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## Solving Clairaut and d’Alembert ﬁrst order diﬀerential equations

July 9, 2022   Compiled on July 9, 2022 at 5:51am

### 1 Introduction

Given the following two forms of ﬁrst order diﬀerential equations\begin {equation} y\left ( x\right ) =x\frac {dy}{dx}+f\left ( \frac {dy}{dx}\right ) \tag {1} \end {equation} And \begin {equation} y\left ( x\right ) =xg\left ( \frac {dy}{dx}\right ) +f\left ( \frac {dy}{dx}\right ) \tag {2} \end {equation} Using $$p\equiv \frac {dy}{dx}\,$$ these equations become

\begin {equation} y\left ( x\right ) =xp+f\left ( p\right ) \tag {1} \end {equation} And the second one becomes\begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {2} \end {equation} The ﬁrst ode above is called the Clairaut ode and the second is called d’Alembert (also called Lagrange ode in some books). In the above $$f$$ and $$g$$ are functions of $$p$$.

The Clairaut ode is a special case of the d’Alembert ode when the function $$g\left ( p\right ) =p$$. Both ode’s must be linear in $$y\left ( x\right )$$ and linear in $$x$$.

In Clairaut ode $$g\left ( p\right )$$ can only be $$p$$. Anything else, even $$g\left ( p\right ) =1$$, means it is not Clairaut, but can be d’Alembert. In the case $$f\left ( p\right ) =0$$ then $$g\left ( p\right )$$ has to be nonlinear in $$p$$ for it to be called d’Alembert otherwise it is separable ode. The function $$f\left ( p\right )$$ in the Clairaut ode $$y=xp+f\left ( p\right )$$ has to be nonlinear in $$p$$ otherwise it becomes linear ode.

#### 1.1 Algorithm for solving Clairaut ODE

Starting with the Clairaut ode as it is easier to solve than d’Alembert. Eq. (1) is\begin {equation} y\left ( x\right ) =xp+f\left ( p\right ) \tag {C1} \end {equation} Taking derivative w.r.t. $$x$$ of the above gives\begin {align*} p & =p+x\frac {dp}{dx}+\frac {df\left ( p\right ) }{dp}\frac {dp}{dx}\\ 0 & =\left ( x+\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx} \end {align*}

The above implies the following two possibilities \begin {align} \frac {dp}{dx} & =0\tag {C2}\\ x+\frac {d}{dp}f\left ( p\right ) & =0 \tag {C3} \end {align}

The complete integral (i.e. the general solution) is obtained from the ﬁrst above and the singular solution (if any) is obtained from the second equation.

The general solution is found by solving for $$p$$ from (C2) and substituting the solution into (C1). $$\frac {dp}{dx}=0$$ implies that $$p=C_{1}$$ where $$C_{1}$$ is some constant to be found. Substituting $$p=C_{1}$$ back in (C1) gives the general solution as \begin {equation} y\left ( x\right ) =C_{1}x+f\left ( C_{1}\right ) \tag {C4} \end {equation} The singular solution is found by solving for $$p$$ from (C3) and substituting the solution $$p$$ back into (C1). This completes the solution for Clairaut ODE.

#### 1.2 Algorithm for solving the d’Alembert ODE

Eq(2) is\begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Taking derivative w.r.t. $$x$$ gives\begin {align} p & =g\left ( p\right ) +x\frac {d}{dp}g\left ( p\right ) \frac {dp}{dx}+\frac {d}{dp}f\left ( p\right ) \frac {dp}{dx}\tag {d2}\\ p-g\left ( p\right ) & =\left ( x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx} \tag {d3} \end {align}

The general solution is found by solving for $$p$$ from (d3) and substituting the result back in (d1). Unless (d3) comes out to be linear or separable, then the way to solve (d3) is by inversion. Rewriting (d3) as\begin {equation} \frac {p-g\left ( p\right ) }{x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) }=\frac {dp}{dx} \tag {d4} \end {equation} The above is now Inverted so that $$x\left ( p\right )$$ becomes the dependent variable and $$p$$ as the independent variable which gives\begin {align} \frac {dx}{dp} & =\frac {x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) }{p-g\left ( p\right ) }\nonumber \\ \frac {dx}{dp}-x\frac {\frac {d}{dp}g\left ( p\right ) }{p-g\left ( p\right ) } & =\frac {\frac {d}{dp}f\left ( p\right ) }{p-g\left ( p\right ) } \tag {d5} \end {align}

The above is always linear ﬁrst order in $$x\left ( p\right )$$ as it is of the form\begin {equation} x^{\prime }+xG\left ( p\right ) =Q\left ( p\right ) \tag {d6} \end {equation} Where $$G\left ( p\right ) =\frac {\frac {d}{dp}g\left ( p\right ) }{p-g\left ( p\right ) }$$ and $$Q\left ( p\right ) =\frac {\frac {d}{dp}f\left ( p\right ) }{p-g\left ( p\right ) }$$. This is easily solved using an integration factor\begin {align} \frac {d}{dp}\left ( xe^{\int G\left ( p\right ) dp}\right ) & =e^{\int G\left ( p\right ) dp}Q\left ( p\right ) \nonumber \\ xe^{\int G\left ( p\right ) dp} & =\int e^{\int G\left ( p\right ) dp}Q\left ( p\right ) +C_{1}\nonumber \\ x\left ( p\right ) & =e^{-\int G\left ( p\right ) dp}\int e^{\int G\left ( p\right ) dp}Q\left ( p\right ) +C_{1}e^{-\int G\left ( p\right ) dp} \tag {d7} \end {align}

Once $$x\left ( p\right )$$ is found from the above as function of $$p$$, then $$p$$ is found by inversion of the solution (d7). This might not be possible and is the hardest step in this method. When it is possible to solve for $$p$$ from (d7) in terms of $$x$$, then substituting this $$p$$ back into (d1) gives the general solution $$y\left ( x\right )$$. If it is not possible to solve for $$p$$ from (d7) as function of $$x$$, then the alternative is to solve algebraically for $$p$$ from the original ode (d1) in terms $$y\left ( x\right )$$ and then replacing $$p$$ in (d7) with the solution. This gives an implicit solution for $$y\left ( x\right )$$ instead of explicit. See example 6,9,10 for such cases.

The singular solution is found by solving for $$p$$ from (d3) when $$\frac {dp}{dx}=0$$. then $$p$$ is constant, say $$p_{0}$$. This constant is found by solving for $$p$$ from $$p-g\left ( p\right ) =0$$ from (d3) after setting $$\frac {dp}{dx}=0$$ and then substituting the solutions $$p$$ (which will be constant) back into (d1).

To show how these methods work, the following ode’s are now solved.

#### 1.3 General method to solve the d’Alembert ODE when f(p)=0

This is special case of the general d’Alembert $$y=xg\left ( p\right ) +f\left ( p\right )$$. The ODE now reduces to\begin {equation} y=xg\left ( p\right ) \tag {sd1} \end {equation} In this case $$g\left ( p\right )$$ must be non-linear in $$p$$ else it becomes separable. For an example, $$y\left ( x\right ) =xp^{2}$$ is d’Alembert. Taking derivative w.r.t. $$x$$ gives\begin {align} y^{\prime } & =g\left ( p\right ) +x\frac {d}{dp}g\left ( p\right ) \frac {dp}{dx}\nonumber \\ p & =g\left ( p\right ) +x\frac {d}{dp}g\left ( p\right ) \frac {dp}{dx}\nonumber \\ p-g\left ( p\right ) & =x\frac {d}{dp}g\left ( p\right ) \frac {dp}{dx} \tag {sd2} \end {align}

When $$\frac {dp}{dx}=0$$, then $$p$$ is constant, say $$p_{0}$$. This constant is found by solving for $$p$$ from $$p-g\left ( p\right ) =0$$ and substituting the result back into (sd1). This gives the singular solution.

The general solution is found by rewriting (sd2) as$\frac {p-g\left ( p\right ) }{x\frac {d}{dp}g\left ( p\right ) }=\frac {dp}{dx}$ And inverting so that $$x\left ( p\right )$$ is now the dependent variable and $$p$$ as the independent variable\begin {align} \frac {dx}{dp} & =\frac {x\frac {d}{dp}g\left ( p\right ) }{p-g\left ( p\right ) }\nonumber \\ \frac {dx}{dp}-x\frac {\frac {d}{dp}g\left ( p\right ) }{p-g\left ( p\right ) } & =0 \tag {sd3} \end {align}

Equation (sd3) is a linear ﬁrst order in $$x\left ( p\right )$$ since it is of the form$x^{\prime }+xG\left ( p\right ) =0$ Where here $$G\left ( p\right ) =-\frac {\frac {d}{dp}g\left ( p\right ) }{p-g\left ( p\right ) }$$. This ode is solved using an integration factor\begin {align} \frac {d}{dp}\left ( xe^{\int G\left ( p\right ) dp}\right ) & =0\nonumber \\ xe^{\int G\left ( p\right ) dp} & =C_{1}\nonumber \\ x & =C_{1}e^{-\int G\left ( p\right ) dp} \tag {sd4} \end {align}

Once $$x$$ is found from the above as function of $$p$$, then $$p$$ is found by inversion of the solution. Substituting this solution $$p$$ back into (sd1) gives the general solution $$y\left ( x\right )$$.

To show how these method work, the following ODE’s are now solved.

### 2 Solved examples

All the ode’s below have the form $$y=xg\left ( p\right ) +f\left ( p\right )$$

 number ode transformed $$g\left ( p\right )$$ $$f\left ( p\right )$$ type $$1$$ $$x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1$$ $$y=xp+\frac {1}{p}$$ $$p$$ $$\frac {1}{p}$$ Clairaut $$2$$ $$y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}$$ $$y=xp-p^{2}$$ $$p$$ $$-p^{2}$$ Clairaut $$3$$ $$y=xy^{\prime }-\frac {1}{4}\left ( y^{\prime }\right ) ^{2}$$ $$y=xp-\frac {1}{4}p^{2}$$ $$p$$ $$-\frac {1}{4}p^{2}$$ Clairaut $$4$$ $$y=x\left ( y^{\prime }\right ) ^{2}$$ $$y=xp^{2}$$ $$p^{2}$$ $$0$$ d’Alembert $$5$$ $$y=x+\left ( y^{\prime }\right ) ^{2}$$ $$y=x+p^{2}$$ $$1$$ $$p^{2}$$ d’Alembert $$6$$ $$\left ( y^{\prime }\right ) ^{2}-1-x-y=0$$ $$y=-x+\left ( p^{2}-1\right )$$ $$-1$$ $$p^{2}-1$$ d’Alembert $$7$$ $$yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x$$ $$y=\frac {1}{p}x+p$$ $$\frac {1}{p}$$ $$p$$ d’Alembert $$8$$ $$y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}$$ $$y=xp^{2}+p^{2}$$ $$p^{2}$$ $$p^{2}$$ d’Alembert $$9$$ $$y=\frac {x}{a}y^{\prime }+\frac {b}{ay^{\prime }}$$ $$y=\frac {x}{a}p+\frac {b}{a}p^{-1}$$ $$\frac {p}{a}$$ $$\frac {b}{ap}$$ d’Alembert $$10$$ $$y=x\left ( y^{\prime }+a\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\right )$$ $$y=x\left ( p+a\sqrt {1+p^{2}}\right )$$ $$p+a\sqrt {1+p^{2}}$$ $$0$$ d’Alembert $$11$$ $$y=x\left ( y^{\prime }\right ) ^{2}$$ $$y=xp^{2}$$ $$p^{2}$$ $$0$$ d’Alembert

Notice in the above table, that the ODE is Clairaut only when $$g\left ( p\right ) =p$$ and d’Alembert otherwise.

#### 2.1 Example 1

$$x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1$$, is put in normal form (by replacing $$y^{\prime }$$ with $$p$$) which gives\begin {equation} y\left ( x\right ) =xp+\frac {1}{p} \tag {1} \end {equation} Comparing to (C1) shows that $$f\left ( p\right ) =\frac {1}{p}$$. Hence (C2,C2) now becomes\begin {align} \frac {dp}{dx} & =0\tag {C1}\\ x+\frac {d}{dp}f\left ( p\right ) & =0 \tag {C2} \end {align}

The general solution is found from (C1) and the singular solution if any from (C2). $$\frac {dp}{dx}=0$$. This implies $$p=C_{1}$$. Substituting this into (1) gives\begin {equation} y\left ( x\right ) =xC_{1}+\frac {1}{C_{1}} \tag {2} \end {equation} The singular solution is found (C2). Since $$f\left ( p\right ) =\frac {1}{p}$$ then (C2) becomes $$x-\frac {1}{p^{2}}=0$$. Hence $$p^{2}=\frac {1}{x}$$ or $$p=\pm \sqrt {\frac {1}{x}}$$. Substituting these back in (1) gives\begin {align} y_{1}\left ( x\right ) & =x\sqrt {\frac {1}{x}}+\sqrt {x}=2\sqrt {x}\tag {3}\\ y_{2}\left ( x\right ) & =-x\sqrt {\frac {1}{x}}-\sqrt {x}=-2\sqrt {x} \tag {4} \end {align}

Eq. (2) is the complete integral and (3,4) are the singular solutions.

#### 2.2 Example 2

$$y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}$$ is put in normal form (by replacing $$y^{\prime }$$ with $$p$$) which gives\begin {equation} y=xp-p^{2} \tag {1} \end {equation} Comparing to (C1) shows that $$f\left ( p\right ) =-p^{2}$$. Hence (C2,C2) now becomes\begin {align} \frac {dp}{dx} & =0\tag {C1}\\ x+\frac {d}{dp}f\left ( p\right ) & =0 \tag {C2} \end {align}

The general solution is found from (C1) and the singular solution if any from (C2). $$\frac {dp}{dx}=0$$. This implies $$p=C_{1}$$. Substituting this into (1) gives\begin {equation} y\left ( x\right ) =xC_{1}+C_{1}^{2} \tag {2} \end {equation} The singular solution is found (C2). Since $$f\left ( p\right ) =-p^{2}$$ then (C2) becomes $$x-2p=0$$. Hence $$p=\frac {x}{2}$$. 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 complete integral and (3) is the singular solution.

#### 2.3 Example 3

$$y=xy^{\prime }-\frac {1}{4}\left ( y^{\prime }\right ) ^{2}$$ is put in normal form (by replacing $$y^{\prime }$$ with $$p$$) which gives\begin {equation} y=xp-\frac {1}{4}p^{2} \tag {1} \end {equation} Comparing to (C1) shows that $$f\left ( p\right ) =-\frac {1}{4}p^{2}$$. Hence (C2,C2) now becomes\begin {align} \frac {dp}{dx} & =0\tag {C1}\\ x+\frac {d}{dp}f\left ( p\right ) & =0 \tag {C2} \end {align}

The general solution is found from (C1) and the singular solution if any from (C2). $$\frac {dp}{dx}=0$$. This implies $$p=C_{1}$$. Substituting this into (1) gives\begin {equation} y\left ( x\right ) =xC_{1}-\frac {1}{4}C_{1}^{2} \tag {2} \end {equation} The singular solution is found (C2). Since $$f\left ( p\right ) =-\frac {1}{4}p^{2}$$ then (C2) becomes $$x-\frac {1}{2}p=0$$. Hence $$p=2x$$. Substituting this back in (1) gives\begin {equation} y_{1}\left ( x\right ) =2x^{2}-x^{2}=x^{2} \tag {3} \end {equation} Eq. (2) is the complete integral and (3) is the singular solution.

#### 2.4 Example 4

$$y=x\left ( y^{\prime }\right ) ^{2}$$ is put in normal form by replacing $$y^{\prime }$$ with $$p$$ which gives\begin {equation} y=xp^{2} \tag {1} \end {equation} This is special case of the general d’Alembert $$y=xg\left ( p\right ) +f\left ( p\right )$$ with $$f\left ( p\right ) =0$$. Comparing with (sd1)\begin {equation} y=xg\left ( p\right ) \tag {sd1} \end {equation} When $$\frac {dp}{dx}=0$$ then $$p=p_{0}$$. A constant. This constant is found from solving $$p-g\left ( p\right ) =0$$ or $$p-p^{2}=0$$. Hence $$p_{0}=0$$ or $$p_{0}=1$$. Substituting each in (sd1) gives two singular solutions\begin {align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =x \end {align*}

The general solution is when $$\frac {dp}{dx}\neq 0$$. Equation (sd3) becomes\begin {align} \frac {dx}{dp}-x\frac {\frac {d}{dp}g\left ( p\right ) }{p-g\left ( p\right ) } & =0\nonumber \\ x^{\prime }+xG\left ( p\right ) & =0 \tag {sd3} \end {align}

Where $$G\left ( p\right ) =-\frac {2p}{p-p^{2}}$$. This is linear ODE in $$x\left ( p\right )$$. Solving it gives $$x=\frac {C_{1}}{\left ( p-1\right ) ^{2}}$$. Solving for $$p$$ gives $$\left ( p-1\right ) ^{2}=\frac {C_{0}}{x}$$ or $$p-1=\pm \sqrt {\frac {C_{0}}{x}}$$ or \begin {align*} p & =1+\sqrt {\frac {C_{0}}{x}}\\ p & =1-\sqrt {\frac {C_{0}}{x}} \end {align*}

For each $$p$$, there is a general solution. Substituting the above in (1) gives\begin {align*} y_{3}\left ( x\right ) & =x\left ( 1+\sqrt {\frac {C_{0}}{x}}\right ) ^{2}\\ y_{4}\left ( x\right ) & =x\left ( 1-\sqrt {\frac {C_{0}}{x}}\right ) ^{2} \end {align*}

Note however, that $$y_{2}=x$$ can be obtained from $$y_{3}\left ( x\right )$$ when $$C_{0}=0$$. Hence $$y_{2}\left ( x\right ) =x$$ is not singular solution. Therefore the ﬁnal solution is\begin {align*} y\left ( x\right ) & =0\\ y\left ( x\right ) & =x\left ( 1+\sqrt {\frac {C_{0}}{x}}\right ) ^{2}\\ y\left ( x\right ) & =x\left ( 1-\sqrt {\frac {C_{0}}{x}}\right ) ^{2} \end {align*}

#### 2.5 Example 5

$$y=x+\left ( y^{\prime }\right ) ^{2}$$ is put in normal form (by replacing $$y^{\prime }$$ with $$p$$) which gives\begin {equation} y=x+p^{2} \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the general d’Alembert form with $$g\left ( p\right ) =1$$ and $$f\left ( p\right ) =p^{2}$$. Eq, (d3) becomes\begin {align} p-g\left ( p\right ) & =\left ( x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx}\nonumber \\ p-1 & =2p\frac {dp}{dx} \tag {d3} \end {align}

When $$\frac {dp}{dx}=0$$, then $$p$$ is constant, say $$p_{0}$$. This constant is found by solving for $$p$$ from $$p-1=0$$. Hence $$p_{0}=1$$. Substituting this in (1) gives the singular solution as $\ y=x+1$ The general solution is found by ﬁnding $$p$$ from (d3). No need here to do the inversion as (d3) is separable already. Solving (d3) gives$p=\operatorname *{LambertW}\left ( Ce^{\frac {x}{2}-1}\right ) +1$ 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. Hence (4) is the only solution.

#### 2.6 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 {equation} y=-x+\left ( p^{2}-1\right ) \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the general d’Alembert form with $$g\left ( p\right ) =-1$$ and $$f\left ( p\right ) =p^{2}-1$$. Eq (d3) becomes\begin {align} p-g\left ( p\right ) & =\left ( x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx}\nonumber \\ p+1 & =2p\frac {dp}{dx} \tag {d3} \end {align}

The singular solution is found when $$\frac {dp}{dx}=0,$$ which implies that $$p$$ is constant. Hence $$p+1=0$$. Solving gives $$p_{0}=-1$$. Substituting this values in (1), gives the singular solution\begin {equation} y\left ( x\right ) =-x \tag {3} \end {equation} The general solution is found by ﬁnding $$p$$ from (d3). No need here to do the inversion as (d3) is separable already. Solving (d3) gives\begin {align*} p & =-\operatorname *{LambertW}\left ( -e^{-\frac {x}{2}-1+\frac {C}{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.

#### 2.7 Example 7

$$yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x$$ is put in normal form (by replacing $$y^{\prime }$$ with $$p$$) which gives\begin {equation} y=\frac {1}{p}x+p \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the general d’Alembert form with $$g\left ( p\right ) =\frac {1}{p}$$ and $$f\left ( p\right ) =p$$. Eq (d3) becomes\begin {align} p-g\left ( p\right ) & =\left ( x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx}\nonumber \\ p-\frac {1}{p} & =\left ( 1-\frac {x}{p^{2}}\right ) \frac {dp}{dx} \tag {d3} \end {align}

The singular solution is found when $$\frac {dp}{dx}=0,$$ which implies that $$p$$ is constant. Hence $$p-\frac {1}{p}=0$$. Solving for this gives $$p=\pm 1$$. Substituting these values in (1) gives the solutions\begin {align} y_{1}\left ( x\right ) & =x+1\tag {3}\\ y_{2}\left ( x\right ) & =-\left ( x+1\right ) \tag {4} \end {align}

The general solution is found by ﬁnding $$p$$ from (d3). Since (d3) is not linear in $$p$$, then inversion is needed. Writing (d3) as$\frac {\frac {p^{2}-1}{p}}{1-\frac {x}{p^{2}}}=\frac {dp}{dx}$ Since this is not linear or separable in $$p$$, it is inverted.\begin {align*} \frac {dx}{dp} & =\frac {\frac {p^{2}-x}{p^{2}}}{\frac {p^{2}-1}{p}}\\ \frac {dx}{dp} & =\frac {p^{2}-x}{p\left ( p^{2}-1\right ) }\\ \frac {dx}{dp}+x\frac {1}{p\left ( p^{2}-1\right ) } & =\frac {p}{\left ( p^{2}-1\right ) } \end {align*}

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 {5} \end {align}

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 (5) 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$x=\frac {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) \sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}+\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac {\frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}}{\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}}$

#### 2.8 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 {equation} y=xp^{2}+p^{2} \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the general d’Alembert form with $$g\left ( p\right ) =p^{2}$$ and $$f\left ( p\right ) =p^{2}$$. Eq (d3) becomes\begin {align} p-g\left ( p\right ) & =\left ( x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx}\nonumber \\ p-p^{2} & =\left ( 2xp+2p\right ) \frac {dp}{dx} \tag {d3} \end {align}

The singular solution is found when $$\frac {dp}{dx}=0,$$ which implies that $$p$$ is constant. Hence $$p-p^{2}=0$$. Solving for this gives $$p=0,p=1$$. Substituting these values in (1) gives the 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 (d3). Since (d3) is not linear in $$p$$, then inversion is needed. Writing (d3) 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*}

#### 2.9 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 {equation} y=\frac {x}{a}p+\frac {b}{a}p^{-1} \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the general d’Alembert form with $$g\left ( p\right ) =\frac {p}{a}$$ and $$f\left ( p\right ) =\frac {b}{a}p^{-1}$$. Eq (d3) becomes\begin {align} p-g\left ( p\right ) & =\left ( x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx}\nonumber \\ p-\frac {p}{a} & =\left ( \frac {x}{a}-\frac {b}{a}p^{-2}\right ) \frac {dp}{dx} \tag {d3} \end {align}

The singular solution is found when $$\frac {dp}{dx}=0,$$ which implies that $$p$$ is constant. In this case $$p\left ( 1-\frac {1}{a}\right ) =0$$. Solving for this gives $$p=0$$. Substituting these values 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 (d3). Since (d3) is not linear in $$p$$, then inversion is needed. Writing (d3) 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*}

#### 2.10 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 {equation} y=x\left ( p+a\sqrt {1+p^{2}}\right ) \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the special d’Alembert form with $$g\left ( p\right ) =p+a\sqrt {1+p^{2}}$$ and $$f\left ( p\right ) =0$$. Eq (sd2) becomes

\begin {align} p-g\left ( p\right ) & =x\frac {d}{dp}g\left ( p\right ) \frac {dp}{dx}\nonumber \\ p-\left ( p+a\sqrt {1+p^{2}}\right ) & =x\frac {d}{dp}\left ( p+a\sqrt {1+p^{2}}\right ) \frac {dp}{dx}\nonumber \\ -a\sqrt {1+p^{2}} & =x\left ( 1+\frac {ap}{\sqrt {1+p^{2}}}\right ) \frac {dp}{dx} \tag {sd2} \end {align}

The singular solution is found when $$\frac {dp}{dx}=0,$$ which implies that $$-a\sqrt {1+p^{2}}=0$$ or $$\sqrt {1+p^{2}}=0$$. This gives no real solution for $$p.$$ Hence no singular solution exist.

The general solution is when $$\frac {dp}{dx}\neq 0$$ in (sd2). Since (sd2) 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*}

#### 2.11 Example 11

$$y=x\left ( y^{\prime }\right ) ^{2}$$ is put in normal form (by replacing $$y^{\prime }$$ with $$p$$) which gives\begin {equation} y=xp^{2} \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the special d’Alembert form with $$g\left ( p\right ) =p^{2}$$ and $$f\left ( p\right ) =0$$. Eq (sd2) becomes\begin {align} p-g\left ( p\right ) & =x\frac {d}{dp}g\left ( p\right ) \frac {dp}{dx}\nonumber \\ p-p^{2} & =2xp\frac {dp}{dx} \tag {sd2} \end {align}

The singular solution is found when $$\frac {dp}{dx}=0,$$ which implies $$p-p^{2}=0$$ or $$p\left ( 1-p^{2}\right ) =0$$. This gives $$p=0$$ or $$p^{2}=1$$. The ﬁrst gives $$y=0$$ and the second gives $$p=1$$ or $$p=-1$$. Therefore $$y=x$$ or $$y=-x$$. But this second solution does not satisfy the ODE. Hence only $$y=0$$ and $$y=x$$ are singular solutions.

The general solution is when $$\frac {dp}{dx}\neq 0$$ in (sd2). Since (sd2) is nonlinear, inversion is needed.

\begin {align*} \frac {p-p^{2}}{2xp} & =\frac {dp}{dx}\\ \frac {dx}{dp} & =\frac {2xp}{-p-p^{2}}\\ \frac {dx}{x} & =\frac {2p}{-p-p^{2}}dp\\ & =\frac {-2}{1+p}dp \end {align*}

Integrating gives$\ln x=-2\ln \left ( 1+p\right ) +c$ Here it is possible to directly solve for $$p$$ in terms of $$x$$. Raising both sides to exponential gives\begin {align} x & =c_{1}\frac {1}{\left ( 1+p\right ) ^{2}}\nonumber \\ \left ( 1+p\right ) ^{2} & =c_{1}\frac {1}{x}\nonumber \\ 1+p & =\pm c_{1}\sqrt {\frac {1}{x}}\nonumber \\ p & =\left ( \pm c_{1}\sqrt {\frac {1}{x}}\right ) -1 \tag {3} \end {align}

Substituting this back in (1) gives the explicit general solution as\begin {align*} y_{1} & =x\left ( c_{1}\sqrt {\frac {1}{x}}-1\right ) ^{2}=x\left ( \frac {1}{x}c_{1}^{2}-2c_{1}\sqrt {\frac {1}{x}}+1\right ) =c_{1}^{2}-2xc_{1}\sqrt {\frac {1}{x}}+x\\ y_{2} & =x\left ( -c_{1}\sqrt {\frac {1}{x}}-1\right ) ^{2}=x\left ( \frac {1}{x}c_{1}^{2}+2c_{1}\sqrt {\frac {1}{x}}+1\right ) =c_{1}^{2}+2xc_{1}\sqrt {\frac {1}{x}}+x \end {align*}

Therefore the solutions are\begin {align*} y_{1}\left ( x\right ) & =c_{1}^{2}-2xc_{1}\sqrt {\frac {1}{x}}+x\\ y_{2}\left ( x\right ) & =c_{1}^{2}+2xc_{1}\sqrt {\frac {1}{x}}+x\\ y_{3}\left ( x\right ) & =x\\ y_{4}\left ( x\right ) & =0 \end {align*}

The last two above are singular solutions.

#### 2.12 Example 12

This ode is an example where $$y$$ does not appear explicitly in the ode. \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 {equation} y=-x-1+p^{2} \tag {1} \end {equation} Comparing to (d1) \begin {equation} y\left ( x\right ) =xg\left ( p\right ) +f\left ( p\right ) \tag {d1} \end {equation} Shows this is the general d’Alembert form with $$g\left ( p\right ) =-1$$ and $$f\left ( p\right ) =-1+p^{2}$$. Eq (d3) becomes\begin {align} p-g\left ( p\right ) & =\left ( x\frac {d}{dp}g\left ( p\right ) +\frac {d}{dp}f\left ( p\right ) \right ) \frac {dp}{dx}\nonumber \\ p+1 & =2p\frac {dp}{dx} \tag {d3} \end {align}

The singular solution is found when $$\frac {dp}{dx}=0,$$ which gives $$p+1=0$$ or $$p=-1$$. 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 (d3). Since (d3) 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$\fbox {x=2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_1}$ 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.

### 4 references

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