) then the general solution becomes \(y\left ( x\right ) =-x\). Hence (3) is a particular solution and not a singular solution. Solution (4) is therefore the only solution.

0.3.7 Example 7

\(yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =\frac {x+p^{2}}{p}\tag {1}\\ & =\frac {1}{p}x+p\nonumber \\ & =xf+g\nonumber \end {align}

Hence \(f=\frac {1}{p},g\left ( p\right ) =p\). Taking derivative w.r.t. \(x\) gives \begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \end {align*}

Using \(f=\frac {1}{p},g=p\) the above simplifies to\begin {equation} p-\frac {1}{p}=\left ( -\frac {x}{p^{2}}+1\right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) in (2) which results in \(Q\left ( p\right ) =0\) or \(p-1=0\) or \(p=1\). Substituting these values in (1) gives the solutions\begin {equation} y_{1}\left ( x\right ) =x+1 \tag {3} \end {equation} The general solution is found by finding \(p\) from (2A). Since (2A) is not linear and not separable in \(p\), then inversion is needed. Writing (2) as\begin {align*} \frac {dx}{dp} & =\frac {1-\frac {x}{p^{2}}}{p-\frac {1}{p}}\\ & =\frac {1}{p-p^{3}}\left ( x-p^{2}\right ) \end {align*}

Hence\[ \frac {dx}{dp}+\frac {x}{p\left ( p^{2}-1\right ) }=\frac {p^{2}}{p\left ( p^{2}-1\right ) }\] This is now linear ODE in \(x\left ( p\right ) \). The solution is \begin {align} x & =\frac {p\sqrt {\left ( p-1\right ) \left ( 1+p\right ) }\ln \left ( p+\sqrt {p^{2}-1}\right ) }{\left ( 1+p\right ) \left ( p-1\right ) }+c_{1}\frac {p}{\sqrt {\left ( 1+p\right ) \left ( p-1\right ) }}\nonumber \\ & =\frac {p\sqrt {p^{2}-1}\ln \left ( p+\sqrt {p^{2}-1}\right ) }{p^{2}-1}+c_{1}\frac {p}{\sqrt {p^{2}-1}} \tag {4} \end {align}

Now we need to eliminate \(p\) from (1,4). From (1) since \(y=\frac {1}{p}x+p\) then solving for \(p\) gives \begin {align*} p_{1} & =\frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\\ p_{2} & =\frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x} \end {align*}

Substituting each \(p_{i}\) in (4) gives the general solution (implicit) in \(y\left ( x\right ) \). First solution is\[ x=\frac {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) \sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}+\sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac {\frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}}{\sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}}\] And second solution is\[ x=\frac {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) \sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}+\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac {\frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}}{\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}}\]

0.3.8 Example 8

\(y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =xp^{2}+p^{2}\tag {1}\\ & =xf+g\nonumber \end {align}

where \(f=p^{2},g=p^{2}\). Taking derivative and simplifying gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \end {align*}

Using values for \(f,g\) the above simplifies to\begin {equation} p-p^{2}=\left ( 2xp+2p\right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=0\) or \(p=1\). Substituting these values in (1) gives the singular solutions\begin {align} y_{1}\left ( x\right ) & =0\tag {3}\\ y_{2}\left ( x\right ) & =x+1 \tag {4} \end {align}

The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\), then inversion is needed. Writing (A2) as\[ \frac {p\left ( 1-p\right ) }{2p\left ( x+1\right ) }=\frac {dp}{dx}\] Inverting gives\begin {align*} \frac {dx}{dp} & =\frac {2\left ( x+1\right ) }{\left ( 1-p\right ) }\\ \frac {dx}{dp}-x\frac {2}{\left ( 1-p\right ) } & =\frac {2}{\left ( 1-p\right ) } \end {align*}

This is now linear \(x\left ( p\right ) \). The solution is\[ x=\frac {C^{2}}{\left ( p-1\right ) ^{2}}-1 \] Solving for \(p\) gives\begin {align*} \frac {C^{2}}{\left ( p-1\right ) ^{2}} & =x+1\\ \left ( p-1\right ) ^{2} & =\frac {C^{2}}{x+1}\\ \left ( p-1\right ) & =\pm \frac {C}{\sqrt {x+1}}\\ p & =1\pm \frac {C}{\sqrt {x+1}} \end {align*}

Substituting the above in (1) gives the general solutions\[ y=\left ( x+1\right ) p^{2}\] Therefore\begin {align*} y\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2} \end {align*}

The solution \(y_{1}\left ( x\right ) =0\) found earlier can not be obtained from the above general solution hence it is singular solution. But \(y_{2}\left ( x\right ) =x+1\) can be obtained from the general solution when \(C=0\). Hence there are only three solutions, they are\begin {align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y_{3}\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2} \end {align*}

0.3.9 Example 9

\(y=\frac {x}{a}y^{\prime }+\frac {b}{ay^{\prime }}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =\frac {x}{a}p+\frac {b}{a}p^{-1}\tag {1}\\ & =xf+g\nonumber \end {align}

Where \(f=\frac {p}{a},g=\frac {b}{a}p^{-1}\).  Taking derivative w.r.t. \(x\) gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \end {align*}

Using values for \(f,g\) the above simplifies to\begin {equation} p-\frac {p}{a}=\left ( \frac {x}{a}-\frac {b}{a}p^{-2}\right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=0\). Substituting this in (1) does not generate any solutions due to division by zero. Hence no singular solution exist.

The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\), then inversion is needed. Writing (2A) as\[ \frac {p\left ( 1-\frac {1}{a}\right ) }{\frac {x}{a}-\frac {b}{a}p^{-2}}=\frac {dp}{dx}\] Since this is nonlinear, then inversion is needed\begin {align*} \frac {dx}{dp} & =\frac {\frac {x}{a}-\frac {b}{a}p^{-2}}{p\left ( 1-\frac {1}{a}\right ) }\\ \frac {dx}{dp}-x\frac {1}{p\left ( a-1\right ) } & =-\frac {b}{a}\frac {1}{p^{3}\left ( 1-\frac {1}{a}\right ) } \end {align*}

This is now linear ode in \(x\left ( p\right ) \). The solution is\begin {equation} x=\frac {b}{(2a-1)p^{2}}+C_{1}p^{\frac {1}{a-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 then 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 {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\\ p_{1} & =\frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x} \end {align*}

Substituting each one of these solutions back in (3) gives two implicit solutions\begin {align*} x & =\frac {b}{(2a-1)\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}}\\ x & =\frac {b}{(2a-1)\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}} \end {align*}

0.3.10 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 \(f=p+a\sqrt {1+p^{2}},g=0\). 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*}

Using values for \(f,g\) the above simplifies to\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*}

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

0.3.11 Example 11

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

Where \(f=1,g=p^{2}\left ( 1-\frac {2}{3}p\right ) \).  Taking derivative w.r.t. \(x\) gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \end {align*}

Using values for \(f,g\) the above simplifies to\begin {equation} p-1=\left ( 2p-2p^{2}\right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is when \(\frac {dp}{dx}=0\) which results in \(p=1\). Substituting this in (1) gives\begin {align*} y & =x-\left ( 1-\frac {2}{3}\right ) \\ & =x+\frac {1}{3} \end {align*}

The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2A) is now separable. Solving for \(p\) gives\begin {align*} p & =-\sqrt {c_{1}-x}\\ p & =\sqrt {c_{1}-x} \end {align*}

Substituting each one of the above solutions of \(p\) in (1) gives \begin {align*} y_{1} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( -\sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( -\sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}

And\begin {align*} y_{2} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( \sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( \sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}

Therefore the solutions are\begin {align*} y & =x+\frac {1}{3}\\ y & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\\ y & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}

0.3.12 Example 12

\begin {align*} \left ( y^{\prime }\right ) ^{2} & =e^{4x-2y}\left ( y^{\prime }-1\right ) \\ \ln \left ( y^{\prime }\right ) ^{2} & =\left ( 4x-2y\right ) +\ln \left ( y^{\prime }-1\right ) \\ 4x-2y & =\ln \left ( y^{\prime }\right ) ^{2}-\ln \left ( y^{\prime }-1\right ) \\ 4x-2y & =\ln \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\\ 2y & =4x-\ln \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\\ y & =2x-\frac {1}{2}\ln \left ( \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\right ) \\ & =2x-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \\ & =xf+g \end {align*}

Where \(f=2,g=-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \).  Taking derivative w.r.t. \(x\) gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \end {align*}

Using values for \(f,g\) the above simplifies to\begin {equation} p-2=\left ( \frac {2-p}{2p^{2}-2p}\right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is when \(\frac {dp}{dx}=0\) which gives \(p=2\). From (1) this gives\[ y=2x-\frac {1}{2}\ln 4 \] The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2) becomes\begin {align*} \frac {dp}{dx} & =\left ( p-2\right ) \left ( \frac {2p^{2}-2p}{2-p}\right ) \\ & =2p\left ( 1-p\right ) \end {align*}

is now separable. Solving for \(p\) gives\[ p=\frac {1}{1+ce^{-2x}}\] Substituting the above solutions of \(p\) in (1) gives \begin {align*} y & =2x-\frac {1}{2}\ln \left ( \frac {\left ( \frac {1}{1+ce^{-2x}}\right ) ^{2}}{\frac {1}{1+ce^{-2x}}-1}\right ) \\ & =2x-\frac {1}{2}\ln \left ( \frac {-e^{4x}}{c\left ( c+e^{2x}\right ) }\right ) \end {align*}

0.3.13 Example 13

\begin {align} y & =\frac {xy^{\prime }+x\left ( y^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2}}{y^{\prime }+1}\nonumber \\ & =\frac {xp+xp^{2}-p^{2}}{p+1}\nonumber \\ & =xp-\frac {p^{2}}{p+1}\tag {1}\\ & =xf+g\nonumber \end {align}

Where \(f=p\) and \(g=-\frac {p^{2}}{p+1}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx} \end {align*}

The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p & =c_{1} \end {align*}

Substituting this in (1) gives the general solution \[ y=xc_{1}-\frac {c_{1}^{2}}{c_{1}+1}\] The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular solutions. \begin {align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\frac {1}{p}\\ & =x-\frac {1}{p^{2}} \end {align*}

Hence \(x-\frac {1}{p^{2}}=0\) or \(p=\pm \frac {1}{\sqrt {x}}\). Substituting these back in (1) gives\begin {align} y_{1}\left ( x\right ) & =xp+\frac {1}{p}\nonumber \\ & =x\frac {1}{\sqrt {x}}+\sqrt {x}\nonumber \\ & =2\sqrt {x}\tag {3}\\ y_{2}\left ( x\right ) & =-x\sqrt {\frac {1}{x}}-\sqrt {x}\nonumber \\ & =-2\sqrt {x}\tag {4} \end {align}

Eq. (2) is the general solution and (3,4) are the singular solutions.

0.3.14 Example 14

\begin {align*} x\left ( y^{\prime }\right ) ^{2}+\left ( x-y\right ) y^{\prime }+1-y & =0\\ x\left ( y^{\prime }\right ) ^{2}+xy^{\prime }-yy^{\prime }+1-y & =0\\ y\left ( -y^{\prime }-1\right ) +x\left ( y^{\prime }\right ) ^{2}+xy^{\prime }+1 & =0 \end {align*}

Solving for \(y\)\begin {align} y & =\frac {-x\left ( y^{\prime }\right ) ^{2}-xy^{\prime }-1}{-y^{\prime }-1}\nonumber \\ & =\frac {-xp^{2}-xp-1}{-p-1}\nonumber \\ & =\frac {xp^{2}+xp+1}{p+1}\nonumber \\ & =x\left ( \frac {p^{2}+p}{p+1}\right ) +\frac {1}{1+p}\nonumber \\ & =xp+\frac {1}{1+p}\nonumber \\ & =xf+g\tag {1} \end {align}

Where \(f=p\) and \(g=\frac {1}{1+p}\). Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx} \end {align*}

The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p & =c_{1} \end {align*}

Substituting this in (1) gives the general solution \begin {equation} y=c_{1}x+\frac {1}{c_{1}+1}\tag {4} \end {equation} The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular solutions. But \begin {align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\left ( \frac {1}{1+p}\right ) \\ & =x-\frac {1}{\left ( p+1\right ) ^{2}} \end {align*}

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

Substituting these values into (1) gives\begin {align} y_{1} & =xp_{1}+\frac {1}{1+p_{1}}\nonumber \\ & =x\left ( \frac {1}{\sqrt {x}}-1\right ) +\frac {1}{1+\left ( \frac {1}{\sqrt {x}}-1\right ) }\nonumber \\ & =\frac {x}{\sqrt {x}}-x+\sqrt {x}\nonumber \\ & =\frac {x\sqrt {x}}{x}-x+\sqrt {x}\nonumber \\ & =2\sqrt {x}-x\tag {5} \end {align}

And substituting \(p_{2}\) into (1) gives\begin {align} y_{1} & =xp_{1}+\frac {1}{1+p_{1}}\nonumber \\ & =x\left ( -\frac {1}{\sqrt {x}}-1\right ) +\frac {1}{1+\left ( -\frac {1}{\sqrt {x}}-1\right ) }\nonumber \\ & =-\frac {x}{\sqrt {x}}-x-\sqrt {x}\nonumber \\ & =\frac {-x\sqrt {x}}{x}-x-\sqrt {x}\nonumber \\ & =-2\sqrt {x}-x\tag {6} \end {align}

There are 3 solutions given in (4,5,6).  One is general  and two are singular.

0.3.15 Example 15

\[ xyy^{\prime }=y^{2}+x\sqrt {4x^{2}+y^{2}}\] Solving for \(y\) gives\begin {align*} {\normalsize y} & {\normalsize =}\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) {\normalsize x}\\ y & =xf+g \end {align*}

Where \(f=\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) \) and \(g=0\). Taking derivative of the above w.r.t. \(x\) gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+xf^{\prime }\frac {dp}{dx}\\ p-f & =xf^{\prime }\frac {dp}{dx} \end {align*}

Using values for \(f\ \)the above simplifies to \begin {equation} p-\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) =\left ( x\frac {d}{dp}\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) \right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) \). Substituting this in (1) does not generate any real solutions (only 2 complex ones) hence will not be used.

The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\), then inversion is needed. Writing (2A) as\begin {align*} \frac {dx}{dp} & =\frac {xf}{p-f}\\ \frac {1}{x}dx & =\frac {f}{p-f}dp \end {align*}

Due to complexity of result, one now needs to obtain explicit result for \(\operatorname {RootOf}\) which makes the computation very complicated. So this is not practical to solve by hand. Will stop here. It is much easier to solve this ode as a homogeneous ode instead which gives the solution as\[ -\frac {\sqrt {4x^{2}+y^{2}}}{x}+\ln \left ( x\right ) =c_{1}\]

0.3.16 Extra example

This ode is an example where \(y\) does not appear explicitly in the ode so not possible to directly solve for \(y\). It is given here to show possible problems with this method.\begin {equation} y^{\prime }=\sqrt {1+x+y} \tag {1A} \end {equation} This ode is squared to first solve for \(y\) which gives\begin {equation} \left ( y^{\prime }\right ) ^{2}=1+x+y \tag {2A} \end {equation} However, here care is needed. To get back to original ode (1A) then (2A) means two possible equations\[ y^{\prime }=\pm \sqrt {1+x+y}\] Hence the solutions obtained using (2A) can be the solution to one of these\begin {align} y^{\prime } & =+\sqrt {1+x+y}\tag {B1}\\ y^{\prime } & =-\sqrt {1+x+y} \tag {B2} \end {align}

Therefore the solution obtained by squaring both sides of (1A), which is done in order to solve for \(y\), must be checked to see if it satisfies the original ode, else it will be extraneous solution resulting from squaring both sides of the ode.

Starting from (2A), in normal form (by replacing \(y^{\prime }\) with \(p\)) it becomes\begin {align} y & =-x-1+p^{2}\tag {1}\\ & =xf+g\nonumber \end {align}

Where \(f=-1,g=-1+p^{2}\). Taking derivative w.r.t. \(x\) gives \begin {align} p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p+1 & =2p\frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=-1\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=-1\). Substituting this in (1) gives the singular solution\begin {equation} y\left ( x\right ) =-x \tag {3} \end {equation} But this solution does not satisfy the ode, hence it is extraneous. The general solution is found by finding \(p\) from (2). Since (2) is nonlinear, then it is inverted which gives\begin {align*} \frac {p+1}{2p} & =\frac {dp}{dx}\\ \frac {dx}{dp} & =\frac {2p}{p+1} \end {align*}

Which is linear in \(x\). Solving gives\begin {equation} x=2p-2\ln \left ( p+1\right ) +c_{1} \tag {4} \end {equation} Instead of inverting this to find \(p\) in terms of \(x\), \(p\) is found from (1) which gives\begin {align*} y+x+1 & =p^{2}\\ p & =\pm \sqrt {y+x+1} \end {align*}

Substituting these solutions in (4) gives implicit solutions as\begin {align*} x & =2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_{1}\\ x & =-2\sqrt {y+x+1}-2\ln \left ( 1-\sqrt {y+x+1}\right ) +c_{1} \end {align*}

But only the first one above satisfies the ode. The second is extraneous. Therefore the final solution is\[ x=2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_{1}\] And no singular solutions exist. If instead of doing the above, \(p\) was found from (4) using inversion, then it will be\[ p=-\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1 \] Substituting this in (1) gives\[ y=-x-1+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1\right ) ^{2}\] But this general solution does not satisfy the original ode.  In general, it is best to avoid squaring both side of the ode in order to solve for \(y\) as this can generate extraneous solutions. Only use this method if the original ode is already given in the form where \(y\) shows explicitly.

0.4 references

1. An elementary treatise on differential equations. By Abraham Cohen. 1906.
2. Applied differential equations, N Curle. 1972
3. Ordinary differential equations, LB Jones. 1976.
4. Elementary differential equations, William Martin, Eric Reissner. second edition. 1961.
5. Differentialgleichungen, by E. Kamke, page 30.
6. Differential and integral calculus by N. Piskunov, Vol II